arXiv:1706.00075v2 [math.GR] 8 Aug 2017 ..CnuayCasso GL of Classes Conjugacy 4.1. .LclCnuayi GL in Conjugacy Local 5. ..Obt fMat of Orbits 4.2. .Sbruso GL of Subgroups 2. Introduction 1. .TeKre fteNtrlHomomorphism Natural the of Kernel Subgroups The Conjugate Locally of 4. Properties 3. 2010 Date ..FnlRslsfrLclCnuayi GL in Conjugacy Local for Results Final 4.6. 1 h SL The case Borel 11. Cases The Cartan of Normalizer 10. The Cases Cartan 9.2. The cases Cartan 9.1. The case Center 9. The Facts 8. Computational 7. Splitting 6. ..LclCnuayi GL in Conjugacy Local 4.5. ..A qiaetCniinfrLclCnuayBtenSubgrou Between Conjugacy Local for Condition Equivalent An 4.3. ..PeiiayRslsfrLclCnuayi GL in Conjugacy Local for Results Preliminary 4.4. ac ,2018. 6, March : ahmtc ujc Classification. Subject Mathematics nSteln’ aeoiain fsbruso GL of subgroups of categorizations Sutherland’s on hm hr r w odtosta oal ojgt subgroups conjugate locally that conditions two are There them. bijection tde oa ojgc mn ugop fGL of subgroups among conjugacy local studies Abstract. casework. utstsy letting satisfy: must ker and ϕ elclycnuaei GL in conjugate locally be ( H ϕ 1 H and ) and 2 ∩ ker f 2 H ker ( ϕ : Z PRFNLPPR UMR21,RVSDAGS 2017 AUGUST REVISED 2016, SUMMER PAPER, FINAL SPUR 2 ϕ H ( Subgroups /p H ∩ ϕ 1 hssuyflyctgrzslclcnuayi GL in conjugacy local categorizes fully study This . 2 ker Z → RJC UGSE YADE SUTHERLAND ANDREW BY SUGGESTED PROJECT ob iia oec te,te nesadtepsiiiisfor possibilities the understand then other, each to similar be to ) 2 case ) ( H ϕ OA OJGC IN CONJUGACY LOCAL 2 Z ( 2 utb oal ojgt nGL in conjugate locally be must ϕ Z /p uhthat such /p GL : H Z ETR TIU CHRISTENSEN ATTICUS MENTOR: 1 k ne ojgto yeeet fGL of elements by conjugation under ) 2 Z 2 and ( ( 2 2 Z ) Z ( ( Z /p /p 2 Z /p H ( h /p Z Z Z 2 2 and .T identify To ). /p Z ) fagroup a of Z YNJN KIM JONG HYUN 04,2F9 03,2J5 11C20. 20J15, 20G35, 20F99, 20E45, ) mn ugop fker of Subgroups among ) 2 → f Z Contents ( ) h GL r ojgt in conjugate are ) 2 1 ( G Z 2 /p ( r adt elclycnuaei hr sa is there if conjugate locally be to said are Z H 2 Z ( ϕ /p 1 Z etentrlhomomorphism, natural the be ) 2 and GL : /p 2 2 Z ( ( Z Z ,where ), Z /p /p GL H 2 mn ugop fker of Subgroups among ) 2 ( 2 Z ( 2 Z G Z Z n oa ojgc among conjugacy local and ) 2 pt ojgto,w choose we conjugation, to up and ) /p ( /p o every for Z p Z H 2 /p sa d rm,building prime, odd an is Z mn ugop of Subgroups among ) 1 ϕ 2 2 ) ϕ and ( Z ( Z → H ∩ ) /p h 2 1 H ( SL( and ) GL ∈ 2 Z 2 Z H /p fGL of hog such through ) so ker of ps 2 p 1 ( Z 2 hspaper This . ϕ Z 9 ) 7 ) ( /p H H 2 ( 1 2 Z Z ∩ must ) /p 5 ) H ker ϕ 2 1 Z ϕ ∩ ϕ ) 10 45 15 36 31 26 23 22 16 16 6 3 3 4 8 7 12. The Exceptional cases 47 13. Conclusion 49 Acknowledgements 49 References 49

2 1. Introduction Given an elliptic curve E over a number field K and an integer n, the action of the absolute Galois group of K on the n-torsion subgroup E[n] (Z/nZ)2 determines a sub- group of Aut(E[n]) GL (Z/nZ). Sutherland [6] gives an efficient≃ algorithm for computing ≃ 2 the images of the Galois representation associated to an elliptic curve in GL2(Z/pZ) that determines the image up to local conjugacy. One needs to understand locally conjugate subgroups to determine the image up to conjugation. [6] categorizes local conjugacy of sub- groups in GL2(Z/pZ), see Theorem 2. A categorization of local conjugacy of subgroups in 2 GL2(Z/p Z) is a first step in extending the categorization in GL2(Z/pZ) to the full p-adic image in GL2(Zp). Locally conjugate subgroups H1 and H2 of a group G form what is known as a Gassman triple (G,H1,H2), which arise in the study of arithmetically equivalent number fields. Such number fields have the same Dedekind zeta function but need not be isomorphic. Gassmann triples of the form (GL2(Z/nZ),H1,H2) can be used to explicitly construct arithmetically equivalent number fields K1 and K2 as subfields of the n-torsion field Q(E[n]) of an elliptic curve E/Q, see [2]. Non-conjugate subgroups H1 and H2 give rise to non-isomorphic number fields K1 and K2. 2 Let ϕ : GL2(Z/p Z) GL2(Z/pZ) be the natural homomorphism and let H1 and H2 be → 2 locally conjugate subgroups of GL2(Z/p Z). The groups H1 ker ϕ and H2 ker ϕ must 2 ∩ ∩ be locally conjugate in GL2(Z/p Z) and ϕ(H1) and ϕ(H2) must be locally conjugate in GL2(Z/pZ). However, the converse is not generally true. Section 4 of this paper identifies 2 local conjugacy in GL2(Z/p Z) among the subgroups of ker ϕ. Using the categorization of local conjugacy in GL2(Z/pZ), we replace H1 and H2 with conjugates so that ϕ(H1) and ϕ(H ) are similar to each other. Moreover, the elements of ϕ(H ) restrict H ker ϕ for 2 i i ∩ i =1, 2. We further replace H1 and H2 with conjugates so that they are in “nice” forms. 2 Theorem 1 completely classifies the conjugacy classes of GL2(Z/p Z). Furthermore, the 2 elements of subgroups of GL2(Z/p Z) are often not difficult to identify. We use this infor- mation to determine when H1 and H2 are not locally conjugate given that ϕ(H1) and ϕ(H2) are locally conjugate and H ker ϕ and H ker ϕ are locally conjugate or when H and 1 ∩ 2 ∩ 1 H2 are locally conjugate but not conjugate. Propositions 10 and 13 yield all of the non-conjugate locally conjugate subgroups, which 2 we refer to as nontrivially locally conjugate subgroups, of GL2(Z/p Z) up to conjugation, see Theorem 4. One can choose generators of such subgroups to resemble each other. In fact, they are expressible in forms resembling the nontrivially locally conjugate subgroups of GL2(Z/pZ).

k 2. Subgroups of GL2(Z/p Z) Throughout this paper, p is an odd prime and ǫ is taken to be some nonsquare in Z/pZ. Let GL2(R), SL2(R) and PGL2(R) denote the general, special and projective linear groups of 2 2 matrices over a ring R. × 3 Define the following subgroups of GL (Z/pkZ) for k 1: 2 ≥ w 0 Z(pk)= GL (Z/pkZ) 0 w ∈ 2    w 0 C (pk)= GL (Z/pkZ) s 0 z ∈ 2    w ǫy C (pk)= GL (Z/pkZ) ns y w ∈ 2    w x B(pk)= GL (Z/pkZ) . 0 z ∈ 2    They are respectively called the center, Cartan-split subgroup, Cartan-nonsplit sub- k k group, and Borel subgroup of GL2(Z/p Z). For H GL2(Z/p Z), let N(H) denote the k ≤ normalizer of H in GL2(Z/p Z). In particular, 0 x N(C (p)) = C (p) GL (Z/pZ) s s ∪ y 0 ∈ 2    0 1 = C (p) C (p) s ∪ 1 0 s   w ǫy N(C (p)) = C (p) GL (Z/pZ) ns ns ∪ y w ∈ 2 − −   1 0 = C (p) C (p). ns ∪ 0 1 ns  −  3. Properties of Locally Conjugate Subgroups This section defines locally conjugate subgroups and discusses some properties of local conjugacy. For a group G and an element g G, let gG denote the conjugacy class of g in G. ∈

Definition 1. Let G be a group and H1,H2 G. If there is a bijection f : H1 H2 such that h and f(h) are conjugate in G for all h ≤H , then H and H are locally→ conjugate ∈ 1 1 2 in G. If H1 and H2 are locally conjugate in G but not conjugate in G, then H1 and H2 are nontrivially locally conjugate in G. Conjugate subgroups of G are always locally conjugate in G. Hence, conjugate subgroups are considered to be “trivially” locally conjugate. Moreover, it is possible for two subgroups H1 and H2 of a group G′, which is in turn a subgroup of G, to be locally conjugate in G but not in G′. It is therefore important to emphasize the parent group in which two subgroups are locally conjugate in. Nevertheless, the parent group will be clear in context even if it is not explicitly stated. Local conjugacy of subgroups in a fixed group G is an equivalence relation. Furthermore, given that subgroups H2 and H3 of G are conjugate, a subgroup H1 of G which is locally conjugate to H2 is conjugate to H2 if and only if H1 is conjugate to H3. With this in mind, 2 we will categorize local conjugacy between subgroups of GL2(Z/p Z) up to conjugation of the subgroups. The following gives an alternate definition to local conjugacy: 4 Proposition 1. Let G be a group and H1,H2 G. Then, H1 and H2 are locally conjugate in G if and only if H C = H C for all≤ conjugacy classes of G. | 1 ∩ | | 2 ∩ | Proof. Suppose that H1 and H2 are locally conjugate in G via f : H1 H2. For every 1 → conjugacy class C of G, f H1 C maps into H2 C. Likewise, f − H2 C maps into H1 C | ∩ ∩ | ∩ ∩ and is the inverse of f H1 C . Thus, H1 C = H2 C . Conversely, suppose| that∩ H C | = ∩H | C| for∩ every| conjugacy class C of G. Choose | 1 ∩ | | 2 ∩ | some bijections fC : H1 C H2 C and define f : H1 H2 as f(h)= fhG (h). Since the conjugacy classes of G partition∩ → G∩, f is a well defined bijection.→ Moreover, h and f(h) are in the same conjugacy class for every h H , and so H and H are locally conjugate.  ∈ 1 1 2 The following two propositions give necessary conditions for local conjugacy on a finite group G in terms of local conjugacy in a normal subgroup of G and quotient groups of G.

Proposition 2. Let G be a group, H1,H2 G and N ⊳ G. If H1 and H2 are locally conjugate in G, then H N and H N are≤ locally conjugate in G. 1 ∩ 2 ∩ Proof. N is the disjoint union of some conjugacy classes of G. Let C be a conjugacy class of G. If C N, then (Hi N) C = Hi C for i =1, 2. Otherwise, (Hi N) C = 0. H N and⊆H N are| therefore∩ ∩ locally| | conjugate∩ | in G by Proposition| 1. ∩ ∩ |  1 ∩ 2 ∩ Proposition 3. Let G,G′ be finite groups, H ,H G subgroups of G and ϕ : G G′ a 1 2 ≤ → surjective homomorphism. If H1 and H2 are locally conjugate in G, then ϕ(H1) and ϕ(H2) are locally conjugate in G′. 1 G Proof. Let C′ be any conjugacy class of G′ and let U = (ϕ− (x)) . We claim that x C′ 1 1 ∈ 1 G ϕ− (C′) = U. If d ϕ− (C′), then ϕ(d) C′, in which case ϕ(d) (ϕ− (ϕ(d)) U. 1 ∈ ∈ 1 G S ∈ ⊆ 1 Therefore, ϕ− (C′) U. Conversely, if d (ϕ− (x)) for some x C′, then d = gyg− ⊆ 1 ∈ ∈ 1 1 for some g G and y ϕ− (x). It follows that ϕ(d) = ϕ(g)ϕ(y)ϕ(g)− = ϕ(g)xϕ(g)− , ∈ 1 ∈ 1 1 and so d ϕ− (C′). Hence, U ϕ− (C′) as desired. In particular, ϕ− (C′) is the union of conjugacy∈ classes of G. ⊆ ker ϕ is the union of conjugacy classes of G because it is normal in G. Moreover, since 1 H1 and H2 are locally conjugate, H1 ker ϕ = H2 ker ϕ . Similarly, H1 ϕ− (C′) = 1 | ∩ | | ∩1 | | ∩ | H2 ϕ− (C′) . Note that ϕ(Hi) C′ = ϕ(Hi ϕ− (C′)), and so Hi ker ϕ has index | ∩ | 1 ∩ ∩ ∩ ϕ(Hi C′) in Hi ϕ− (C′). Thus, ϕ(H1) C′ = ϕ(H2) C′ and so ϕ(H1) and ϕ(H2) are| locally∩ conjugate| ∩ by Proposition 1.| ∩ | | ∩ | 

4. The Kernel of the Natural Homomorphism ϕ : GL (Z/p2Z) GL (Z/pZ) 2 → 2 For the rest of this paper, let ϕ denote the natural homomorphism GL (Z/p2Z) 2 → GL2(Z/pZ). An element κ of ker ϕ is of the form κ = I + Ap, where A is identifiable as an element of Mat2(Z/pZ) and A uniquely determines κ. We will refer to A as the p-part of κ and define p(κ) to be A. Note that ker ϕ is isomorphic to the 4 dimensional Z/pZ vector space because (I +A1p)(I + A2p)= I +(A1 + A2)p for all A1, A2 Mat2(Z/pZ). ∈ 2 Suppose that H1 and H2 are subgroups of GL2(Z/p Z) that are locally conjugate in 2 GL2(Z/p Z). By Proposition 2, H1 ker ϕ and H2 ker ϕ are also locally conjugate in 2 ∩ ∩ GL2(Z/p Z). This section determines the subgroups of ker ϕ which are locally conjugate in 2 GL2(Z/p Z). 2 Lemma 1 below determines when two elements of ker ϕ are conjugate in GL2(Z/p Z). 5 Lemma 1. Let κ1, κ2 ker ϕ, A1 = p(κ1) and A2 = p(κ2). Then, κ1 and κ2 are conjugate 2 ∈ in GL2(Z/p Z) if and only if A1 and A2 are conjugate by an element of GL2(Z/pZ). 2 1 Proof. If κ1 = I +A1p is conjugate to κ2 = I +A2p via g GL2(Z/p Z), i.e. g(I +A1p)g− = 1 ∈ I + A2p, then I + gA1g− p = I + A2p and so A1 is conjugate to A2 via ϕ(g). Conversely, if A1 is conjugate to A2 are conjugate via some g′ GL2(Z/pZ), then I + A1p is conjugate to 1 ∈ I + A p via any g ϕ− (g′).  2 ∈ Furthermore, Lemma 2 below yields a well defined group action of GL2(Z/pZ) on ker ϕ in 1 1 which g GL (Z/pZ) sends κ ker ϕ togκ ˆ gˆ− , whereg ˆ is any element of ϕ− (g). ∈ 2 ∈ 2 Lemma 2. Let g,g′ GL2(Z/p Z) with ϕ(g) = ϕ(g′), i.e. g and g′ are congruent modulo ∈ 1 1 p. For any κ ker ϕ, gκg− = g′κg′− . ∈ Proof. Let A = p(κ) and let g′ = g + Bp for some B Mat2(Z/pZ). It is not difficult to see 1 1 1 1 ∈ that g′− = g− g− Bg− p. Therefore, − 1 1 g′κg′− = g′(I + Ap)g′− 1 1 1 =(g + Bp)(I + Ap)(g− g− Bg− p) − 1 1 1 = I +(Bg− + gAg− Bg− )p − 1 = I + gAg− 1 = g(I + Ap)g− 1 = gκg− .  Lemma 3 restricts the possible combinations of ϕ(H) and H ker ϕ for subgroups H of 2 ∩ GL2(Z/p Z). 2 1 Lemma 3. If H GL2(Z/p Z), h ϕ(H) and κ H ker ϕ, then hκh− ker ϕ. Furthermore, H ker≤ ϕ is fixed under conjugation∈ by h.∈ ∩ ∈ ∩ Proof. This is because H ker ϕ is a normal subgroup of H.  ∩ 2 k 4.1. Conjugacy Classes of GL2(Z/p Z). For k 1, the similarity classes of Mat(Z/p Z) k ≥ k are defined as the orbits of Mat(Z/p Z) under conjugation by elements of GL2(Z/p Z). The k k similarity classes of Mat(Z/p Z) extend the conjugacy classes of GL2(Z/p Z) in that the conjugacy classes are themselves similarity classes. k [1, Theorem 2.2] yields a way to categorize the similarity classes of Mat2(Z/p Z) for k k .k 1. Just as in [1, Section 1.2], fix a section Z/pZ ֒ Z/p Z with image K1 Z/p Z Further≥ fix compatible sections Z/plZ Z/pkZ for 1 →l

Table 1. Representatives of the Similarity Classes of Mat2(Z/pZ)

Representative det trace w 0 0 w

4.3. An Equivalent Condition for Local Conjugacy Between Subgroups of ker ϕ. Table 4.2 yields the following observation:

Lemma 4. The similarity class of a nonscalar element M of Mat2(Z/pZ) is uniquely deter- mined by trace(M), det(M) Z/pZ. ∈ We define χ below to use Lemma 4 as a way to find an equivalent condition for local conjugacy between subgroups of ker ϕ.

Definition 2. For a subgroup H of ker ϕ and for t, d Z/pZ, let χ(H,t,d)= k H k = I + Ap where trace(A)= t, det(A)= d . ∈ |{ ∈ | }| Let H1 and H2 be subgroups of ker ϕ. Say that H1 and H2 have equal trace-determinant distribution if χ(H ,t,d)= χ(H ,t,d) for all t, d Z/pZ. 1 2 ∈

Proposition 4. Let H1 and H2 be subgroups of ker ϕ. Then, H1 and H2 are locally conjugate 2 2 2 in GL2(Z/p Z) if and only if H1 Z(p ) = H2 Z(p ) and H1 and H2 have equal trace- determinant distribution. ∩ ∩

2 2 Proof. If H1 and H2 are locally conjugate, then H1 Z(p ) = H2 Z(p ) because every 2 ∩ ∩ element of Z(p ) is the sole member of its conjugacy class. Moreover, H1 and H2 must have equal trace-determinant distribution by Lemma 4. 2 2 Conversely, if H1 Z(p ) = H2 Z(p ) and H1 and H2 have equal trace-determinant distribution, then for∩ all t, d Z/pZ∩, ∈ k H Z(p2) k = I + Ap where trace(A)= t, det(A)= d |{ ∈ 1 \ | }| = k H Z(p2) k = I + Ap where trace(A)= t, det(A)= d . |{ ∈ 2 \ | }|

By Lemmas 1 and 4, H1 and H2 are locally conjugate.  7 4.4. Preliminary Results for Local Conjugacy in GL2(Z/pZ) among Subgroups of ker ϕ. The definition of locally conjugate subgroups yields the following result: Lemma 5. If H ,H ker ϕ are locally conjugate in GL (Z/pZ), then dim H = dim H . 1 2 ≤ 2 1 2 Proof. Locally conjugate subgroups are in bijection and ϕ is a finite dimensional vector space over the finite field Z/pZ.  Lemma 6 categorizes local conjugacy of finite cyclic subgroups.

Lemma 6. Let G be a group and let H1,H2 be finite locally conjugate subgroups of G. If H1 is cyclic, then H2 is cyclic and H1 and H2 are conjugate.

Proof. Say that h1 generates H1. There is some h2 H2 which is conjugate to h1 in G. The orders of h and h are equal, H and H are finite and∈ H = H , and so h generates H . 2 1 1 2 | 1| | 2| 2 2 Therefore, H2 is conjugate to H1.  Lemma 7. The subgroups of ker ϕ of dimension 0 or 1, i.e. the cyclic subgroups, are conjugate in GL2(Z/pZ) to one of the following subgroups of ker ϕ: (1) I h i 0 1 (2) I + p 0 0     1 1 (3) I + p 0 1     1 0 (4) I + p , where d Z/pZ 0 d ∈     0 ǫ (5) I + p 1 0     1 ǫc p 1 (6) I + p , where c Z/pZ and 0

Proof. Let H be a subgroup of ker ϕ generated by h = I + Ap for some A Mat2(Z/pZ). H can be replaced with a conjugate such that A is one of the matrices in Table∈ 4.2. The categorization of cyclic subgroups of ker ϕ is finished by determining an alternative generator of H and conjugating H if necessary. w 0 1 0 Suppose that A = . If w = 0, then H = I . Otherwise, H = I + p . 0 w h i 0 1       w 1 0 1 Suppose that A = . If w = 0, then H = I + p . Otherwise, H is 0 w 0 0       1 1 1 1 alternatively generated by I + w p, and so H is conjugate to I + p . 0 1 0 1       w 0 Suppose that A = , where 0 w

4.6. Final Results for Local Conjugacy in GL2(Z/pZ) among Subgroups of ker ϕ. Since dim(T ) = 3 and dim(ker ϕ) = 4, any subgroup of ker ϕ with at least 2 dimensions must have nontrivial intersection with T . In particular, letting H be a subgroup of ker ϕ, if dim(H) = 2, then dim(H T ) 1 and if dim(H) = 3, then dim(H T ) 2. ∩ ≥ ∩ ≥ Lemma 10. The subgroups of ker ϕ of dimension 2 that are not subgroups of T are conjugate 2 in GL2(Z/p Z) to one of the following: 10 0 1 0 0 (1) H = I + p,I + p 1 0 0 1 1       0 1 0 0 (2) H = I + p,I + p 2 0 0 0 1       0 1 1 0 (3) H = I + p,I + p , where d Z/pZ is not 1. 3,d 0 0 0 d ∈ −       1 0 0 1 (4) H = I + p,I + p , where c Z/pZ. 4,c 0 1 c 1 ∈   −     1 0 0 0 (5) H = I + p,I + p 5 0 1 0 1   −     0 ǫ 1+ a ǫb (6) H6,a,b = I + p,I + − p , where a, b Z/pZ. 1 0 b 1 a ∈     −  

In particular, H and H are nontrivially locally conjugate. For d,d′ Z/pZ where d,d′ = 2 3,0 ∈ 6 1, H3,d and H3,d′ are nontrivially locally conjugate if d = d′ and dd′ =1. For a1, a2, b1, b2 − 2 2 6 2 2 ∈ Z/pZ, H6,a1,b1 and H6,a2,b2 are conjugate if a1 ǫb1 = a2 ǫb2. All other pairs of distinct subgroups listed above are not locally conjugate.− −

Proof. Let H be a 2 dimensional subgroup of ker ϕ that is not a subgroup of T . H T has dimension 1. Replace H with a conjugate so that H T is one of the subgroups of T as∩ listed 0 1∩ 1 0 0 ǫ in Lemma 7, i.e. H T = u , where u = I + p, I + p, or I + p. ∩ h i 0 0 0 1 1 0    −    0 1 Suppose that u = I + p. Choose a nonidentity element h H to be of the form 0 0 ∈   a 0 h = I + p, i.e. H = u, h . Since dim(H T ) = 1, a + d =0. If a = 0, then d = 0. h c d h i ∩ 6 6   1 can be replaced with h d so that H is still u, h and d =1. If c = 0 as well, then H = H . h i 2 c 0 1 Otherwise, H is conjugate to H via . If a = 0, then h can be replaced with h a so 1 0 1 6   that H = u, h , a = 1 and a + d =0. If c = 0, then H = H3,d. If c = 0, then H is conjugate h c i 1 6 6 d+1 d+1 to H1 via − . 0 1   1 0 Suppose that u = I + p. Choose a nonidentity element h H to be of the form 0 1 ∈  −  0 b 1 I + p. Since trace(p(h)) = d = 0, replacing h with h d makes d = 1. If b = c = 0, c d 6   1 0 then H = H . If b = 0, then H is conjugate to H c via . Otherwise, b = 0 and 5 6 4, b 0 b   0 1 c = 0, but conjugating H via reduces H to the case where b = 0. 6 1 0 6   11 0 ǫ Suppose that u = I+ p. Choose a nonidentity element h H so that trace(p(h)) = 1 0 ∈   1+ a b′ b′+ǫc′ 2, i.e. h is of the form h = I + p. Letting b = − ǫ , compute c′ 1 a 2  −  b ǫc ′+ ′ b ǫc 2ǫ ′+ ′ 1+ a b′ 0 ǫ − hu− 2ǫ = I + p I + p c′ 1 a 1 0   −       b′+ǫc′ 1+ a b′ 0 2 = I + p I + b +ǫc − p c′ 1 a ′ ′ 0   −    − 2ǫ   b′ ǫc′ 1+ a −2 = I + b +ǫc p − ′ ′ 1 a   2ǫ −   1+ a bǫ = I + p. b 1− a  −  1+ a bǫ Replacing h with I + p shows that H = H . b 1− a 6,a,b  −  It remains to determine local conjugacy among the listed subgroups. If H and H′ are locally conjugate and among the subgroups listed, then H T and H′ T must be locally ∩ ∩ conjugate by Lemma 8. Thus, H T and H′ T are equal due to the the way in which ∩ ∩ they were chosen in the beginning of the proof. In particular, H1,H2,H3,d are not locally conjugate to H4,c,H5,H6,a,b and H4,c,H5 are not locally conjugate to H6,a,b. For an element h ker ϕ, we will respectively call det(p(h)) and trace(p(h)) simply the determinant and trace∈ of h for the rest of the proof. The elements of H2 all have zero determinant, and so H2 is not locally conjugate to H1 or 0 x H where d = 0. Elements of H are of the form I + p and elements of H are of 3,d 6 2 0 y 3,0   y x 0 x the form I + p where x, y Z/pZ. By Lemmas 1 and 4, I + p is conjugate to 0 0 ∈ 0 y     y x I + p, and so H and H are locally conjugate. Suppose, for contradiction, that H 0 0 2 3,0 2   1 and H are conjugate, say via g GL (Z/pZ). In this case, H T = H T = g(H T )g− 3,0 ∈ 2 2∩ 3,0∩ 2∩ 0 1 because T is normal. Using that H T = H T = I + p , it is not difficult 3,0 ∩ 2 ∩ 0 0  1   to see that g must be upper triangular. However, H2 = gH2g− in this case. Hence, H2 and H3,0 are nontrivially locally conjugate. This fully categorizes local conjugacy of H2 with the other subgroups. An element of H1 with trace 1 can have any determinant, whereas a trace 1 element of d H3,d can only have determinant (d+1)2 . H1 and H3,d are therefore not locally conjugate by Proposition 4. This fully categorizes local conjugacy of H1 with the other subgroups. H is not locally conjugate to H where d = 0 because H has only elements of 3,0 3,d 6 3,0 determinant 0 whereas H3,d has elements of nonzero determinant. Moreover, H3, 1 is not locally conjugate to H where d = 1 because the latter has elements of nonzero− trace, 3,d 6 − whereas the former does not. Let d1,d2 Z/pZ such that d1,d2 =0, 1. For t Z/pZ and ∈ t 6 − ∈ di+1 x i = 1, 2, the trace t elements of H3,di are of the form I + tdi p where x Z/pZ. 0 di+1 ∈ 12   2 t di Such an element has determinant (di+1)2 . Thus, H3,d1 and H3,d2 are locally conjugate exactly 2 2 d1 d2 when H3,d1 Z(p ) = H3,d2 Z(p ) and 2 = 2 by Proposition 4. The latter ∩ ∩ (d1+1) (d2+1) condition is equivalent to 0= d (d + 1)2 d (d + 1)2 1 2 − 2 1 =(d d 1)(d d ), 1 2 − 2 − 1 i.e. d1 = d2 or d1d2 = 1. One can check that the latter condition implies the former condition.

Hence, H3,d1 and H3,d2 are locally conjugate exactly when d1 = d2 or when d1d2 = 1. Suppose, for contradiction, that H and H are conjugate but d = d . Let g 3,d1 3,d2 1 6 2 ∈ 1 0 1 GL (Z/pZ) satisfy gH g− = H . Using that H T = H T = I + p , 2 3,d1 3,d2 3,d1 ∩ 3,d2 ∩ 0 0 1     one can deduce that g must be upper triangular, but then gH3,d1 g− = H3,d1 , which contra- dicts H = H . Hence, H and H are nontrivially locally conjugate if d = d and 3,d1 6 3,d2 3,d1 3,d2 1 6 2 d1d2 = 1. This fully categorizes local conjugacy of H3,d with the other subgroups. x 1 The trace 1 elements of H are of the form I + p for x Z/pZ, and such an 4,c c 1 x ∈ 2 2 −  1 2 1 element has determinant x + x c. Compute x + x c = x 2 c + 4 , and so 2 − 1− − − − − − x + x c takes the value c + 4 exactly once and all other values in Z/pZ exactly 2 or 0 − − Z− Z
times. Therefore, if c,c′ /p are distinct, then H4,c and H4,c′ are not locally conjugate. Note that H Z(p2)=∈ I , whereas H Z(p2) = I . Thus, H and H are not locally 4,c ∩ h i 5 ∩ 6 h i 4,c 5 conjugate. This fully categorizes local conjugacy of H4,c and H5 with the other subgroups. 2 2 2 2 √ǫ ǫ Suppose a1, b1, a2, b2 Z/pZ satisfy a ǫb = a ǫb . Conjugating H6,a,b via − − ∈ 1− 1 2− 2 √ǫ ǫ ∈ −  GL2(Fp2 ) results in the group √ǫ 0 1 a + b√ǫ I + p,I + p .1 0 √ǫ a b√ǫ 1   −   −   α 0 Further conjugating this group by GL (F 2 ) results in 0 δ ∈ 2 p   √ǫ 0 1 (a + b√ǫ) α I + p,I + δ p . 0 √ǫ (a b√ǫ) δ 1   −   − α   Therefore, H6,a1,b1 is conjugate to H6,a2,b2 via 1 √ǫ ǫ − a + b √ǫ 0 √ǫ ǫ 2 2 , −√ǫ− ǫ 0 a + b √ǫ −√ǫ− ǫ −   1 1  −  which is a scalar multiple of 2 2 (a1 + a2) (b1 + b2) ǫ 2(a2b1 a1b2) 2(a2b−1 a1b2) 2 − 2 . − (a + a ) (b + b ) ǫ  ǫ 1 2 − 1 2  2 H6,a1,b1 and H6,a2,b2 are thus conjugate in GL2(Z/p Z). 1+a ǫb 2 2 + ǫx The trace 1 elements of H6,a,b are of the form I + b − 1 a p, whose determinant 2 + x −2 1 a2+ǫb2 2 1 a2+ǫb2  is − ǫx . This expression takes the value − exactly once and all values of Z/pZ 4 − 4 1See Lemma 14 13 exactly two or zero times. Therefore, if a , b , a , b Z/pZ satisfy a2 ǫb2 = a2 ǫb2, then 1 1 2 2 ∈ 1 − 1 6 2 − 2 H6,a1,b1 and H6,a2,b2 are not locally conjugate. This fully categorizes local conjugacy of H6,a,b with the other subgroups.  Lemma 11. The subgroups of ker ϕ of dimension 3 that are not subgroups of T are conjugate 2 in GL2(Z/p Z) to one of the following: 1 0 0 1 0 0 (1) H = I + p,I + p,I + p 1 0 1 0 0 1 1   −       1 0 0 1 0 0 (2) H = I + p,I + p,I + p 2 0 1 0 0 0 1   −       1 0 0 1 0 0 (3) H = I + p,I + p,I + p , where c Z/pZ with 0 3,c 0 1 1 0 c 1 ∈ ≤ p 1   −       c − . ≤ 2 1 0 0 ǫ 0 0 (4) H = I + p,I + p,I + p , where c Z/pZ with 0 4,c 0 1 1 0 c 1 ∈ ≤ p 1   −       c − . ≤ 2 No two distinct subgroups among these are locally conjugate. Proof. Let H be a 3 dimensional subgroup of ker ϕ that is not a subgroup of T . H T has dimension 2. Replace H with a conjugate so that H T is one of the subgroups∩ of T as listed in Lemma 9. In any of these cases, a third basis∩ element h of H can be chosen to be 0 0 1 of the form h = I + p, where d = 0. Replace h with h d so that d = 1. c d 6   1 0 0 1 Suppose that H T = I + p,I + p . If c = 0, then H is conjugate to ∩ 0 1 0 0 6   −     c 0 H via . Otherwise, H = H . 1 0 1 2   1 0 0 1 p 1 Suppose that H T = I + p,I + p . If0 c − , then H is H . ∩ 0 1 1 0 ≤ ≤ 2 3,c   −     1 0 Otherwise, H is conjugate to H3, c via . − 0 1  −  1 0 0 ǫ p 1 Suppose that H T = I + p,I + p . If0 c − , then H is H . ∩ 0 1 1 0 ≤ ≤ 2 3,c   −     1 0 Otherwise, H is conjugate to H4, c via . − 0 1 It remains to determine local conjugacy − among the listed subgroups. Similarly as in Lemma 10, if H and H′ are among the listed subgroups, then H T = H′ T . In particular, ∩ ∩ H1 and H2 are not locally conjugate to H3,c and H4,c and H3,c is not locally conjugate to H4,c. H and H are not locally conjugate because H Z(p2)= I , whereas H Z(p2) = I . 1 2 1 ∩ h i 2 ∩ 6 h i This fully categorizes local conjugacy of H1 and H2 with the other subgroups. p 1 2 H3,0 and H3,c where 0 < c −2 are not locally conjugate because H3,0 Z(p ) = 2 ≤ p 1 ∩ 6 H Z(p ). Let c ,c Z/pZ be distinct such that 0 < c ,c − . Elements h H 3,c ∩ 1 2 ∈ 1 2 ≤ 2 i ∈ 3,ci x y are of the form hi = I + p where x, y, z Z/pZ. The trace of such an y + ciz x + z ∈  − 14 element is z. Note that scaling x, y, z by r Z/pZ multiplies the trace(p(hi)) by r and 2 ∈ det(p(hi)) by r . Therefore, local conjugacy of H3,ci is determined by the determinants of its trace 1 elements and H Z(p2). Fixing z = 1, we compute det(p(h )) = x2 +x y2 c y. 3,ci ∩ i − − − i For each d Z/pZ, the number of solutions (x, y) to d = det(p(hi)) is the number of solutions 2 ∈2 2 1+ci 1 (x′,y′) to x′ + y′ = d + 4 , where x′,y′ Z/pZ are parametrized as x′ = x 2 and ci − ∈ − y′ = y + 2 . Z Z 2 2 p+1 Z Z For nonzero K /p , x′ and K y′ each take 2 distinct values over x′ /p and ∈ − 2 2 ∈ y′ Z/pZ respectively. By the pigeonhole principle, x′ + y′ = K has at least one solution ∈ 2 2 (x′,y′). In particular, ( x′, y′) yields at least two distinct solutions to x′ + y′ = K even ± ± 2 2 if one of x′ and y′ is 0. If p 1 (mod 4), then the equation x′ + y′ = K is equivalent ≡ 2 to (x′ + jy)(x′ jy) = K where j Z/pZ satisfies j = 1. The number of solutions to 2 2 − ∈ − x′ + y′ = K in this case is therefore p 1. 2 − 1+ci 2 2 The number of solutions to det(p(hi)) = 4 is the number of solutions to x′ + y′ = 0, 2 2 which is 1 if p 3 (mod 4) and 2p 1 if p 1 (mod 4). Since c1 = c2, the number of 2 ≡ 1+c1 − ≡ 6 solutions to det(p(h2)) = 4 is at least 2 and exactly p 1 if p 1 (mod 4). H3,c1 and − ≡  H3,c2 are thus not locally conjugate. Similarly, H4,c1 and H4,c2 are not locally conjugate. Local conjugacy in ker ϕ can be summarized as follows: Proposition 5. Let H ,H GL (Z/p2Z) be nontrivially locally conjugate in GL (Z/p2Z). 1 2 ≤ 2 2 H1 and H2 are conjugate, in some order, to the following subgroups for some d Z/pZ such that d = 1: ∈ 6 ± 1 0 0 1 d 0 0 1 I + p,I + p , I + I + p . 0 d 0 0 0 1 0 0             Proof. dim H1 = dim H2 and dim(H1 T ) = dim(H2 T ) by Lemmas 5 and 8. The claim follows from Lemmas 7, 9, 10 and 11. ∩ ∩ 

5. Local Conjugacy in GL2(Z/pZ)

Dickson [3] classifies the subgroups of GL2(Z/pZ) based on their images in PGL2(Z/pZ):

Proposition 6. Let p be an odd prime and let G be a subgroup of GL2(Z/pZ) with image H in PGL2(Z/pZ). If G contains an element of order p then G B(p) or SL2(Z/pZ) G. Otherwise, one of the following holds: ⊆ ⊆

(1) H is cyclic and a conjugate of G lies in Cs(p) or Cns(p). (2) H is dihedral and a conjugate of G lies in N(Cs(p)) or N(Cns(p)), but no conjugate of G lies in Cs(p) or Cns(p). (3) H is isomorphic to A4,S4 or A5 and no conjugate of G lies in N(Cs(p)) or N(Cns(p)). Proof. See [5, Section 2] or [7, Lemma 2].  Sutherland [6] uses this classification to identify local conjugacy among the subgroups of GL2(Z/pZ). Theorem 2. Let H ,H GL (Z/pZ) be nontrivially locally conjugate in GL (Z/pZ). H 1 2 ≤ 2 2 1 and H2 are, in some order, conjugate to the following groups: 1 1 1 1 D, , D′, , 0 1 0 1       15 1 0 1 0 1 − 2 w 0 where D C (p), D′ = D and D = D′, i.e. there is some D ≤ s 1 0 1 0 6 0 z ∈       z 0 such that D. 0 w 6∈   Proof. See [6, Lemma 3.6, Corollary 3.30]. 

6. Splitting A special case of the Schur-Zassenhaus Theorem, which is stated below in Theorem 3, 2 gives a sufficient condition for certain pairs of subgroups of GL2(Z/p Z) to be conjugate. Recall that a Hall subgroup of a finite group is a subgroup whose order is relatively prime to its index. Theorem 3 (Schur-Zassenhaus). If K is an abelian normal Hall subgroup of a finite group G, then there is a splitting ψ : G/K G which is unique up to conjugation. → Proof. See [4, Theorem 7.39, 7.40]  2 Proposition 7. Suppose that H1,H2 GL2(Z/p Z). If H1 ker ϕ = H2 ker ϕ, ϕ(H1) = ϕ(H ) and p does not divide ϕ(H ) , then≤ H and H are conjugate∩ in GL∩(Z/p2Z). 2 | i | 1 2 2 Proof. For i =1, 2, consider the short exact sequence 1 1 ker ϕ ϕ− (ϕ(H )) ϕ(H ) 1. → → i → i → Since ker ϕ is a 4 dimensional Z/pZ vector space, ker ϕ is abelian and ker ϕ = p4. Moreover, | | p does not divide ϕ(Hi) by assumption, and so the Schur-Zassenhaus Theorem yields a | 1| splitting ψ : ϕ(Hi) ϕ− (ϕ(Hi)), which is unique up to conjugation. Similarly, there is a splitting ψ : ϕ(H ) → H that arise from the short exact sequence i i → i 1 H ker ϕ H ϕ(H ) 1. → i ∩ → i → i → 1 1 Since Hi is a subgroup of ϕ− (ϕ(Hi)), ψi and ψ are both splittings into ϕ− (ϕ(Hi)). ψi 1 and ψ are thus conjugate in ϕ− (ϕ(Hi)) and by extension, ψ1 is conjugate to ψ2 via some 1 g ϕ− (ϕ(Hi)). One can express Hi as the internal semidirect product Hi =(Hi ker ϕ) ⋊ ∈ 1 1 ∩ ϕ(Hi). Conjugating the expression for H1 yields gH1g− =(g (H1 ker ϕ) g− ) ⋊ ϕ(H2). By 1 1 ∩ Lemma 3, g(H1 ker ϕ)g− = H1 ker ϕ because ϕ(g) ϕ (ϕ− (ϕ (Hi))) = ϕ(H1). Since ∩ ∩ 1 ∈ ϕ(H1) = ϕ(H2) by assumption, gH1g− = H2 and, so H1 and H2 are conjugate to each other. 

7. Computational Facts This section lists algebraic computations and facts resulting from such computations that are used in previous sections. 2 From this point on and unless stated otherwise, R will denote the ring (Zp[√ǫ])/(p Zp[√ǫ]), which is isomorphic to (Z/p2Z)[√ǫ]. The elements of R are identifiable as the sums a + b√ǫ where a, b Z/p2Z. Letϕ ˜ denote the natural homomorphism ∈ ϕ˜ : GL (R) GL (Z [√ǫ]/pZ [√ǫ]) GL (F 2 ). 2 → 2 p p ≃ 2 p In particular,ϕ ˜ is an extension of ϕ.

2 2 z 0 ′ w 0 Since D Cs(p ), D if and only if D by Lemma 16. ≤ 0 w ∈ 0 z ∈   16  2 p There is a splitting (Z/pZ)× (Z/p Z)× given by x x , where x (Z/pZ)× and x is any lift of x in Z/p2Z. This map→ is well defined because7→ (x + ap)p x∈p (mod p2) for all x, a Z/p2Z by the bionamial theorem. It is a splitting as xp x (mod≡ p). Similarly, there ∈ p2 ≡ is a splitting (Z/pZ[√ǫ])× F×2 R× given by x x where x F×2 and x is any lift of ≃ p → 7→ ∈ p x in R. Let S denote the image of the map (Z/pZ[√ǫ])× R×. For x F 2 , we will often abuse → ∈ p notation and let x also denote the lift of x Fp2 in S. In particular, given w,x,y,z,a,b,c,d Z/pZ, write ∈ ∈ w 0 a b + p 0 w c d     w 0 a b + p 0 z c d     0 x a b + p y 0 c d     to denote some elements of GL (Z/p2Z); w,x,y,z as written above are elements of S Z/p2Z. 2 ∩ Likewise, given w,x,y,z,a,b,c,d F 2 , write ∈ p w 0 a b + p 0 w c d     w 0 a b + p 0 z c d     0 x a b + p y 0 c d    

to denote some elements of GL2(R).

Lemma 12. For w (Z/pZ)× and a,b,c,d Z/pZ, ∈ ∈ p 1 w 0 a b − 1 a b + p = I p 0 w c d − w c d        and p w 0 a b w 0 + p = . 0 w c d 0 w        For w R× and a,b,c,d R, ∈ ∈ p2 1 w 0 a b − 1 a b + p = I p 0 w c d − w c d        and p2 w 0 a b w 0 + p = . 0 w c d 0 w    17    w 0 a b Proof. Since multiplicatively commutes with , the Binomial Theorem ap- 0 w c d plies.      w 0 Corollary 1. Let H be a subgroup of GL (Z/p2Z). Then, H contains h = + 2 0 w   a b w 0 p, where w (Z/pZ)× and a,b,c,d Z/pZ, if and only if H contains h = c d ∈ ∈ 1 0 w     a b and h = I + p. 2 c d   1 1 a b Proof. Suppose that h H. h H by Lemma 12 and hh− = I + p. Therefore, ∈ 1 ∈ 1 w c d   1 w a b (hh− ) = I + p = h , and so h H. 1 c d 2 2 ∈   1 a b Conversely, if h , h H, then H contains h h w = h I + 1 p = h.  1 2 ∈ 1 2 1 w c d     Lemma 13. For w, z (Z/pZ)× and a,b,c,d Z/pZ such that w = z, ∈ ∈ 6 p 1 w 0 a b − a 0 + p = I + w p 0 z c d −0 d       − z  and p w 0 a b w 0 0 b + p = + p 0 z c d 0 z c 0          For w, z R× and a,b,c,d Z/pZ such that w = z, ∈ ∈ 6 p2 1 w 0 a b − a 0 + p = I + w p 0 z c d −0 d       − z  and p2 w 0 a b w 0 0 b + p = + p 0 z c d 0 z c 0          Proof. Suppose that w, z (Z/pZ)× and a,b,c,d Z/pZ. By expanding, ∈ ∈ p 1 p 1 p 2 k p 2 k w 0 a b − w 0 − − w 0 a b w 0 − − + p = + p 0 z c d 0 z 0 z c d 0 z        k=0       ! p X 2 p 2 k p 2 k − aw − bw z − − = I + p 2 k k p 2 p. cw − − z dz − k X=0   p 2 k p 2 k p 2 p 2 k k Since w z (mod p), k−=0 w z − − and k−=0 w − − z are both geometric series evalu- ating to6≡ 0. Therefore, P P p 1 p 2 w 0 a b − aw − 0 + p = I + − p 2 p 0 z c d 0 dz −       −  a 0 = I + w p. −0 d  − z  18 From here,

p w 0 a b w 0 0 b + p = + p 0 z c d 0 z c 0         

can be immediately calculated. The claims made for w, z R× and a, b, c, d, R such that w = z can be proved similarly. ∈ ∈  6 w 0 a b Corollary 2. Let H be a subgroup of GL (R). H contains h = + p, 2 0 z c d     where w, z (Z/pZ)× and a,b,c,d Z/pZ with w = z, if and only if H contains h1 = a ∈0 w 0 0∈ b 6 I + w p and + p −0 d 0 z c 0  − z      Proof. This is immediate from Lemma 13. 

Lemma 14. For w,y,a,b,c,d Z/pZ, ∈ 1 − b b √ǫ ǫ a b √ǫ ǫ 1 (a + d)+ ǫ + c √ǫ (a d)+ ǫ + c √ǫ (1) − − − − = 2 b − − b √ǫ ǫ c d √ǫ ǫ (a d)+ ǫ c √ǫ (a + d)+ ǫ c √ǫ −    −  1  − −
− −
 √ǫ ǫ w ǫy √ǫ ǫ − w + √ǫy 0 (2) =

−√ǫ− ǫ y w −√ǫ− ǫ 0 w √ǫy −    −  1  −  √ǫ ǫ w ǫy √ǫ ǫ − 0 w √ǫy (3) = −√ǫ− ǫ y w −√ǫ− ǫ w + √ǫy −0 −  − −  −    w 0 a b Lemma 15. Let R be a ring. If GL (R) and Mat (R), then 0 z ∈ 2 c d ∈ 2     1 w 0 a b w 0 − a b w = z . 0 z c d 0 z c z d        w  0 x a b Lemma 16. Let R be a ring. If GL (R) and Mat (R), then y 0 ∈ 2 c d ∈ 2     1 x 0 x a b 0 x − d c = y . y 0 c d y 0 b y a        x  In particular,

1 0 1 a b 0 1 − d c = . 1 0 c d 1 0 b a         Lemma 17. For a,b,c,d Z/pZ and n Z, ∈ ∈ n c(n 1)n (a+d+c(n 1))(n 1)n n 1 2 − − − − 1 1 a b 1 n an + 2 2 c k=0 k + bn + p = + c−(n 1)n p 0 1 c d 0 1 cn dn + −         2 P  19 Proof. Expand n n n 1 k n 1 k 1 1 a b 1 1 − 1 1 a b 1 1 − − + p = + p 0 1 c d 0 1 0 1 c d 0 1 k        X=0       n 1 1 n − 1 k a b 1 n 1 k = + p 0 1 0 1 c d 0− 1 − k   X=0       n 1 1 n − a + ck (a + ck)(n 1 k)+ b + dk = + p 0 1 c c(n −1 −k)+ d k   X=0  − −  n 1 1 n − a + ck a(n 1) ak + ck(n 1) ck2 + b + dk = + p. 0 1 c− − c(n 1) −ck +−d k   X=0  − −  n 1 (n 1)n − − Since p is an odd prime, k=0 k = 2 , and so n c(n 1)n (a+d+c(n 1))(n 1)n n 1 2 − − − − 1 1 a b P 1 n an + 2 2 c k=0 k + bn + p = + c−(n 1)n p 0 1 c d 0 1 cn dn + −         2 P  

Lemma 18. Let R be a ring. For a,b,c,d R, ∈ 1 1 1 a b 1 1 − a + c a + b c + d = . 0 1 c d 0 1 c − c +− d        −  1 1 Lemma 19. Let H GL (Z/p2Z) with t = ϕ(H). Suppose that k = I + ≤ 2 0 1 ∈   a b p H. c d ∈   0 1 (1) If a = d, then I + p H. 6 0 0 ∈   0 1 1 0 (2) If c =0, then I + p,I + p H. 6 0 0 0 1 ∈    −  1 1 1 Proof. (1) By Lemma 3, tkt− H. Let k′ = tkt− k− , which must be in H as well. Using Lemma 18, compute ∈

1 1 a + c a + b c + d a b k′ = tkt− k− = I + − − p I p c c + d − c d   −       c a c + d = I + p. 0 − − c  −  0 1 If c = 0, then a power of k′ is I + p because a = d by assumption. If c = 0, 0 0 6 6   0 1 then I + p H because 0 0 ∈   1 1 0 2c tk′t− k′− = I + p 0− 0   20 is an element of H. 0 1 1 1 c a c + d (2) By 1, I + p H. Furthermore, since tkt− k− = I + − − p H, 0 0 ∈ 0 c ∈    −  c 0 1 0 I + p H as well. Thus, I + p H. 0 c ∈ 0 1 ∈  −   −  

1 1 0 1 Lemma 20. Let H GL (Z/p2Z) with t = ϕ(H). If p> 3, then I + p ≤ 2 0 1 ∈ 0 0 ∈ H.    

1 1 a b Proof. There is some element h H of the form + p. By Lemma 17, the ∈ 0 1 c d pth power of h is    

c(p 1)p (a+d+c(p 1))(p 1)p p 1 2 − − − − p 1 p ap + 2 2 c k=0 k + bp h = + c−(p 1)p p 0 1 cp dp + −    2 P  p 1 2 1 p 0 c − k = + k=0 p. 0 1 0− 0    P 

p 1 2 (p 1)p(2p 1) 0 1 Since p> 3, − k = − − , which is 0 modulo p. Therefore, I + p H,  k=0 6 0 0 ∈   P Lemma 21. For a, b, c, d, β Z/pZ, ∈ 1 1 a b 0 β 1 1 a b + β + p I + p = + p. 0 1 c d 0 0 0 1 c d              Lemma 22. For a,b,c,d,α,δ Z/pZ, ∈ 1 1 a b α 0 1 1 a + α b + δ + p I + p = + p. 0 1 c d 0 δ 0 1 c d + δ              Lemma 23. For a,b,c,d,α,γ,δ Z/pZ, ∈ 1 1 a b α 0 1 1 a + α + γ b + δ + p I + p = + p. 0 1 c d γ δ 0 1 c + γ d + δ              1 1 a b w 0 Lemma 24. Let H GL (Z/p2Z). If τ = + H and h = H ≤ 2 0 1 c d ∈ 0 z ∈ where w z (mod p), then a = d or there is some element ofH of the form   6≡ α 0 I + p 0 δ   where α, δ Z/pZ are unequal. ∈ 21 Proof. Using Lemmas 15 and 17, compute w w c z +1 z w w w w w ( ) 1 z 1 z a b z 1 z a z + 2 hτh− τ − = + z p − + − ∗ w w p 0 1 c d 0 1 w w c( z +1) z w c d + ! !        − z − z 2 = I + + p c∗z ∗ ∗c w ∗  w ∗ − z ∗ = I + p. c z ∗w ∗  w − z ∗ 1 0 If c = 0 and w = 1, then c
z w = 0 and so I + p H by Lemma 19. If c = 0, 6 z 6 ± w − z 6 0 1 ∈  −  then compute

w w w w w 1 z 1 z a b z 1 z a z hτh− τ − = + p + p 0 1 0 d 0− 1 −0 ∗d w         − z   a a w + b w a w = I + z z + z p 0 − d −0 ∗d w    − z  a 1 w = I + z p. 0− d 1 ∗ w 
− z  0 1 If a = d, then I + p H by Lemma
19 and since w = 1 by assumption, I + 6 0 0 ∈ z 6 w   a 1 z 0 w w − w p H with a 1 z = d 1 z . 0 d 1 z ∈ − 6 −  w
−  If = 1, then compute

z −
w 1 1 1 a b 1 1 a hτh− τ − z = + p + p 0− 1 c− d 0 1 c d∗   −        2a c = I + p. 0− 2d∗ c  −  2a c 0 If a = d, then I + − p H with 2a c =2d c.  6 0 2d c ∈ − 6 −  −  8. The Center case 2 This section categorizes local conjugacy for subgroups H of GL2(Z/p Z) whose images under ϕ lie in Z(p). To do so, we first understand the structure of such H. Let σ : Z(p) → 2 w 0 w 0 Z(p ) be the splitting given by where w (Z/pZ)× and w S is the 0 w 7→ 0 w ∈ ∈ lift of w. For all h ϕ(H), σ(h) is an element of Hby Lemma 12. It is not difficult to see that H is the direct∈ product σ(ϕ(H)) (H ker ϕ) by Corollary 1. × ∩ 2 Lemma 25. Let H1,H2 GL2(Z/p Z) such that ϕ(Hi) Z(p) for i = 1, 2. Then, H1 ≤ 2 ≤ and H2 are nontrivially locally conjugate in GL2(Z/p Z) if and only if ϕ(H1)= ϕ(H2) and H ker ϕ and H ker ϕ are nontrivially locally conjugate in GL (Z/p2Z). 1 ∩ 2 ∩ 2 Proof. If H1 and H2 are nontrivially locally conjugate, then ϕ(H1) and ϕ(H2) are locally conjugate in GL2(Z/pZ) by Proposition 3. Since ϕ(Hi) Z(p), ϕ(H1) must equal ϕ(H2). 22 ≤ 2 Furthermore, H1 ker ϕ and H2 ker ϕ are locally conjugate in GL2(Z/p Z) by Proposition 2. Suppose, for contradiction,∩ that∩ H ker ϕ and H ker ϕ are conjugate in GL (Z/p2Z). 1 ∩ 2 ∩ 2 Replace H1 with a conjugate so that H1 ker ϕ = H2 ker ϕ. This conjugation preserves ϕ(H ). Since ϕ(H ) Z(p) and Z(p) = ∩p 1, H and∩H are conjugate by Proposition 7. 1 i ≤ | | − 1 2 This contradicts that H1 and H2 are nontrivially locally conjugate. Hence, H1 ker ϕ and H ker ϕ are nontrivially locally conjugate. ∩ 2 ∩ Conversely, suppose that ϕ(H1) = ϕ(H2) and H1 ker ϕ and H2 ker ϕ are nontrivially locally conjugate in GL (Z/p2Z). Since H = σ(ϕ(H∩ ) (H ker ϕ∩), and since σ(ϕ(H )) 2 i 1 × i ∩ 1 consists only of scalar matrices, it is not difficult to see that H1 and H2 are locally conjugate. They are not conjugate because any conjugation from H1 to H2 yields a conjugation from H ker ϕ to H ker ϕ. Hence, H and H are nontrivially locally conjugate.  1 ∩ 2 ∩ 1 2

9. The Cartan cases 2 This section categorizes the subgroups, up to conjugation, of GL2(Z/p Z) whose images under ϕ are subgroups of Cs(p) or Cns(p) but not subgroups of Z(p). The notation below will make the Cartan split and Cartan nonsplit cases similar to each other.

2 Definition 3. Let H GL2(R). Say that H is of type Cs if H GL2(Z/p Z) and ≤ 2 ≤ ϕ(H) Cs(p). Say that H is of type N(Cs) if H GL2(Z/p Z) and ϕ(H) N(Cs(p)). ≤ ≤1 ≤ √ǫ ǫ − Let H′ be the conjugate of H via − − . Say that H is of type Cns if H′ √ǫ ǫ ≤ 2 −  2 GL (Z/p Z) and ϕ˜(H′) C (p). Say that H is of type N(C ) if H′ GL (Z/p Z) and 2 ≤ ns ns ≤ 2 ϕ˜(H′) N(C (p)). ≤ ns Say that H is in diagonalized form if H is of type N(Cs) or of type N(Cns).

Suppose H GL2(R) is of type N(Cns). The elements ofϕ ˜(H) must be of the form w + √ǫy ≤0 0 w √ǫy or of the form by Lemma 14. From now on, 0 w √ǫy w + √ǫy −0  −    √ǫ ǫ let K denote the conjugate of ker ϕ via . Note that K is a 4 dimensional Z/pZ −√ǫ− ǫ vector space just as ker ϕ is. By Lemma 14,− 

1 0 0 1 √ǫ 0 0 √ǫ K = I + p,I + p,I + p,I + p . 0 1 1 0 0 √ǫ √ǫ 0       −  −   Note that H kerϕ ˜ is a subgroup of K when H is of type N(C ). On the other hand, ∩ ns H kerϕ ˜ is a subgroup of ker ϕ when H is of type N(Cs). ∩ 2 Lemma 26 and Corollaries 3 and 4 below show that local conjugacy in GL2(Z/p Z) be- tween two subgroups H ,H of GL (Z/p2Z) such that ϕ(H ) N(C (p)) is equivalently 1 2 2 i ≤ ns determined by local conjugacy between H1′ and H2′ in GL2(R), where Hi′ is the conjugate of √ǫ ǫ H via . i −√ǫ− ǫ −  2 Lemma 26. If g1,g2 GL2(Z/p Z) are conjugate via g GL2(R), then g1 and g2 are ∈ 2 ∈ conjugate via some g′ GL2(Z/p Z), whose value is only dependent on g. ∈ 23 Proof. Express g in the form α + α √ǫ β + β √ǫ g = 1 2 1 2 , γ + γ √ǫ δ + δ √ǫ  1 2 1 2  where α , β ,γ , δ Z/p2Z for i =1, 2. Since the entries of g and g are in Z/p2Z, i i i i ∈ 1 2 αi βi αi βi h1 = h2 . γi δi γi δi     αi βi 1 αi βi Therefore, if is invertible for i =1 or 2, then h2 = g′h1g′− , where g′ = . γi δi γi δi Otherwise, αδ β γ= 0 for i =1, 2, in which case   i i − i i det(g)=(α δ β γ )+(α δ β γ )ǫ +(α δ + α δ β γ β γ )√ǫ 1 1 − 1 1 2 2 − 2 2 1 2 2 1 − 1 2 − 2 1 =(α δ + α δ β γ β γ )√ǫ. 1 2 2 1 − 1 2 − 2 1 α1 + α2 β1 + β2 Since g is invertible, α1δ2 + α2δ1 β1γ2 β2γ1 = 0. Let g′ = . Note that − − 6 γ1 + γ2 δ1 + δ2   g′ is invertible because

det(g′)=(α δ + α δ + α δ + α δ ) (β γ + β γ + β γ + β γ ) 1 1 1 2 2 1 2 2 − 1 1 1 2 2 1 2 2 =(α δ β γ )+(α δ β γ )+(α δ + α δ β γ β γ ) 1 1 − 1 1 2 2 − 2 2 1 2 2 1 − 1 2 − 2 1 =(α δ + α δ β γ β γ ) 1 2 2 1 − 1 2 − 2 1 =0. 6 1 Furthermore, g′g1 = g2g′, and so g2 = g′g1g′− as desired.  The idea behind Lemma 26 can be immediately extended to the following corollaries:

2 Corollary 3. Two subgroups of GL2(Z/p Z) that are conjugate in GL2(R) are conjugate in 2 GL2(Z/p Z). 2 Corollary 4. Two subgroups of GL2(Z/p Z) that are locally conjugate in GL2(R) are locally 2 conjugate in GL2(Z/p Z).

Lemma 27. Let H GL2(R) be in diagonialized form. Suppose that there is some h H w≤ 0 a b ∈ such that ϕ˜(h) = where w = z. Then, H has some element k = I + p if 0 z 6 c d     a 0 0 b and only if H has both l = I + p and m = I + p. 0 d c 0     Proof. Since k = lm, all three of l,m,k are in H if two of them are. In particular, if l, m, H, then k H. ∈ Conversely,∈ assume that k H. Use Lemma 15 to compute ∈ w 1 1 0 b 1 hkh− k− = I + z p. c z 1 0−  w −
 1 1 0 2b
1 1 If w = z, then hkh− k− = I + − p. Moreover, m is a power of hkh− k− , and − 2c 0 so m H. Thus, l H as well. −  ∈ ∈ 24 1 1 0 0 Now assume that w = z. If b = 0, then hkh− k− = I + p. If H is of 6 ± c z 1 0  w −  type N(Cns) as well, then c = 0 by Lemma 14, and there is nothing to prove. Otherwise, H z 2
is of type N(Cs), in which case w 1 is in Z/p Z and nonzero modulo p, and so a power of 1 1 − hkh− k− is m and we are done. The case where c = 0 is similar. Assume that b, c = 0. Use Lemma 15 to compute 6 w w 1 1 1 0 b z 1 z h(hkh− k− )h− = I + p c z 1 w −0  w − z
 1 1 1 1 1 to see that hkh− k− and h(hkh− k− )h− are linearly
independent as Z/pZ-vectors. More- over, both have p-parts whose diagonal entries are 0. Note that the subspaces of K and ker ϕ consisting of the matrices whose p-parts have 0 as their diagonal entries are both 2 dimensional. Therefore, m H as desired.  ∈ For H GL2(R) in diagonalized form, define ∆H as the set of elements of H kerϕ ˜ whose p-parts are≤ diagonal. In other words, ∩ a 0 I + p H ker ϕ if H is of type N(Cs) ( 0 d! ∈ ∩ ) ∆H =   a 0  I + p H K if H is of type N(Cns). ( 0 d! ∈ ∩ )  Similarly, define   0 b I + p H ker ϕ if H is of type N(Cs) ( c 0! ∈ ∩ ) ∆H⊥ =   0 b  I + p H K if H is of type N(Cns). ( c 0! ∈ ∩ )  Lemma 27 then yields the following:  Corollary 5. Let H GL2(R) be in diagonalized form. Suppose that there is some h H w ≤0 ∈ such that ϕ˜(h)= where w = z. 0 z 6   (1) The groups ∆ and ∆⊥ together generate H kerϕ ˜. H H ∩ (2) Suppose that w = z. If H is of type N(Cs), then ∆H⊥ is one of the following: 0 16 ± 0 0 (a) I + p,I + p 0 0 1 0       0 1 (b) I + p 0 0     0 0 (c) I + p 1 0 (d) I .    h i If H is of type N(Cns), then ∆H⊥ is one of the following: 0 1 0 √ǫ (a) I + p,I + p 1 0 √ǫ −0 (b) I .      h i 25 Proof. (1) This is immediate from Lemma 27. 0 b (2) Suppose that H is of type N(C ). If there is some k = I + p ∆⊥ where s c 0 ∈ H w   1 0 b z w b, c = 0, then hkh− = I + p is linearly independent to k because bc = 6 c z 0 z 6  w  0 1 0 0 bc z . Thus, ∆ = I + p,I + p . If no such k is in ∆ , then ∆ is w H⊥ 0 0 1 0 H⊥ H⊥       0 1 0 0 one of I + p , I + p and I . 0 0 1 0 h i         0 b Suppose that H is of type N(C ). If there is some nonidentity k = I + p, ns c 0   then one of b or c is nonzero. Since b and c are of the form α + √ǫδ and α √ǫδ for some α, δ Z/pZ by Lemma 14, b and c are both nonzero. Similarly as in− the last ∈ 1 paragraph, hkh− and k are linearly independent, and so ∆H⊥ must be 2-dimensional 0 1 0 √ǫ and hence equal to I + p,I + p . 1 0 √ǫ −0       

9.1. The Cartan Cases. This section categorizes the subgroups H of GL2(R) of type Cs or of type C such thatϕ ˜(H) Z(p) up to conjugation to understand local conjugacy among ns 6≤ such subgroups. For H GL2(R) let DH denote the group consisting of all elements of H that are diagonal matrices.≤ Proposition 8. Let H GL (R) be of type C or of type C such that ϕ˜(H) Z(p). ≤ 2 s ns 6≤ There is a conjugate H′ of H satisfying the following:

(1) H′ is of type Cs if H is of type Cs and H′ is of type Cns if H is of type Cns (2) ϕ˜(H′)=ϕ ˜(H) (3) H′ is the internal semidirect product ∆⊥ ⋊ DH H′ ′

Proof. Suppose that H is of type Cs, i.e. ϕ(H) is a subgroup of Cs(p). The image of Cs(p) in PGL2(p) is cyclic and so Cs(p) is generated by two elements, one of which is of the form w 0 w 0 where w = z and the other of which is of the form 0 . By Corollaries 1 and 0 z 6 0 w0     w 0 w 0 0 b 2, H has elements of the form h = 0 and h = + p. Note that H 0 0 w 0 z c 0  0     w 0 is generated by h, h and H ker ϕ. Let h′ = and let H′ = h′, h ,H ker ϕ . 0 ∩ 0 z h 0 ∩ i   We show that H′ ker ϕ = H ker ϕ. Clearly, H′ ker ϕ H ker ϕ, and so it suffices to ∩ ∩ ∩ ⊇ ∩ n show that H′ ker ϕ H ker ϕ. The elements of H′ are of the form m = m where ∩ ⊆ ∩ i=1 i mi = h′, mi = h0, or mi H ker ϕ for each i. By Lemmas 2 and 3, if mi H ker ϕ, then ∈ ∩ ∈ Q∩ there is some m′ H ker ϕ such that m h′ = h′m′ . Moreover, h is a scalar matrix and hence i ∈ ∩ i i 0 commutes with h′ and all elements of H ker ϕ, and so m is alternatively of the form m = n1 n2 n3 ∩ 2 h′ h0 i=1 ki, where ki H ker ϕ for each i. Due to the way that w,z,w0 (Z/p Z)× are ∈ ∩ n1 n2 n1 n2 ∈ 3 chosen based on w,z,w0 (Z/pZ)×, h′ h = I whenever ϕ(h′) ϕ(h0) = I. Therefore, Q ∈ 0 if m H′ ker ϕ, then m H ker ϕ, and so H′ ker ϕ = H ker ϕ as desired. ∈ ∩ ∈ ∩ ∩ ∩ 3Recall S as defined in Section 7 26 Since ϕ(H)= ϕ(H′) and C (p) is indivisible by p, H and H′ are conjugate by Proposition | s | 7. Note that H′ ker ϕ is generated by ∆H and ∆⊥ by Corollary 5. Thus, H′ is generated by ∩ ′ H′ h′, h , ∆H and ∆⊥ . It is not difficult to see that ∆⊥ is a normal subgroup of H′, that H′ = 0 ′ H′ H′ (∆⊥ )( h′, h0, ∆H ) and that ∆⊥ h′, h0, ∆H = I . Therefore, H′ = ∆⊥ ⋊ h′, h0, ∆H . H′ h ′ i H′ ∩h ′ i h i H′ h ′ i Furthermore, DH = h′, h0, ∆H , and so H′ = ∆⊥ ⋊ DH . ′ h ′ i H′ ′ The case where H is of type Cns is similar.  Before proceeding, we introduce a definition which will be useful for understanding the 2 structure of nontrivially locally conjugate subgroups of GL2(Z/p Z).

Definition 4. Let R be a ring and let D1 and D2 be two subgroups of GL2(R) consisting only w 0 of diagonal matrices. Say that D and D are diagonal swaps if D exactly 1 2 0 z ∈ 1   z 0 0 1 when D . Equivalently, D and D are conjugate to each other via . 0 w ∈ 2 1 2 1 0     It will turn out that up to conjugation, pairs of nontrivially locally conjugate subgroups 2 2 of GL2(Z/p Z) are generated by some equal generators along with two subgroups of Cs(p ) which are unequal diagonal swaps. Proposition 9. Let H ,H GL (R) both be of type C or of type C . Suppose that H 1 2 ≤ 2 s ns 1 and H are locally conjugate in GL (R), ϕ˜(H ) Z(p), ϕ˜(H )=ϕ ˜(H ) and H = ∆⊥ ⋊D i . 2 2 i 6≤ 1 2 i Hi H (1) The groups DH1 and DH2 are equal or are diagonal swaps.

(2) If H1 and H2 are nontrivially locally conjugate and DH1 = DH2 , then they are of type 0 1 0 0 C and one of ∆ and ∆ is I + p and the other is I + p. s H⊥1 H⊥2 0 0 1 0       

Proof. (1) Suppose that H1 and H2 are both of type Cs. Elements of Hi are expressible

as the product kh for unique k ∆⊥ and h D i . Fix h D i so that ϕ(h) = ∈ Hi ∈ H ∈ H w 0 w 0 , where w, z (Z/pZ)× are unequal. Express h in the form h = + 0 z ∈ 0 z     a 0 0 β p. For k ∆⊥ , express k in the form k = I + p. Compute 0 d ∈ Hi γ 0     0 β w 0 a 0 kh = I + + p γ 0 0 z 0 d         w 0 a βz = + p 0 z γw d     and trace(kh)=(w + z)+(a + d)p = trace(h) det(kh)=(w + ap)(z + dp) = det(h). By Theorem 1, kh and h are in the same conjugacy class, i.e. the conjugacy class of kh does not depend on k. The above calculation also shows that the only matrices z 0 d 0 that could be elements of D and conjugate to h are h itself and + p. Hi 0 w 0 a 27     w 0 z 0 Suppose that there is some ϕ(H ) such that ϕ(H ). Since 0 z ∈ 1 0 w 6∈ 1     w 0 a 0 H = ∆⊥ ⋊ D , H by Corollary 2. If I + H , then 1 H1 H1 0 z ∈ 1 0 d ∈ 1    

w 0 a 0 w 0 aw 0 I + p = + p 0 z 0 d 0 z 0 dz          

z 0 w 0 aw 0 is an element of H . Since ϕ(H ), + p H , and so 1 0 w 6∈ 2 0 z 0 dz ∈ 2       a 0 I + p H as well. Therefore, D D and D D by symmetry. 0 d ∈ 2 H1 ⊆ H2 H2 ⊆ H1   w 0 z 0 Now suppose that for all ϕ(H ), ϕ(H ) as well. For any 0 z ∈ 1 0 w ∈ 1     a 0 w 0 aw 0 w 0 aw 0 I + p H , + p H , and so + p or 0 d ∈ 1 0 z 0 dz ∈ 1 0 z 0 dz           z 0 dz 0 a 0 + p is in H . In the former case, I + p H . In the latter, 0 w 0 aw 2 0 d ∈ 2       d 0 1 0 I + p H . If ∆ is 2 dimensional, i.e. it is generated by I + p and 0 a ∈ 2 H1 0 1     1 0 1 0 1 0 I + p, then ∆ contains both I + p and I + p, and so 0 1 H2 0 1 0 1  −     −  a 0 ∆ = ∆ . If ∆ is 1 dimensional, then say that it is generated by I + p. H1 H2 H1 0 d   a 0 d 0 In particular, ∆ is not 2 dimensional. If I + p ∆ , then I + p H2 0 d 6∈ H2 0 a ∈     0 1 ∆ . In this case, let H be the conjugate of H via . By Lemma 16, H2 2′ 2 1 0   a 0 I + p ∆ , and so ∆ = ∆ . If ∆ is 0 dimensional, then so is ∆ , 0 d ∈ H2′ H1 H2′ H1 H2   concluding the case where H1 and H2 are both of type Cs. The case where H1 and H2 are both of type Cns is similar.

(2) Suppose that H1 and H2 are nontrivially locally conjugate and DH1 = DH2 . Further suppose that H1 and H2 are both of type Cs. By Proposition 2 and Lemma 5,

dim(∆H⊥1 ) = dim(∆H⊥2 ). Note that dim(∆H⊥i ) is not 2 or 0 because H1 = H2 otherwise. 0 bi Hence, dim(∆H⊥i ) = 1. Say that I + p generates ∆H⊥i . In particular, bi or ci ci 0 28  α 0 is nonzero. By Lemma 15, there is some diagonal matrix i such that 0 δi   0 1 I + if bici = 0 and bici is a square  1 0! 6   0 ǫ 1 I + if bici is not a square −  1 0 αi 0 0 bi αi 0  ! I + p =  0 δi ci 0 0 δi  0 1         I + if bi =0 0 0! 6   0 0 I + if ci =0.  1 0 6  !  By the categorization of subgroups of ker ϕ up to conjugacy in GL (Z/p2Z) and  2 techniques, particularly those involving Proposition 4, used to obtain these catego- rization as discussed in Sections 4.4, 4.5 and 4.6, b1c1 and b2c2 must be both zero, αi 0 both nonzero square or both nonsquares. Replace Hi by its conjugate via . 0 δi   DHi is preserved in this conjugation because diagonal matrices multiplicatively com-

mute. Therefore, for H1 and H2 to be nontrivially locally conjugate, one of DH1 and 0 1 0 0 D must be I + p and the other must be I + p . This must H2 0 0 1 0 have have been true before the conjugation as well.    

Now suppose that H1 and H2 are both of type Cns. Note that ∆H⊥i consists only 0 a √ǫc of elements of the form I + p. Similarly as in the type C case, a + √ǫc −0 s   0 ai + √ǫci dim(∆H⊥1 ) = dim(∆H⊥2 ) = 1. Say that I + p generates ∆H⊥i . ai √ǫci 0  −  a + √ǫc 0 At least one of a and c is nonzero modulo p, and so 2 2 is i i 0 a + √ǫc  1 1 a + √ǫc 0 invertible. Conjugating H by 2 2 yields H by Lemma 15, and 1 0 a + √ǫc 2  1 1 so H1 and H2 are conjugate to begin with, a contradiction. Hence, H1 and H2 must both be of type Cs.  2 Corollary 6 below categorizes local conjugacy for the subgroups of GL2(Z/p Z) whose images under ϕ are contained in Cns(p) but not Z(p). 2 Corollary 6. Let H1,H2 GL2(Z/p Z) with ϕ(Hi) Cns(p) but ϕ(Hi) Z(p). If H1 and ≤ 2 ≤ 6≤ 2 H2 are locally conjugate in GL2(Z/p Z), then they are conjugate in GL2(Z/p Z).

Proof. It suffices to show that H1 and H2 are conjugate in GL2(R) by Corollary 3. Replace √ǫ ǫ H with its conjugate via so that H is of type C . Using Proposition 8, i −√ǫ− ǫ i ns −  further replace Hi with a conjugate such that Hi is still of type Cns,ϕ ˜(Hi) is preserved

and Hi = ∆H⊥i ⋊ DHi . Proposition 9 shows that H1 and H2 are conjugate in GL2(R) as desired.  29 2 Proposition 10 below categorizes local conjugacy for the subgroups of GL2(Z/p Z) whose images under ϕ are contained in Cs(p). 2 Proposition 10. Let H1,H2 GL2(Z/p Z) with ϕ(Hi) Cs(p). Then, H1 and H2 are nontrivially locally conjugate if≤ and only if they are conjugate≤ to the groups 0 1 0 1 D,I + p and D′,I + p 0 0 0 0         2 in some order, where D and D′ are subgroups of Cs(p ), D and D′ are diagonal swaps, and D = D′. 6 Proof. Suppose that H1 and H2 are nontrivially locally conjugate. By Proposition 3, ϕ(H1) and ϕ(H2) are locally conjugate. By Theorem 2, ϕ(H1) and ϕ(H2) are conjugate, and so H2 can be replaced with a conjugate so that ϕ(H1)= ϕ(H2). If ϕ(Hi) Z(p), then H1 ker ϕ and H2 ker ϕ must be nontrivially locally conjugate. Proposition≤ 5 asserts that H ∩ ker ϕ and H ∩ ker ϕ are, in some order, conjugate to 1 ∩ 2 ∩ 1 0 0 1 d 0 0 1 I + p,I + p and I + p,I + p . 0 d 0 0 0 1 0 0             where d = 1. The discussion in the beginning of Section 8 concludes that Hi = σ(ϕ(Hi)) 6 ± w 0 × (H ker ϕ), where σ : Z(p) Z(p2) is the splitting which maps Z(p) to i ∩ → 0 w ∈   w 0 2 w0 0 Z(p ). Since Z(p) (Z/pZ)×, σ(ϕ(Hi)) is cyclic. Say that generates 0 w ∈ ≃ 0 w0     σ(ϕ(Hi). H1 and H2 are, in some order, conjugate to w 0 1 0 0 1 w 0 d 0 0 1 0 ,I + p,I + p and 0 ,I + p,I + p 0 w 0 d 0 0 0 w 0 1 0 0  0       0      by Lemma 25 and Proposition 5. Let

w0 0 1 0 w0 0 d 0 D = ,I + p and D′ = ,I + p . 0 w 0 d 0 w 0 1  0     0    1 0 1 0 1 − Note that D′ = D and that D = D′. Furthermore, H and H are 1 0 1 0 6 1 2 conjugate to     0 1 0 1 D,I + p and D′,I + p 0 0 0 0         in some order as desired. Now assume that ϕ(H ) Z(p). Through Proposition 8, replace H and H so that i 6≤ 1 2 ϕ(H1) and ϕ(H2) are still equal, ϕ(Hi) is still a subgroup of Cs(p) and Hi = ∆H⊥i ⋊ DHi .

By Proposition 9, H2 can be replaced with a conjugate, if necessary, so that DH1 = DH2 as 0 1 well. Proposition 9 further asserts that one of ∆ and ∆ is I + p and that H⊥1 H⊥2 0 0     0 0 0 1 the other is I + p . Thus, H and the conjugate of H via are 1 0 1 2 1 0       0 1 0 0 D ,I + p and D ,I + p Hi 0 0 Hi 1 0         30 0 1 in some order. Conjugating the second of these groups by yields 1 0   1 0 1 0 1 − 0 1 D ,I + p . 1 0 Hi 1 0 0 0 *      + 1 0 1 0 1 − Note that H and H are conjugate if D = D . It is also not difficult 1 2 Hi 1 0 Hi 1 0     to see that H1 and H2 are nontrivially locally conjugate otherwise.  9.2. The Normalizer of Cartan Cases. This section categorizes the subgroups, up to 2 conjugation, of GL2(Z/p Z) whose images via ϕ are subgroups of N(Cs(p)) or N(Cns(p)) that are not contained in Cs(p) or Cns(p).

Lemma 28. Let H GL2(R) be of type N(Cs(p)) or N(Cns(p)). Suppose that ϕ˜(H) ≤ w 0 contains some element of the form where w = z and another element of the form 0 z 6   0 x . y 0   (1) Suppose that H is of type N(Cs(p)). Then, ∆H is generated by both, one or neither 1 0 1 0 of I + p and I + p. 0 1 0 1    −  (2) Suppose that H is of type N(Cns(p)). Then, ∆H is generated by both, one or neither 1 0 √ǫ 0 of I + p,I + p. 0 1 0 √ǫ    −  (3) Suppose that H is of type N(Cs(p)). Then, ∆H⊥ is generated by both, one or neither 0 x 0 x of I + p and I + p. y 0 y 0   −  (4) Suppose that H is of type N(Cns(p)). Then, ∆H⊥ is generated by both, one or neither 0 x 0 x√ǫ of I + p and I + p. y 0 y√ǫ 0   −  (5) If w = z, then dim(∆⊥ )=2 or 0. 6 ± H a 0 d 0 Proof. (1) If I + p H, then I + p H by Lemma 16. Suppose that 0 d ∈ 0 a ∈     a 0 dim(∆ ) = 1. Note that ∆ must only contain elements of the form I + p H H 0 a   a 0 or only contain elements of the form I + p. 0 a − (2) This follows from the same argument as the last part. x 0 b 0 c y (3) If I + p H, then I + y p H be Lemma 16. Suppose that c 0 ∈ b x 0 ∈     2 x 2 y dim(∆H⊥ ) = 1 and assume that b or c is nonzero. In this case, c y = b x . Since x and y are nonzero and at least one of b and c is nonzero, both b and c are nonzero. x b Thus, y = c . Since b,c,x,y Z/pZ, we are done. ± ∈ 31 0 b (4) Likewise, suppose that dim(∆⊥ ) = 1 and take I + p H such that b or c is H c 0 ∈ nonzero. Note that x,y,b,c are of the forms   x = α + β√ǫ y = α β√ǫ − b = γ + δ√ǫ c = γ δ√ǫ − for some α,β,γ,δ Z/pZ by Lemma 14. If x = b , then cx = by, in which case ∈ y c (αγ ǫβδ)+( αδ + βγ)√ǫ =(γ δ√ǫ)(α + β√ǫ) − − − =(γ + δ√ǫ)(α β√ǫ) − =(αγ ǫβδ)+(αδ βγ)√ǫ. − − 0 x Thus, αδ = βγ, i.e. x is in Z/pZ. In this case, ∆ is generated by I + p. b H⊥ y 0 x b x√ǫ   If = , then similarly compute αγ = ǫβδ. Note that Z/pZ, and so ∆⊥ is y − c b ∈ H 0 x√ǫ generated by I + p. y√ǫ 0 − (5) This is due to the last two parts and Lemma 15. 

Lemma 29. Let H ,H GL (R) be locally conjugate and both of type N(C ) or both of 1 2 ≤ 2 s type N(Cns) but not of types Cs or Cns. If one of H1 or H2 has an element of the form w 0 where w = z, then H and H are conjugate. 0 z 6 ± 1 2   Proof. By Propositions 3 and 6 and Theorem 2,ϕ ˜(H1) andϕ ˜(H2) are conjugate. Replace H2 with a conjugate so thatϕ ˜(H1)=ϕ ˜(H2). Note that if Hi is of type N(Cns), then such 1 √ǫ ǫ √ǫ ǫ − a conjugation can be done by an element of GL (Z/pZ) . In −√ǫ− ǫ 2 −√ǫ− ǫ particular, H kerϕ ˜ is still a subgroup of K after− the conjugation. −  2 ∩ By Lemma 28, ∆H⊥i is 2 or 0 dimensional. If ∆H⊥1 is 2 dimensional but ∆H⊥2 is 0 dimensional, then ∆H1 must be 0 dimensional and ∆H2 must be 2 dimensional. However, ∆H2 would then have elements of nonzero trace whereas all of the elements of ∆H1 has 0 trace, which is a contradiction. Hence, ∆H⊥1 = ∆H⊥2 . Lemma 28 also asserts that ∆Hi is generated by both, one 1 0 a 0 1 0 or neither of I + p and a matrix of the form I + p. Note that I + p 0 1 0 a 0 1    −    is an element of H1 if and only if it is an element of H2. Therefore, H1 kerϕ ˜ = H2 kerϕ ˜. ∩ 1 ∩ √ǫ ǫ − Recall that if H is of type N(C ), then its conjugate via is a subgroup i ns −√ǫ− ǫ −  of N(Cns(p)). Since N(Cs(p)) and N(Cns(p)) both have orders which are indivisible by p, Proposition 7 shows that H1 and H2 are conjugate.  Proposition 11. Let H ,H GL (R) be locally conjugate and both of type N(C ) or both 1 2 ≤ 2 s of type N(Cns) but not of types Cs or Cns. H1 and H2 are conjugate. 32 Proof. If all of the diagonal elements ofϕ ˜(Hi) are scalar, then there is some r R× such 0 x ∈ that x = r for all ϕ˜(H ). Furthermore, whether or not H is of type N(C ), it is y y 0 ∈ i i s   1 √r √r − not difficult to see that r is a square in R. Replace H by its conjugate via . i 1 1  −  All elements ofϕ ˜(Hi) are then diagonal and it is not difficult to see that Hi is in fact of type Cs or of type Cns. This possibility was already considered in Section 9. Assume that some diagonal elements ofϕ ˜(Hi) are nonscalar. By Lemma 29, it suffices to show that H1 and H2 are conjugate given that all nonscalar w 0 diagonal elements ofϕ ˜(H ) are of the form . In this case, there is some r R× i 0 w ∈  −  0 x such that x = r for all ϕ(˜H ). Moreover, y ± y 0 ∈ i   1 1 w 0 0 x w 0 − 0 x − = I 0 w y 0 0 w y 0 −  −     −    and so I ϕ˜(H ). Similarly as in Lemma 29,ϕ ˜(H ) andϕ ˜(H ) are conjugate by Proposi- − ∈ i 1 2 tions 3 and 6 and Theorem 2. Replace H2 with a conjugate so thatϕ ˜(H1)=ϕ ˜(H2). If Hi is of type N(Cns), then the replacement can be done in a way such that H2 kerϕ ˜ is still a subgroup of K after the conjugation. ∩ w 0 Suppose that there is some ϕ˜(H ) which is not conjugate to any element of 0 w ∈ i  −  0 x w 0 H of the form . By Corollary 2, H has an element h of the form h = + i y 0 i i i 0 w    −  0 bi α β p. For k = I + p H1, compute ci 0 γ δ ∈     w 0 αw βw + b h k = + 1 p 1 0 w γw + c δw  −  − 1 −  w 0 αw An element of H which is conjugate to h k must be of the form + p 2 1 0 w ∗δw  −   ∗ −  w 0 δw α 0 α 0 δ 0 or − + − ∗ p. Thus, if I+ p H1 then I+ p or I+ p 0 w αw 0 δ ∈ 0 δ 0 α    ∗        1 0 is an element of H . By Lemma 28, ∆ is generated by both, one or neither of I + p 2 Hi 0 1   a 0 and I + p where a =1 or √ǫ. Thus, ∆ = ∆ and so dim(∆ ) = dim(∆ ). If 0 a H1 H2 H⊥1 H⊥2  −  dim(∆H⊥i )=2 or 0, then ∆H⊥1 = ∆H⊥2 , in which case H1 = H2. 0 x Fix an element ofϕ ˜(H ) and suppose that dim(∆⊥ ) = 1. Fix g H to be of y 0 i Hi i ∈ i   0 x ai bi the form gi = + p. An element of H2 which is conjugate to g1 must be an y 0 ci di     1 0 x′ 0 x′ 0 x element ofϕ ˜− for some ϕ˜(Hi). In particular, xy = det = y′ 0 y′ 0 ∈ − y 0    33   0 x′ y y′ 0 x′ 0 x 0 jx det = x′y′ and x = x , and so is one of ± or ± y′ 0 − ± ′ y′ 0 y 0 jy 0 ± ∓ where j is a square root of 1. If H is of type N(C ), thenj Z/pZ. Moreover, using − i s ∈ Lemma 14 shows that j Z/pZ if Hi is of type N(Cns). 6∈ 0 jx Suppose that one of ± is an element ofϕ ˜(Hi). Since I ϕ˜(Hi), both of jy 0 − ∈ ∓  them are elements ofϕ ˜(Hi). Moreover,

0 jx 0 1 j 0 ± y = ± , jy 0 1 0 0 j ∓   x   ∓ 

0 x′ 0 jx′ and so for all ϕ˜(Hi), ± are elements ofϕ ˜(Hi). Note that Lemma y′ 0 ∈ jy′ 0   ∓  j 0 15 shows thatϕ ˜(H ) is preserved under conjugation via . Further suppose that i 0 1   ∆⊥ = ∆⊥ . If H is of type N(C ), then say that, without loss of generality, ∆⊥ is H1 6 H2 i s H1 0 x 0 x generated by I + p and that ∆ is generated by I + p using Lemma 28. y 0 H⊥2 y 0   −  j 0 Replace H with its conjugate via . Since j Z/pZ, ∆⊥ and ∆⊥ are now equal. 2 0 1 ∈ H1 H2   By Proposition 7, H1 and H2 are conjugate. Similarly, if Hi is of type N(Cns), then one can j 0 replace H with its conjugate via . Since j√ǫ is an element of Z/pZ, ∆ and ∆ 2 0 1 H⊥1 H⊥2   are now equal and so H1 and H2 are conjugate by Proposition 7. 0 jx α β Suppose that neither of ± is inϕ ˜(Hi). For k = I + p H1, compute jy 0 γ δ ∈ ∓   

0 x ai + γx bi + δx g1k = + p. y 0 ci + αy di + βy     0 x If ∆ is generated by I + p, then trace(g k) is of the form (a + d + 2nxyu)p H⊥i y 0 1 i i   where n Z/pZ and u = 1 or √ǫ, in particular, when β = nxu and γ = nyu. Otherwise, ∈ trace(g1k) can only have trace (a1 +d1)p. Since the only elements ofϕ ˜(Hi) that are conjugate 0 x 0 x to are ± , one can see that ∆⊥ = ∆⊥ . Thus, H1 and H2 are conjugate y 0 y 0 H1 H2   ±  w 0 by Proposition 7. This concludes the case where there is some ϕ˜(H ) which is 0 w ∈ i  −  0 x not conjugate to any element ofϕ ˜(H ) of the form . i y 0   w 0 Now assume that every ϕ˜(H ) is conjugate to an element ofϕ ˜(H ) of the 0 w ∈ i i  −  0 x w 0 form . In particular, xy = w2. Replace H with its conjugate via x so that y 0 i 0 x    w  0 w 0 x ϕ˜(H ) instead of . w 0 ∈ i y 0     34 1 0 Suppose that H is of type N(C (p)). By Lemma 28, some matrices among I + p, i s 0 1   1 0 0 1 0 1 I + p, I + p and I + p together generate H ker ϕ. Since I + 0 1 1 0 1 0 i ∩  −    −  1 0 p is an element of H if and only if it is an element of H , H ker ϕ and H ker ϕ 0 1 1 2 1 ∩ 2 ∩   1 0 0 1 0 1 have the same number of matrices among I + p, I + p and I + p. 0 1 1 0 1 0  −    −  Suppose that Hi ker ϕ has only one among the three matrices. If H1 ker ϕ has I + 1 0 ∩ 0 1 ∩ p and that H ker ϕ has I + p, then replace H with its conjugate via 0 1 2 ∩ 1 0 2  −    1 1 . Compute 1 1 −  1 1 1 w 0 1 1 − 0 w = 1 1 0 w 1 1 w −0 −   −  −  −  1 1 1 0 w 1 1 − w 0 = 1 1 w 0 1 1 0 w −    −   −  and recall that I ϕ(Hi). Therefore H1 ker ϕ = H2 ker ϕ and ϕ(H2) is preserved − ∈ ∩ 1∩ 0 after the conjugation. Now say that H ker ϕ has I + p and that H ker ϕ has 1 ∩ 0 1 2 ∩  −  0 1 1 0 I + p. Whether or not I + p ϕ(H ), it is not difficult to see that 1 must 1 0 0 1 ∈ i − − be a square in Z/pZ using Proposition 4. Again, say that j Z/pZ is a square root of 1. 0 jw ∈ − If neither of is in ϕ(H ), then compute jw ±0 i ∓  1 1 j w 0 1 j − 0 w = 1 j 0 w 1 j w 0  −   −   −    1 1 j 0 w 1 j − 0 wj = . 1 j w 0 1 j wj 0  −     −  −  1 j Thus, replacing H with its conjugate via makes all diagonal elements of ϕ(H ) 2 1 j 2  −  0 jx scalar, which is a case that was already discussed. If ± ϕ(Hi), then further jy 0 ∈ compute ∓ 

1 1 j 0 j 1 j − 1 0 = . 1 j j 0 1 j 0 1  −  −   −   −  1 j Replacing H with its conjugate via therefore makes H ker ϕ and H ker ϕ 2 1 j 1 ∩ 2 ∩  −  equal and preserves ϕ(H2). Thus, H1 and H2 are conjugate, no matter which of the three matrices they have. 35 1 0 0 1 Suppose that H ker ϕ has two among the matrices I + p, I + p and i ∩ 0 1 1 0  −    0 1 1 0 0 1 1 0 I + p. If I + p,I + p H ker ϕ and I + p,I + 1 0 0 1 1 0 ∈ 1 ∩ 0 1 −   −     −  0 1 p H ker ϕ, then 1 is a square modulo p. Similarly as before, we can assume 1 0 ∈ 2 ∩ − −  0 jw 1 j that ± ϕ(Hi). Replacing H2 with its conjugate via makes H1 jw 0 ∈ 1 j ∩ ∓   −  ker ϕ and H2 ker ϕ equal and preserves ϕ(H2), and so H1 and H2 are conjugate. If I + 1 0 ∩ 0 1 0 1 0 1 p,I + p H ker ϕ and I + ,I + p H ker ϕ, then 0 1 1 0 ∈ 1 ∩ 1 0 1 0 ∈ 2 ∩ compute −  −    −  1 1 1 0 1 1 1 − 0 1 = . 1 1 1 0 1 1 1 0 −  −  −  −  1 1 Replacing H with its conjugate via makes H ker ϕ = H ker ϕ and preserves 2 1 1 1 ∩ 2 ∩ −  ϕ(H2), and so H1 and H2 are conjugate. 1 0 0 1 0 1 If H ker ϕ has all three of I + p,I + p and I + p, then i ∩ 0 1 1 0 1 0 H ker ϕ = H ker ϕ, in which caseH and− H are conjugate. −   1 ∩ 2 ∩ 1 2

10. The Borel case 2 This section categorizes the subgroups, up to conjugation, of GL2(Z/p Z) whose images 1 1 under ϕ are subgroups of B(p) containing . 0 1 As with Lemma 20, the types of such subgroups  that can arise differ between the case when p> 3 and the case when p = 3. We consider the case where p> 3 first. 2 For H GL2(Z/p Z), let CH denote the image of ϕ(H) Cs(p) under the splitting ≤ w 0 w 0 ∩ .(C (p) ֒ C (p2) which maps C (p) to C (p2 s → s 0 z ∈ s 0 z ∈ s     1 1 Lemma 30. Suppose that p > 3. Let H GL (Z/p2Z) with ϕ(H) B(p) and ≤ 2 ≤ 0 1 ∈ ⋊   ϕ(H). There is a conjugate H′ of H such that H′ = τ,H′ ker ϕ CH′ , where τ 1 1 a 0 h ∩ i ∈ GL (Z/p2Z) is a matrix of the form + p for some a, c Z/pZ. 2 0 1 c a ∈     Proof. [6, Lemma 3.3] shows that ϕ(H) is expressible as the internal semidirect product 1 1 ⋊ ϕ(H) C (p) . Suppose that ϕ(H) C (p) Z(p). For every element h 0 1 h ∩ s i ∩ s ≤   l 1 1 of H, ϕ(h) is expressible as ϕ(h) = z for some l Z and z ϕ(H) C (p). 0 1 ∈ ∈ ∩ s   w 0 w 0 Say that z = , in which case H by Corollary 1. Fix any τ H 0 w 0 w ∈ ∈    36  1 1 w 0 w 0 such that ϕ(τ) = . h is then expressible as τ l k = τ lk for some 0 1 0 w 0 w k H ker ϕ. Moreover,  it is not difficult to see that τ,H  ker ϕ is normal in H and that∈ τ,H∩ ker ϕ C = I . Therefore, H is theh internal∩ semidirecti product H = h ∩ i ∩ H h i 1 1 a0 b0 τ,H ker ϕ ⋊ CH . Express τ as τ = + p. Calculate h ∩ i 0 1 c d0     det(τ)=1+(a + d γ)p and trace(τ)=2+(a + d )p 0 0 − 0 0 1 1 a 0 By Theorem 1, τ is conjugate to τ = + p, where a = a0+d0 . Say that ′ 0 1 c a 2 1 2    1  τ ′ = gτg− for g GL (Z/p Z) and let H′ = gHg− . One can then express H′ as the ∈ 2 internal semidirect product τ,H′ ker ϕ ⋊ C . h ∩ i H′ Now assume that ϕ(H) Cs(p) Z(p). By Proposition 8, replace H with a conjugate so ∩ 6≤ 1 that ϕ(H) Cs(p) is preserved and H ϕ− (Cs(p))= ∆H⊥ ⋊ DH . Say that the conjugation ∩ 2 ∩ is done via g GL2(Z/p Z). The conjugation preserves ϕ(H) Cs(p), a diagonal subgroup 2 ∈ ∩ of GL2(Z/p Z) which does not lie in the center. It is not difficult to show that ϕ(g) is therefore an element of N(Cs(p)). Furthermore, if ϕ(H) is not a subgroup of B(p), then 1 0 1 0 1 − ϕ(H) is a subgroup of B(p). If necessary, replace H with such a conjugate 1 0 1 0     1 so that ϕ(H) B(p) and H ϕ− (Cs(p)) is still ∆H⊥ ⋊ DH . In particular, the image of the splitting C (p)≤֒ C (p2) restricted∩ to ϕ(H) C (p) maps isomorphically onto C . Similarly s → s ∩ s H as before, H is the internal semidirect product τ,H ker ϕ ⋊ CH for any τ H satisfying ϕ(τ)= t. h ∩ i ∈ 0 1 Since p > 3 by assumption, Lemma 20 asserts that I + p H. Suppose that τ 0 0 ∈   1 1 a b is of the form τ = + p. By Lemma 21, there is some τ ′ H of the form 0 1 c a ∈     1 1 a 0 τ ′ = + p, in which case H = τ ′,H ker ϕ ⋊ C and we are done. 0 1 c a h ∩ i H     1 1 a b Now assume that τ is of the form τ = + p where a = d. By Lemma 24, 0 1 c d 6     α 0 there is some I + p H where α = δ. Moreover, 0 δ ∈ 6   a d α−δ a d − α 0 − 1 1 a α α−δ τ I + p = + − − ∗ a d p 0 δ 0 1 c d δ α−δ        − −  a d a d by Lemma 22. One can further show that a α α−δ = d δ α−δ . By Lemma 21, there is − − − − 1 1 a′ 0 some τ ′ H of the form τ ′ = + p, which yields the desired result.  ∈ 0 1 c′ a′     1 1 Lemma 31. Suppose that p> 3. Let H GL (Z/pZ) such that ϕ(H) B(p) and ≤ 2 ≤ 0 1 ∈ ϕ(H). H ker ϕ is one of the following:   ∩ a 0 0 1 (1) I + p,I + p for some a, d Z/pZ 0 d 0 0 ∈       37 1 0 0 0 0 1 (2) I + p,I + p,I + p 0 0 0 1 0 0         a 0 1 0 0 1 (3) I + p,I + p,I + p for some a, c Z/pZ such that c =0 c a 0 1 0 0 ∈ 6 − (4) ker ϕ.       0 1 Proof. By Lemma 20, I + p H. Suppose that there is some element k H of the 0 0 ∈ ∈   a b 1 0 form k = I + p such that c = 0. By Lemma 19, I + p H, and so I + c d 6 0 1 ∈ a+d    −  2 0 a′ 0 1 0 0 1 a+d p H as well. Therefore, H contains I + p,I + p,I + p , c ∈ c a′ 0 1 0 0  2      −     a+d a′ 0 1 0 0 1 where a′ = , and so H = I + p,I + p,I + p or H = ker ϕ. 2 c a′ 0 1 0 0     −     a b Now assume that every k H is of the form k = I + p. It is not difficult to see ∈ 0 d that   a 0 0 1 H = I + p,I + p , where a, d Z/pZ 0 d 0 0 ∈       or 1 0 0 0 0 1 H = I + p,I + p,I + p , 0 0 0 1 0 0          1 1 Proposition 12. Suppose that p > 3. Let H GL (Z/p2Z) with ϕ(H) . H ≤ 2 ≤ 0 1 2   is conjugate to some H′ GL2(Z/p Z) which is of the form H′ = τ,H′ ker ϕ , where one of the following holds: ≤ h ∩ i 1 1 (1) H′ ker ϕ is ker ϕ and τ = . ∩ 0 1   (2) H′ ker ϕ is one of ∩ 1 0 1 0 0 1 (a) I + p,I + p,I + p where γ =0 γ 1 0 1 0 0 6 − (b) T        and τ is one of 1 1 (a) 0 1   1 1 1 0 (b) + p 0 1 0 1     α β (3) H′ ker ϕ does not contain any elements of the form I + p where γ =0 and ∩ γ δ 6 τ is one of   1 1 (a) 0 1   1 1 1 0 (b) + p 0 1 0 1     38 1 1 a 0 (c) + p for some a Z/pZ 0 1 1 a ∈     1 1 a 0 (d) + p for some a Z/pZ. 0 1 ǫ a ∈ .     1 1 a 0 Proof. Let H′ be a conjugate of H where H′ = τ,H′ ker ϕ with τ = + p h ∩ i 0 1 c a     1 1 using Lemma 30. If H′ ker ϕ = ker ϕ, then clearly H′, so assume that H′ ker ϕ = ∩ 0 1 ∈ ∩ 6 ker ϕ.   α 0 1 0 0 1 Suppose that H′ ker ϕ = I + p,I + ,I + p where γ = 0. ∩ γ α 0 1 0 0 6     −     Alternatively, we can choose either α =1 or H′ ker ϕ = T . By using Lemmas 23, 22 and ∩ 1 1 a′ 0 21 in that order, one sees that H′ has an element τ ′ of the form τ ′ = + p. 0 1 0 a′     If a′ = 0, then we are done. Otherwise,

1 1 1 0 a 1 a τ ′ ′ = ′ + p 0 1 0 1     a 0 1 1 1 0 by Lemma 17. Replace H with its conjugate via ′ . By Lemma 15, + p ′ 0 1 0 1 0 1       is an element of H′. Therefore, H′ 1 1 1 0 + p,H′ ker ϕ 0 1 0 1 ∩      1 0 1 0 0 1 and H′ ker ϕ is T or I + p,I + ,I + p for some γ′ = 0. ∩ γ′ 1 0 1 0 0 6     −     Now further assume that H′ ker ϕ is not of the form ∩ α 0 1 0 0 1 H′ ker ϕ = I + p,I + ,I + p ∩ γ α 0 1 0 0     −     where γ = 0. By Lemma 31, H′ ker ϕ does not contain any elements of the form I + α β 6 ∩ p where γ = 0. If c = 0, then proceed just as in the last paragraph to replace H′ γ δ 6   1 1 1 1 1 0 with a conjugate of the form H′ = τ,H′ ker ϕ where τ is or + p. h ∩ i 0 1 0 1 0 1       √c 0 Assume that c = 0. If c is a square, then replace H′ with its conjugate via 1 so 6 0  √c  1 √c a 0 that + p H′ by Lemma 15. By Lemma 17, raising this element to the 0 1 √c a ∈ 1     √c th power yields a matrix of the form

1 1 a′ + ∗ p, 0 1 1 a′     39 1 1 a′ 0 and so + p H′. If c is a nonsquare, then one can similarly replace H′ with 0 1 1 a′ ∈     1 1 a′ 0 a conjugate so that a matrix of the form + p H′ and so H′ is therefore 0 1 ǫ a′ ∈ of the desired form.      1 1 Lemma 32. Let H ,H GL (Z/p2Z) such that ϕ(H ). Suppose that H and 1 2 ≤ 2 0 1 ∈ i 1 H are locally conjugate and that H is of the form τ ,H ker ϕ ⋊ C for some τ 2 i h i i ∩ i Hi i ∈ 1 1 1 ϕ− . Then, τ ,H ker ϕ and τ ,H ker ϕ are locally conjugate and C 0 1 h 1 1 ∩ i h 2 2 ∩ i H1   0 1 and C are equal or conjugate via . H2 1 0   Proof. Since H1 and H2 are locally conjugate, there is a bijection f : H1 H2 in which corresponding elements are conjugate. By Theorem 1, the only elements of→B(p) that are 1 1 1 n conjugate to are of the form where n = 0. f must therefore map the 0 1 0 1 6     elements of τ1,H1 ker ϕ into τ2,H2 ker ϕ and vice versa, and so these two groups are locally conjugate.h ∩ i h ∩ i w x Similarly, the only elements of B(p) that are conjugate to p where x = 0 are of 0 w 6   w x′ the form where x′ = 0. Furthermore, the only elements of B(p) that are conjugate 0 w 6   w x w x z x to p where w = z are of the form ′ or ′ . Subgroups of C (p) are 0 z 6 0 z 0 w s generated  by at most two elements, one of which can be chosen to be in Z(p). It is then not  difficult to see that CH1 and CH2 are equal or diagonal swaps. 1 n a b Lemma 33. The conjugacy class of an element + p in GL (Z/p2Z), where 0 1 c d 2 0

2 Proposition 13. Suppose that p> 3. Let H1,H2 GL2(Z/pZ/p Z) be nontrivially locally 1 1 ≤ conjugate with ϕ(H ). H and H are conjugate to 0 1 ∈ i 1 2   τ, k, D , τ, k, D′ h i h i in some order, where τ is one of 1 1 (1) 0 1   1 1 1 0 (2) + p 0 1 0 1     1 1 a 0 (3) + p for some a Z/pZ 0 1 1 a ∈     40 1 1 a 0 (4) + p for some a Z/pZ, 0 1 ǫ a ∈     k is one of

(1) I 0 0 (2) I + p 1 0  

D is a subgroup of C (p), D and D′ are diagonal swaps but D = D′. s 6

Proof. Replace Hi with a conjugate in the form specified in Lemma 30. In particular, Hi = 1 1 ai 0 τi,Hi ker ϕ ⋊ CHi for some τi of the form τi = + p. By Lemma 32, h ∩ i 0 1 ci ai     CH1 and CH2 are equal or diagonal swaps, i.e. ϕ(H1) and ϕ(H2) are equal or diagonal swaps. First consider the case where ϕ(Hi) Cs(p) = I , i.e. ϕ(Hi) = t . Replace H1 and H with conjugates listed in Proposition∩ 12. In particular,h i H = τ ,Hh i ker ϕ for a 2 i h i i ∩ i τi listed in Proposition 12. By Proposition 2, H1 ker ϕ and H2 ϕ are locally conjugate and so dim(H ker ϕ) = dim(H ker ϕ) by Lemma∩ 5. If dim(∩ H ker ϕ) = 4, then 1 ∩ 2 ∩ i ∩ Hi ker ϕ = ker ϕ, in which case H1 and H2 are equal. Assume∩ that dim(H ker ϕ) 3. If dim(H ker ϕ) = 3, then H ker ϕ is one of i ∩ ≤ i ∩ i ∩ 1 0 1 0 0 1 (1) I + p,I + p,I + p where γ =0 γ 1 0 1 0 0 6 − (2) T        1 0 0 1 0 0 (3) I + p,I + p,I + p . 0 0 0 0 0 1        

No two among these three are locally conjugate by Proposition 4. Suppose that Hi ker ϕ = 1 0 1 0 0 1 ∩ I + p,I + p,I + p where γi = 0. Further suppose, for contra- γi 1 0 1 0 0 6     −     1 1 1 1 diction, that τ = τ . By Proposition 12, one can set τ = and τ = + 1 6 2 1 0 1 2 0 1     1 0 p without loss of generality. Note that H = τ ⋊ (H ker ϕ). τ must be conju- 0 1 i h 1i i ∩ 2   n gate to some τ1 k1 where k1 H1 ker ϕ, but this is cannot happen by Lemma 33. Hence, 1 1 ∈ ∩ γ 0 τ = τ . If τ = , then H is in fact conjugate to H via 1 . Now suppose that 1 2 i 0 1 1 2 0 γ    2 1 1 1 0 τ = + p. The elements of H are of the form τ ni k for some k H ker ϕ. i 0 1 0 1 i i i i ∈ i ∩     1 0 Moreover, it is not difficult using Lemma 33 to see that multiplying I + p or 0 1  −  0 1 I + p to k preserves the conjugacy class of τ ni k . Thus, local conjugacy of H is de- 0 0 i i i i   mi ni 1 0 termined by the conjugacy classes of the elements τi I + p . Such an element γi 1 41     is of the form

mi ni 1 0 1 ni ni 0 mi 0 τi I + p = + p I + p γi 1 0 1 0 ni miγi mi              1 n n 0 m + m n γ m n = i + i + i i i i i i p 0 1 0 ni miγi mi       1 n m + n + m n γ m n = i + i i i i i i i p. 0 1 miγi mi + ni     In the case that n = 0 (mod p), Lemma 33 asserts that its conjugacy class is determined by i 6 the values of 2mi +2ni +miniγi and miniγi. Alternatively, the conjugacy class is determined by the values of mi + ni and miniγi. Parametrize m1 as m1 = rn1. Note that m1 + n1 =(r + 2 1)n1 and m1n1γ1 = rn1γ1, and so there must be some m2 and n2 such that m2 +n2 =(r+1)n1 2 γ1 2 2 γ1 and m2n2 = rn , i.e. there is a solution to the quadratic equation x (r+1)n1x+rn = 1 γ2 − 1 γ2 0. This happens only when the discriminant, which is (r + 1)2n2 4rn2 γ1 , is a square in 1 1 γ2 2 γ1 − Z/pZ. When n1 =0, (r + 1) 4r must be a square. Completing the square shows that 6 − γ2 γ γ 2 γ 2 (r + 1)2 4r 1 = r + 1 2 1 +1 1 2 1 . − γ − γ − − γ 2   2   2  This value needs to be a square for all r Z/pZ. Therefore, for all squares s Z/pZ, 2 ∈ 2 ∈ s +1 1 2 γ1 is also a square. Let c = 1 1 2 γ1 and suppose that c = 0. The − − γ2 − − γ2 6 case where s = 0 shows that c must be a square, say c = t2. In this case, s/t2 + 1 is a square for all squares s Z/pZ. However, s/t2 spans all squares in Z/pZ, which is a contradiction. ∈ 2 γ1 Hence, 1 1 2 =0 and so γ1 = γ2. H1 and H2 are therefore equal. − − γ2 If Hi ker ϕ = T, then one can show that τ1 = τ2 similarly as above, in which case ∩ H1 = H2. For the remaining possibilities of Hi ker ϕ, Hi ker ϕ has no element of the form I + α β ∩ ∩ 1 1 p with γ = 0. It is then not difficult to see that if τ is of the form τ = + γ δ 6 1 1 0 1     a1 0 1 1 a2 0 p where c =0, 1 or ǫ, then τ2 is of the form τ2 = + using Lemma c a1 0 1 c a2 33.      1 0 0 1 0 0 If H ker ϕ = I + p,I + p,I + p , then τ can be chosen to be i ∩ 0 0 0 0 0 1 i         1 1 0 0 of the form + p. Thus, H = H . 0 1 c 0 1 2     α 0 0 1 Suppose that H ker ϕ = I + p,I + p for some α, δ Z/pZ which 1 ∩ 0 δ 0 0 ∈       are not both 0. Since H1 ker ϕ and H2 ker ϕ are locally conjugate, H2 ker ϕ = α 0 0 1 ∩ δ ∩0 0 1 ∩ I + p,I + p or I + p,I + p by Proposition 5. Fur- 0 δ 0 0 0 α 0 0             1 n ther suppose that α = δ. If c = 1, then replace H by its conjugate via , where 6 − 1 0 1 42   α δ n = a − . Doing so preserves H ker ϕ and takes τ to 1 α+δ 1 ∩ 1 1 1 n 1 1 a 0 1 n − 1 1 a + n n2 + 1 p = + 1 − p. 0 1 0 1 1 a1 0 1 0 1 1 a1 n             − 

a1+n α 1 1 0 0 Note that = , and so H1 has an element τ ′ of the form τ ′ = + p. Thus, a1 n δ 1 1 0 1 1 0 −     1 1 a1 0 H1 = τ1′ ⋊ (H1 ker ϕ). Similarly, if τ1 starts out as + p, then H1 can be h i ∩ 0 1 ǫ a1     replaced by a conjugate so that H1 ker ϕ is preserved and H1 = τ1′ ⋊ (H1 ker ϕ), where 1 1 0 0 ∩ 1 1 1 0 h i ∩ τ = + p. If τ = + , then replace H with its conjugate 1′ 0 1 ǫ 0 1 0 1 0 1 1         0 0 δ α via I + p where γ = − . Doing so preserves H ker ϕ, while making H = γ 0 δ+α 1 ∩ 1   1 1 ⋊ (H ker ϕ). Similarly replace H with a conjugate. If H ker ϕ = H ker ϕ, 0 1 1 ∩ 2 1 ∩ 2 ∩   then H1 and H2 are conjugate. Otherwise, it is not difficult to see that H1 and H2 are nontrivially locally conjugate. Now suppose that α = δ. Local conjugacy of Hi is determined by the conjugacy classes − 1 1 1 1 of the elements of τ due to Lemma 33. Therefore, if τ = , then τ = and h ii 1 0 1 2 0 1     1 1 1 0 1 1 1 0 if τ = + p, then τ = + p. Lemma 33 further shows that if 1 0 1 0 1 2 0 1 0 1         1 1 a1 0 1 1 a1 0 1 τ1 = + p or + p, then τ1 is conjugate to τ2 or to τ2− . Any 0 1 1 a1 0 1 ǫ a1        1 such conjugation must be via an element of ϕ− (B(p)), which preserves Hi ker ϕ. Again, if H ker ϕ = H ker ϕ, then H and H are conjugate. Otherwise, it is∩ not difficult to 1 ∩ 2 ∩ 1 2 see that H1 and H2 are nontrivially locally conjugate. 0 1 If H ker ϕ = I + p , then H = τ , and so H and H are conjugate. This i ∩ 0 0 i h ii 1 2     concludes the case where ϕ(Hi) Cs(p) = I ; i.e. we have showed that H1 and H2 are ∩ h i 2 conjugate to groups of the desired form. The case where ϕ(Hi) Cs(p) Z(p ) is similar. 2 ∩ ≤ Now assume that ϕ(Hi) Cs(p) Z(p ). Recall that Hi = τi,Hi ker ϕ ⋊ CHi . If ∩ 6≤ h ∩ i 1 1 H ker ϕ = ker ϕ, then H and H are of the desired form; in particular, let τ = , i ∩ 1 2 0 1   0 0 1 0 k = I + p and D = C ,I + p . 1 0 Hi 0 1       1 1 ai ci 0 Suppose that Hi ker ϕ = T . If ci = 0, then observe that Hi has + − p ∩ 6 0 1 0 ai     1 0 0 1 by Lemma 23. One can multiply this matrix by powers of I + p and I + p to 0 1 0 0  −    1 1 ai′ 0 see that Hi has a matrix of the form + p. Thus, whether or not ci = 0, Hi 0 1 0 a′    i has a matrix of this form. By Lemmas 32 and 33, a1 and a2 must be both 0 or both nonzero. 1 1 1 a′i 1 0 Suppose that ai′ = 0. The th power of this matrix has the form + p. 6 a′i 0 1 0 1 43     a 0 Replacing H by its conjugate via i′ preserves H ker ϕ and C and makes H have i 0 1 i ∩ Hi i   1 1 1 0 + p as an element. Therefore, H and H are of desired form; in particular, 0 1 0 1 1 2     1 1 1 1 1 0 let τ = or + p, k = I and D = C . 0 1 0 1 0 1 Hi       w 0 Since ϕ(H ) C (p) contains some element of the form where w = z, H ker ϕ i ∩ s 0 z 6 i ∩ is not of the form   1 0 1 0 0 1 I + p,I + p,I + p γ 1 0 1 0 0     −     for γ = 0 by Corollary 2. 6 α β Suppose that H ker ϕ has no element of the form I + p with γ = 0. By Lemmas i ∩ γ δ 6   32 and 33, c1 and c2 must be both 0, both nonzero squares, or both nonsquares. Suppose a 0 that c = 0. If a = 0, then replacing H with its conjugate via i makes H have i i 6 i 0 1 i   1 1 1 0 1 1 + p while preserving H ker ϕ and C . If a = 0, then H . 0 1 0 1 i ∩ Hi i 0 1 ∈ i       Suppose that c1 and c2 are both nonzero squares. In this case, Hi has 1 ci 1 1 1 a 0 √ 1 a′ + i p = √ci + i ∗ p 0 1 ci ai 0 1 √ci a′         i 1 1 √ci ai′ 0 for some ai′ Z/pZ and so Hi also has + p. Replacing Hi with its ∈ 0 1 √ci ai′     √ci 0 1 1 ai′ 0 conjugate via preserves Hi ker ϕ and CHi and makes Hi have + p. 0 1 ∩ 0 1 1 ai′       If c is a nonsquare, then one can similarly replace H′ with a conjugate such that H′ ker ϕ and i i i∩ 1 1 ai′ 0 CHi′ are preserved and Hi has an element of the form + p. From here, assume 0 1 ǫ a′    i 1 0 1 0 1 0 ai′ 0 1 1 that τi has one of the following: , + p,I + p, + 0 1 0 1 0 1 1 a′ 0 1        i   a 0 i′ p. ǫ a′  i 1 0 0 1 0 0 1 1 Suppose that H ker ϕ = I + p,I + p,I + p . H has + i∩ 0 0 0 0 0 1 i 0 1           0 0 1 1 p. Therefore, H1 and H2 are of the desired form; in particular, let τ = , ci 0 0 1     1 1 0 0 1 1 0 0 1 0 0 0 + p or + p, k = I and D = C ,I + p,I + p . 0 1 1 0 0 1 ǫ 0 Hi 0 0 0 1               α 0 0 1 Suppose that H ker ϕ is of the form H ker ϕ = I + p,I + p . 1 ∩ 1 ∩ 0 δ 0 0       1 1 ai 0 Express τi in the form + p where ci = 0, 1 or ǫ and ai = 0 or 1 if ci = 0. 0 1 ci ai     44 w 0 Since ϕ(H ) C (p) Z(p2), ϕ(H ) has an element of the form where w = z. Just i ∩ s 6≤ i 0 z 6 as in Lemma 24, compute   1 − w w 0 w 0 z τ τ − = I + z ∗w ∗ p. 0 z 0 z ci      w − z ∗ By the assumption on H1 ker ϕ, ci = 0 or every element of ϕ
(Hi) Cs(p) is of the form w 0 ∩ 1 1 1 0 ∩ . Suppose that c = 0. If τ = + p for i = 1 or 2, then another 0 w i 0 1 0 1  ±      1 0 1 1 computation in Lemma 24 shows that I + p H . In this case, H 0 1 ∈ i 0 1 ∈ i     1 0 0 1 and H ker ϕ = I + p,I + p for both i = 1, 2. Whether or not τ = i ∩ 0 1 0 0 i       1 1 1 0 1 1 + p, H and H are of the desired form; in particular, let τ = , k = I 0 1 0 1 1 2 0 1     2   and D = Hi ker ϕ Cs(p ),CHi . h ∩ ∩ i 1 0 Suppose that c = 0. Yet another computation in Lemma 24 shows that I + p H i 6 0 1 ∈ i   1 0 0 1 given that 2a = c . If2a = c for i =1or i = 2, then H ker ϕ = I + p,I + p i 6 i i 6 i i∩ 0 1 0 0       for both i = 1 and 2, and so H1 and H2 are of the desired form; in particular, let τ = 1 1 0 0 2 + p, k = I and D = Hi ker ϕ Cs(p ),CHi . If2ai = ci for both i = 1 and 0 1 ci 0 h ∩ ∩ i     ci 1 1 2 0 2, then H1 and H2 are of the desired form as well; in particular, let τ = + ci p, 0 1 ci    2  k = I and D = H ker ϕ C (p2),C .  h i ∩ ∩ s Hi i 2 For p = 3, there are 40 pairs of nontrivially locally conjugate subgroups of GL2(Z/p Z) up to conjugation. They can be expressed similarly as the pairs of nontrivially locally conjugate 2 subgroups of GL2(Z/p Z) for p> 3 as described in Proposition 13. In particular, the pairs are expressible in the form

τ, k, D , τ, k, D′ , h i h i 1 1 where τ satisfies ϕ(τ) = , k ker ϕ and D and D′ are unequal diagonal swaps. 0 1 ≤ The main difference between the cases where p = 3 and p > 3 is in Lemma 20: whereas 0 1 1 1 I + p H if p> 3 and H is a subgroup of GL (Z/p2Z) with ϕ(H), then 0 0 ∈ 2 0 1 ∈     0 1 I + p H, but if p = 3, then this is not necessarily true. 0 0 ∈  

11. The SL2(Z/pZ) case 2 This section categorizes the subgroups, up to conjugation, of GL2(Z/p Z) whose images via ϕ contain SL2(Z/pZ). 2 Lemma 34. Suppose that p > 3. Let H GL2(Z/p Z). If SL2(Z/pZ) ϕ(H), then T H. ≤ ≤ ≤ 45 0 0 Proof. A computation similar to that of Lemma 20 shows that I + p H. By Lemma 1 0 ∈ 19, T H.    ≤ 2 Lemma 35. Suppose that p > 3. If H GL2(Z/p Z) with ϕ(H) = SL2(Z/pZ), then 2 1 ≤ H = SL2(Z/p Z) or ϕ− (SL2(Z/pZ)). 1 1 Proof. By Lemma 34, T H. Therefore, H has an element τ of the form τ = + ≤ 0 1   a 0 p. Since p > 3, there is some w Z/pZ such that w2 = 1. In particular, H 0 a ∈ 6   w 0 0 b 0 1 has an element of the form + p by Corollary 2. Since I + p,I + 0 1 c 0 0 0  w      0 0 w 0 p H, H has as well. If a = 0, then a computation in Lemma 24 shows 1 0 ∈ 0 1 6    w  1 0 1 that I + p, and so ker ϕ H. In this case, H = ϕ− (SL (Z/pZ)). Similarly, H has 0 1 ≤ 2   1 1 a′ 0 an element of the form + p. If a′ = 0, then ker ϕ H. If a = a′ = 0, then 0 1 0 a′ 6 ≤     1 1 1 0 , H and so SL (Z/p2Z) H. 0 1 1 1 ∈ 2 ≤     2 1 It now suffices to show that if SL2(Z/p Z) = H, then H = ϕ− (SL2(Z/pZ)). Choose 2 6 some h H that is not in SL2(Z/p Z). Since ϕ(H)=SL2(Z/pZ), det(h) 1 (mod p). On ∈ 2 ≡ 2 the other hand, det(h) = 1 because h SL2(Z/p Z). Choose some s SL2(Z/p Z) such 6 1 6∈ 1 ∈ 1 that ϕ(s) = ϕ(h). Note that sh− ker ϕ, and det(sh− ) = 1, and so sh− is of the form ∈ 6 a b 1 I + p where a + d = 0. Hence, ker ϕ H, in which case H = ϕ− (SL (Z/pZ)).  c d 6 ≤ 2   2 Proposition 14. Suppose that p> 3. Let H1,H2 GL2(Z/p Z). If H1 and H2 are locally conjugate with SL (Z/pZ) ϕ(H ), then H = H .≤ 2 ≤ i 1 2 Proof. Since ϕ(H1) and ϕ(H2) are locally conjugate by Proposition 3, they are conjugate, and in fact equal, by Theorem 2. Clearly, ϕ(Hi)=SL2(Z/pZ) ⋊ D, where D is a subgroup of the 1 0 1 0 group GL2(Z/pZ) . Say that generates D. By Corollary 2 and since 0 d ∈ 0 d0      0 1 0 0 1 0 1 I + p,I + p Hi, Hi has . Therefore, Hi = (Hi ϕ− (SL2(p))) ⋊ 0 0 1 0 ∈ 0 d0 ∩       1 0 1 1 . By Lemma 35, Hi ϕ− (SL2(Z/pZ)) = SL2(Z/pZ) or ϕ− (SL2(Z/pZ)). H1 0 d0 ∩ ∩ 1  1 ϕ− (SL (Z/pZ)) and H ϕ− (SL (Z/pZ)) are thus equal, and so H = H .  2 2 ∩ 2 1 2 2 For p = 3, the subgroups of GL2(Z/p Z) whose images under ϕ are SL2(Z/pZ) are conju- gate to one of the following: 7 6 7 4 (1) , 4 4 6 4     1 1 1 0 (2) , = SL (Z/p2Z) 0 1 7 1 2     46 4 1 7 0 (3) , 0 1 4 1     1 1 4 0 (4) , = ϕ 1(SL (Z/pZ)) 0 1 1 1 − 2     7 6 7 4 (5) , 1 7 6 4     1 6 4 7 (6) , . 7 7 6 4     No two distinct subgroups among these are locally conjugate. Therefore, the same result as Proposition 14 holds for p = 3.

12. The Exceptional cases 2 This section determines local conjugacy in GL2(Z/p Z) for the remaining cases.

2 2 Lemma 36. Let H be a subgroup of GL2(Z/p Z). If the image of H in PGL2(Z/p Z) is isomorphic to A ,S or A , then H ker ϕ is one of the following: 4 4 5 ∩ (1) I h i 1 0 (2) I + p 0 1 (3) T    (4) ker ϕ

Proof. Let H denote the image of H in PGL2(Z/pZ). Note that H contains a subgroup which is isomorphic to A4. Therefore, H has an element h1 of order 3 and an element h2 of order 2 which do not commute. In fact, h1h2 has order 3. Furthermore, H is generated by h1 and h2. Choose h1, h2 ϕ(H) such that the images of h1 and h2 in H are h1 and h2 respectively. By Proposition∈ 6, ϕ(H) has no element of order p. In particular, ϕ(H) has no w 1 element which is conjugate to a matrix of the form where w (Z/pZ)×. Thus, h 0 w ∈ 1   is conjugate to an element in Cs(p) or an element in Cns(p). Assume for the rest of the proof that h1 is conjugate to an element in Cs(p). The case where h1 is conjugate to an element of Cns(p) works similarly. Replace H with a conjugate so that h1 is replace with a matrix of w 0 α β the form C (p). Note that w = z. Express h as h = . Compute 0 z ∈ s 6 ± 2 2 γ δ     α2 + βγ (α + δ)β h2 = . 2 (α + δ)γ βγ + δ2   2 2 Since h2 is an element of PGL2(Z/pZ) of order 2, α = δ and (α + δ)β =(α + δ)γ =0. If α 0 0 β α = δ, then h = or . However, H would be cyclic or dihedral, which is a 2 0 α γ 0     α β contradiction to Proposition 6. Hence, α = δ = 0 and we express h as h = , − 6 2 2 γ α 47  −  where not both β and γ are 0. Suppose that γ = 0. Compute αw βw 3 (h h )3 = 1 2 0 αz  −  α3w3 = . 0 α∗3z3  −  3 3 3 3 3 3 Since h1h2 has order 3, α w = α z . However, w = z because h1 has order 3, which is a contradiction. Hence, γ = 0 and− similarly, β = 0. 6 6 a b An argument similar to that of Lemma 27 shows that I + H if and only if c d ∈   a 0 0 b 0 b I + p,I + p H. Moreover, if I + p H with b and c nonzero, 0 d c 0 ∈ c 0 ∈       w 0 b 0 b 1 0 b then I + p and h I + p h− = I + z p are linearly independent c 0 1 c 0 1 c z 0        w  0 1 0 0 because w = z, in which case I + p,I + p H. 6 ± 0 0 1 0 ∈     0 b Suppose that H has an element of the form I + p where b or c is nonzero. Compute c 0   cαβ + bαγ cβ2 bα2 2 2 − p 0 b 1 cα + bγ cαβ bαγ h I + p h− = I + − − −  . 2 c 0 2 α2 βγ     − − cβ2 bα2 and cα2 + bγ2 are in the same ratio as b and c, i.e. c(cβ2 bα2)= b( cα2 + bγ2), − − − −0 b when b2γ2 = c2β2. If this holds, then b and c are both nonzero. Otherwise, I + p and c 0 2 2    0 cβ bα  2 2 − p  cα + bγ 0  0 1 I + − 2 are linearly independent. In either case, I + p,I + α βγ 0 0 − −   0 0 p H. Moreover, 1 0 ∈   αγ α2 2 − p 0 1 1 γ αγ h I + p h− = I +  −  2 0 0 2 α2 βγ     − − 1 0 and since α and γ are nonzero, I + p H. 0 1 ∈  −  a 0 Suppose that there is some I + p H such that a = d. Compute 0 d ∈ 6   aα2 + dβγ aαβ dαβ − 2 p a 0 1 aαγ dαγ aβγ dα h I + p h− = I +  − −  . 2 0 d 2 α2 βγ     − − 0 b Since α = 0 and β and γ are nonzero, H has an element of the form I + p where b 6 c 0 and c are nonzero.   48 From the last two paragraphs, it is not difficult to see that H ker ϕ is one of the four groups as claimed. ∩  Proposition 15. Let H ,H GL (Z/p2Z) be locally conjugate and suppose that the image 1 2 ≤ 2 of Hi in PGL2(Z/pZ) is isomorphic to A4, S4 or A5. H1 and H2 are conjugate.

Proof. By Proposition 2, H1 ker ϕ and H2 ker ϕ are locally conjugate. Lemma 36 shows that H ker ϕ = H ker ϕ. Moreover,∩ ϕ(H )∩ and ϕ(H ) are locally conjugate by Proposition 1∩ 2∩ 1 2 3. They must be conjugate by Proposition 6 and Theorem 2. Replace H2 with a conjugate so that ϕ(H1) = ϕ(H2). All of the subgroups of ker ϕ which are listed in Lemma 36 are 2 normal in GL2(Z/p Z) and so H2 ker ϕ is preserved by the conjugation. Note that ϕ(H ) divides 12(p ∩ 1), 24(p 1) or 60(p 1) because the image of H in | i | − − − i PGL2(Z/pZ) is isomorphic to A4, S4 or A5 and the kernel of the natural homomorphism GL2(Z/pZ) PGL2(Z/pZ) is Z(p), which has order p 1. Therefore, if p > 5, then p does not divide→ ϕ(H ) [6, Table 2] shows that the image− of H in PGL (Z/pZ) cannot be | i | i 2 isomorphic to A4, S4 or A5 if p = 3 and cannot be isomorphic to A5 if p = 5. In any case, p does not divide ϕ(H ) and so H and H are conjugate by Proposition 7.  | i | 1 2 13. Conclusion Theorem 4 summarizes the categorization of pairs of nontrivially locally conjugate sub- 2 groups of GL2(Z/p Z). 2 Theorem 4. The pairs of nontrivially locally conjugate subgroups of GL2(Z/p Z) are, up to conjugation, those listed in Propositions 10 and 13 and for p =3, the ones described at the end of Section 10. Acknowledgements I would like to thank my mentor Atticus Christensen for his invaluable insight on ap- proaching the main problem, his daily help on the complicated details towards solving it, and his guidance in constructing this paper. I would also like to thank Professor David Jerison, Professor Ankur Moitra, and Dr. Andrew Sutherland for their encouragement and experienced advice on working with mathematics. In particular, Dr. Sutherland motivated this project and guided me on it as my supervisor for the MIT Undergraduate Research Op- portunities Program. I give my final thanks to Dr. Slava Gerovitch and MIT Mathematics for making the Summer Program in Undergraduate Research possible. References

[1] N. Avni, U. Onn, A. Prasad, and L. Vaserstein. Similarity class of 3x3 matrices over a local principal ideal ring, eprint arXiv:0708.1608, (2007). [2] B. de Smit, Generating arithmetically equivalent number fields with elliptic curves, Lecture Notes in Computer Science, Vol. 1423 (1998), 392–399. [3] L.E. Dickson, Linear groups with an exposition of Galois field theory, Cosimo Classics 2007 reprint of original publication by Leipzig B.G. Teubner, 1901. [4] J.J. Rotman, An introduction to the theory of groups, Graduate Texts in Mathematics. Springer-Verlag New York, 4th ed., 1995. [5] J. Serre, Proprits galoisiennes des points d’ordre fini des courbes elliptiques, Inventiones mathematicae, Vol. 15 (1972), 259–331. [6] A.V. Sutherland, Computing images of Galois representations attached to elliptic curves, Forum of Math- ematics, Sigma, Vol. 4 (2016), 4th ed.. 49 [7] H.P.F. Swinnerton-Dyer, On l-adic representations and congruences for coefficients of modular Forms, Lecture Notes in Mathematics, Vol. 350 (1973), 1–55.

50