Groups eric auld Original: February 5, 2017 Updated: August 15, 2017

Contents

What is #Grk,npFpq?

I. Group Actions

1.1. Semi-direct products that differ by postcomposition with Aut AutpNq are isomorphic (unfinished). 1.2. Number of subgroups of finite index (unfinished). Maybe has something to do with the notion that G Ñ Sn ö gives isic stabilizers? (if that’s true) 1.3. k fixes gH iff k is in gHg´1.

kgH “ gH ú g´1kgH “ H g´1kg P H k P gHg´1

1.4. Corollary. The stabilizer of gH in K œ rG : Hs is K X gHg´1. 1.5. Corollary. The fixed cosets of H œ rG : Hs comprise the normalizer of H. The fixed points of H œ rG : Hs are those such that

StabH pgHq “ H H X gHg´1 “ H.

1.6. [?, p. 3] If I have a G map X Ñ Y of finite sets, where Y is transitive, then #Y | #X.

1.6.1. Remark. In particular this map must be surjective, because I can get to any y from any other y by multiplication.

The action of G œ X is G œ rG : Hs for some finite index subgroup. Look where eH goes...then label Y as G œ rG : Ks where K is the stabilizer of where eH goes. Always Stabϕx ą Stabx, so we know that K ą H.

1 Proof 2: We show that all fibers are the same size. If y1 “ gy, then

f ´1rys ÑÐ f ´1ry1s x ÞÑ g ¨ x g´1 ¨ ξ ÞÑ ξ

1.7. [?, p. 23] Normal subgroups play nice with Sylow p subgroups. More explicitly, if

P P Sylp G, and N Ÿ G, then

P X N P Sylp N and PN{N P SylppG{Nq.

In both cases, we show the group is a p group, then show its index isn’t divisible by p (so it contains a Sylow p subgroup).

Proof of first result: rN : N X P s “ rNP : P s, and the latter isn’t divisible by p. Proof of second result: The index thing is easy to see because the index of PN{N in G{N is the index of PN in G, which isn’t divisible by p.

1.7.1. Lemma. PN{N is a p group.

k Proof 1: Let p P PN{N, then pp “ e, so the order of p divides pk. Proof 2: PN{N – P {N X P .

1.8. [?, p. 3] When N is normal, nppNq | nppGq and nppG{Nq | nppGq. Note that

G œ SylppGq and G œ SylppNq

both transitive. (For the second, even the action of N on SylppNq is transitive!) Now we have surjective maps of G sets

Sylp G Ñ Sylp N Sylp G Ñ SylppG{Nq K ÞÑ K X NK ÞÑ KN{N gKg´1N gKg´1gNg´1 gKNg´1 gKg´1 ÞÑ gpK X Nqg´1 gKg´1 ÞÑ “ “ N N N so we get the divisibility relations on the size of the sets. 1.9. [?, p. 5] A transitive action breaks into equal orbits under a normal subgroup. Given two orbits Nx, Ny, get a set bijection Nx Ñ Ny which is “multiplication by g”.

nx ÞÑ gnx “ ϕpnqgx “ ϕpnqy where ϕ P AutpNq is “conjugation by g”. This map is clearly invertible.

2 1.9.1. Example [?, p. 5]. If G ă Sp acts transitively on 1, 2, . . . , p, then any normal subgroup is either transitive or trivial.

´1 1.10. Stabgx “ g Stabx g . For suppose gx “ y. Then h stabilizes y iff h ¨ pgxq “ gx pg´1hgq ¨ x “ x

´1 So Stabx “ g Staby g. 1.10.1. Example. If G œ X is a finite transitive action, and P is a Sylow p subgroup P P of the stabilizer of x, then NGP acts transitively on X . Let x, y P X . Among those g P sending x y, we look for one in NGP . Since x, y P X , P ă Stabx and P ă Staby.A g sending x y conjugates Stabx to Staby, so it conjugates P ă Stabx to some Sylow p 1 1 P ă Staby. The group P may not be P , but they’re conjugate in Staby, so

´1 1 hph “ P h P Staby hP h´1 “ gP g´1

´1 ´1 ´1 So h g P NGP and h gP g h “ P 1 h´ g : x y

1.11. Corollary. The kernel of a transitive action is the normal core of any stabilizer. For the intersection of all the stabilizers is the kernel, and it’s also their normal core, because they’re all conjugate.

1.11.1. Example. The kernel of the action G œ rG : Hs is Core H. So if H is normal, H acts trivially on its cosets, and there’s a group structure on the cosets.

1.12. Corollary. G œ rG : Hs is equivalent to G œ rG : Ks iff H is conjugate to K. 1.12.1. Lemma. A transitive action G œ X is equivalent to G œ rG : Ks ðñ K is one of the stabilizer subgroups.

Therefore, G œ rG : Hs is equivalent to G œ rG : Ks exactly when K is one of the stabilizer subgroups.

so if gHg´1 K, then K is the Stab “ g Stab g´1 “ gx x stabilizer of gH.

1.13. Corollary. A subgroup of maximal size is normal. In other words, of index the smallest prime dividing |G|. Let G act on the conjugate subgroups of M by conjugation. The # of conjugate subgroups doesn’t exceed the # of cosets, and it divides |G|, so it must be either 1 or the index. If it’s the index p, consider what the map

G Ñ Sp is.

3 It can only have image size p, because the image size must divide both |Sp| and |G|. Then the kernel (normal core of the stabilizer “ NGH “ H) is the same size as M...X. 1.14. There is a G-map rG : Hs Ñ rG : Ks iff H is conjugate to a subgroup of K. For ´1 look where eH lands, on gK, say. We know H “ StabeH ă StabgK “ gKg . 1.14.1. Remark. Equivalent is: H œ rG : Ks has a fixed point. That is to say, H is a ´1 ´1 subgroup of StabgK for some g. And StabgK “ g StabeK g “ gKg . 1.15. Maps G : H Ñ X pick out H-fixed points of X (unfinished). 1.16. Maps X Ñ rG : Hs where X is transitive pick out H orbits of X with stabilizers inside H (unfinished). See this email https://mail.google.com/mail/u/0/#inbox/ 15dc813436b0a36b

1.17. An action on a finite set X embeds G{ ker into the symmetric group on |X| letters.

II. Basic Results

2.1. The centers of GLnpkq and SLnpkq are the scalar matrices. We show that just commuting with the elementary shearing matrices (which are in SLn) forces the matrix to be scalar.

a11 a12 1 1 1 1 a11 a12 a a 0 1 0 1 a a ˆ 21 22˙ ˆ ˙ ˆ ˙ ˆ 21 22˙ Add one of column Add one of row 2 1 to column 2 to row 1

a11 a12 ` a11 a11 ` a21 a12 ` a22 a a ` a a a ˆ 21 22 21˙ ˆ 21 22 ˙

This shows that a21 “ 0 and we can do that for any off-diagonal entry. Also, a12 ` a22 “ a12 ` a11 so all diagonal entries are equal. 2.2. [?] The invertible upper triangular matrices are solvable. 2.2.1. When I multiply UT matrices, the diagonal entries just multiply. So the inverse matrix has 1{a, 1{b, etc. on the diagonal. 2.2.2. The commutator is the matrices with ones on the diagonal.. The previous comment implies that a commutator has ones on the diagonal. To show it’s the entire commutator, note that pones on diagonalq is a normal subgroup (because diagonal entries multiply), and every right coset has a (unique) representative a diagonal matrix,

1 1 1 a1 a d f a1 1 a d f 1 1 a2 b e a2 1 b e » fi “ » fi » 1 fi a3 c a3 1 c — a ffi — a ffi — 1 ffi — 4ffi — 4ffi — ffi – fl – fl – fl 4 and those representatives form an abelian group pF ˆqn. So the commutator contains these matrices. 2.2.3. When the diagonal is ones, the superdiagonal adds. Because 1 a 1 a1 1 a ` a1 1 b 1 b1 1 b ` b1 » fi » fi “ » fi 1 c 1 c1 1 c ` c1 — 1ffi — 1ffi — 1 ffi — ffi — ffi — ffi – fl – fl – fl plus a bunch of crap on the higher diagonals. 2.2.4. The next commutator is the matrices with zero superdiagonal. The commutator is in there by the previous point. Now we claim every left coset has a unique representative with only a superdiagonal. 1 a d f 1 a 1 d1 f 1 1 b e 1 b 1 e1 “ » 1 cfi » 1 cfi » 1 fi — 1ffi — 1ffi — 1 ffi — ffi — ffi — ffi – fl – fl – fl Since everything is invertible, there’s a unique matrix that could be on the right, and it must have zero superdiagonal since superdiagonals add. Therefore every left coset has a unique representative with only a superdiagonal, and the quotient is three copies of the additive group of the field, which is abelian. 2.2.5. When the superdiagonal is zero, the super-superdiagonal adds. 2.2.6. The next commutator has zeros on the super-superdiagonal. And so on. So the derived series will certainly terminate in the identity subgroup, once it kills all the diagonals. Our solvable series has quotients pF ˆqn, pF `qn´1, pF `qn´2,..., pF `q2,F `. 2.2.7. [?] Another solvable series for invertible upper triangular matrices (unfinished). The previous one was the standard series. In this one, we first go to the commutator, then go to the group 1 c 1 b » 1 afi — 1ffi — ffi Note this contains the subgroup zeros– on diagonalfl which is the commutator, so if it’s normal, the quotient is abelian. 2.2.8. This is normal. Think of conjugation as changing basis. The matrices in my subgroup are exactly those that leave alone the first n ´ 1 basis vectors. If I conjugate by a UT matrix with ones on the diagonal, I am replacing my basis with a basis like e1, e2 ` k21e1, e3 ` k32e2 ` k31e1, etc. My matrix will still leave alone the first n ´ 1 of those basis vectors.

5 2.2.9. The quotient is isomorphic to stuff like

1 a b 1 c » 1 fi — 1ffi — ffi – fl 2.3. The normalizer of the diagonal matrices are the generalized permutation matrices. The generalized permutation matrices are those with only one nonzero entry in each column ˆ ˆ and row. They are the wreath product k o Sn, where Sn acts on n. That means I take k n ˆ and let Sn act on it by permuting the components, and take p k q ¸ Sn. ś To show this is the normalizer, I think about conjugating as changing basis. In the general ś case, I’m taking a diagonal matrix with all different eigenvalues, so these directions are all specified, and I can’t change basis by anything except swapping the order and multiplying by scalars.

2.4. The commutator of GLnpkq is SLnpkq. (With one exception.)

2.4.1. Lemma. The basic shearing matrices generate SLnpkq. Think about diagp.5, 2q and how you’d write this as a product of shears. It’s not clear a priori. 2.4.2. I can always get a one on the corner. Notice that this solves my problem, since after that, I can shear away all the other entries in the first row and column, and proceed by induction. The way I do this is to look at the first column (say), and through shears make sure I have nonzero entries in the corner and one other place...that’s enough to get a one on the corner. So to do this, I just find one nonzero entry, and add it to the corner.

2.4.3. [?, p. 2] The elementary shearing matrices are commutators of SL, except SL2pF3q and SL2pF2q. In SL2pF3q, they are actually commutators of elements in GL2pF3q, so we’re OK: SL2pF3q is still the commutator, but we haven’t shown it’s perfect. (It’s not.) For GL2pF2q “ SL2pF2q, it just doesn’t work: this group is S3 which has an abelian quotient by its three-cycle, which must then be the commutator. (Needs to be finished...)

2.5. [?, p. 7] SL2pZq is generated by a shear and a square root of ´1 (unfinished) 2.6. Any maximal p-subgroup of a finite group is a Sylow p-subgroup. For let P be a p-subgroup, and let P act on the Sylow p-subgroups S1,...,Sn by conjugation. There are one of them mod p, so there is some orbit size not divisible by p. Since all orbit sizes divide

|P |, there must be a fixed point Si.

2.6.1. P ă Si. For P is in the normalizer of Si, since P fixes Si, so PSi is a subgroup of the normalizer of Si of order |P | ¨ |Si|{|P X Si| divisible by p, so it must be Si itself. 2.7. A subgroup has no more conjugate subgroups than it has cosets. For the number of

conjugate subgroups is rG : NGHs, which is less than rG : Hs.

6 2.8. If N Ÿ G, subgroups of G{N are subgroups of G containing N, and normal iff normal. 2.9. If N and G{N are coprime, it’s a semidirect product (unfinished). http://people. brandeis.edu/~igusa/Math101b/SZL.pdf Maybe just treat the case where they are solv- able? Why does that work? 2.10. A section splits a group as a semi-direct product. A retract splits it as a product. We look at 0 / A / B / B{A / 0 the inclusion of a normal subgroup. Saying we have a section

s Ó 0 / A / B / B{A / 0

is saying that B{A and A are subgroups of B that generate B, and A X B{A “ e. So we have a semidirect product. On the other hand, suppose I have a retract

r Ò 0 / A / B / B{A / 0

j 2.10.1. ker r ÝÑ B Ñ B{A is an iso. For kerprq is a normal subgroup of B which meets A trivially, and so that # kerprq ¨ #A “ #B.

i

Ð r 0 / ker r / B / A / 0 j

So B “ A ˆ kerprq and B{A “ ker r.

2.11. The inner automorphisms are a normal subgroup of Aut G. For if ϕg is the automorphism that is conjugation by g, and σ is another automorphism, then

´1 ´1 ´1 ´1 ´1 ´1 ´1 pσ ϕgσqh “ σ pg pσhqgq “ σ pg q ¨ h ¨ σ pgq “ pσgqhpσ gq

´1 so σ ϕgσ “ ϕσ´1g

2.12. All the subgroups of Q8 are normal. For the ones of index 2 have to be, and the one of order 2 is the centralizer.

2.13. [?] The Borel subgroup is self-normalizing in GLn (unfinished). 2.14. Finite groups have composition series, and any two are equivalent. Finding a composition series for a finite group G is easy: I find a maximal normal subgroup N, then switch my focus from G to N, and do the same. If we want to extend to infinite order stuff, it is worth noting that

7 2.14.1. If N and G{N have composition series, then G does. To show that two composition series are equivalent, we show that 2.14.2. Any two subnormal series have equivalent refinements.

Definition. Two subnormal series are equivalent when they have the same quotients, up to permutation.

To show the claim, suppose we have 0 ă A ă B ă C ă D ă G and 0 ă A1 ă B1 ă C1 ă G subnormal. Lay out a grid of intersections

G X GG X DG X CG X BG X AG X e “ e C1 X GC1 X DC1 X CC1 X BC1 X AC1 X e “ e B1 X GB1 X DB1 X CB1 X BB1 X AB1 X e “ e A1 X GA1 X DA1 X CA1 X BA1 X AA1 X e “ e

This gives us a bifiltration. To make our refinements, we basically want to snake from UL to BR in the two obvious ways. To get from A1 to B1, I go

A1,A1 ` B1 X A, A1 ` B1 X B,...,A1 ` B1 X G “ B1

2.14.3. B1pC1 X Aq “ pC1 X AqB1. This is true when one subgroup normalizes the other, and here C1 X A normalizes B1 since C1 does. (So we aren’t dangerous in using additive notation.)

2.14.4. This series is subnormal. We ask if A1 ` pB1 X Dq conjugates A1 ` pB1 X Cq to itself. Since abc “a pbcq, we can consider conjugation by A1 and B1 X D separately. The group B1 X D works: it normalizes A1, and it normalizes C, and it normalizes B1. As for A1, it clearly normalizes A1, and

2.14.5. A1 conjugates B1 X C to itself, up to something in A1. If x P B1 X C, then ax “ axa´1 “ xpx´1axqa´1 and x´1ax P A1, x P B1 X C. Now why are the two ways of snaking equivalent? We want a bijection between the quotients. What could A1 ` pB1 X D correspond to? A1 ` pB1 X Cq C ` D X B1 Only the other one involving these subgroups, . C ` D X A1 A1 ` pB1 X D C ` D X B1 2.14.6. [?, p. 177] [?, p. 8] (Zassenhaus Lemma) – . Consider A1 ` pB1 X Cq C ` D X A1

8 the diagram A ` B X E1 D1 ` E1 X B

E1 X B

A ` B X D1 D1 ` E1 X A

B X D1 ` E1 X A

2.14.7. Claim. These are both instances of the diamond lemma.

2.15. N X H is normal in H.

III. The Symmetric and Alternating Groups

3.1. An is the stuff with an even number of even-length cycles. That is, an even number of odd cycles. :-)

3.2. Reminder. Sn is generated by transpositions. For I can write a cycle pabcq as

pabcdq “ p1aqp1bqp1cqp1dqp1aq. or pabcdq “ pabqpacqpadq or pabcdq “ pdcqpcbqpbaq

3.3. [?] If p is prime, Sp is generated by a p cycle and a transposition. If we have p12 ¨ ¨ ¨ pq and p1kq, then by conjugation I can make pipi ` kqq. Note that p1kq conjugated by kpi ` kq is p1pi ` kqq. Now by primeness, no matter what k is, starting at 1 and adding k’s gets me everything. So I get all transpositions starting with 1. Then I have everything.

3.3.1. Remark. Any Sn is generated by p12 ¨ ¨ ¨ nq and p12q. But for instance, S4 isn’t generated by p1234q and p13q. That’s what?

Remark. In [?, p. 4], it is shown that pabq and p12 ¨ ¨ ¨ nq generate Sn iff gcdpb ´ a, nq “ 1

3.4. [?] An is generated by three-cycles. (Of course we mean when n ě 3.) 3.4.1. Any pair of transpositions can be written in three-cycles. We already know pabqpacq “ pbaqpacq “ pcabq “ pabcq, so that disposes of the case where they share an entry. If they don’t,

pabqpcdq “ pabqpbcqpbcqpcdq “ pcbaqpdcbq

9 Now the main claim is proven, because everything in An necessarily breaks down into an even number of transpositions, every pair of which we can write as three-cycles.

3.5. [?, p. 1] Corollary. For n ě 5, An is generated by stuff like p12qp34q.

Remark. By this, we mean products of transpositions that don’t share any elements. (Ob- viously it’s generated by products of any two transpositions.)

It suffices to show that these generate all the three-cycles. The trick is to break up pabcq “ pcbqpbaq with two irrelevant entries. (Here is where we need n ě 5):

pabcq “ pcbqpbaq “ pcbqpdeqpdeqpbaq

3.6. [?, p. 131] A conjugacy class in Sn splits into two in An iff it’s a product of cycles of distinct odd length. Let g P Sn. Sn acts transitively on the conjugacy classes of an

element g by conjugation, and # Conjpgq “ rSn : Stabgs “ rSn : CSn pgqs. The size of the

conjugacy class in An is rAn : CAn pgqs “ rAn : CSn pgq X Ans. If CSn pgq lies entirely in An, we cut the size in half. If not, then

CSn pgqA4

and the two conjugacy

A4 CSn pgq classes are of the same size.

CSn pgq X A

So now we just need to ask when g commutes with no odd permutations: then the conjugacy class will split into two.

3.6.1. Lemma. If an element in Sn has cycle type distinct integers, then it only commutes with the subgroup generated by its cycles. (Notice this subgroup is always in the centralizer.) This because if I have something like

p1234qp56qp7q in S7,

and I want τστ ´1 “ σ, that just swaps the entries remember, and I can’t have an element in one cycle going to one in another and have it remain the same, becuase the cycles have different lengths. And I can’t do anything with the fixed point (if there is one, and there can be only one). Now I just have to see that a cycle doesn’t commute with anything involving only it’s

own entries, except its own powers. CSn pσq “ xσy for σ an n-cycle.

3.6.2. Lemma. g P An commutes with no odd permutations iff its cycle type is distinct odd integers. If it has any cycles of even length (that is, cycles which are odd), it commutes

10 with them, so it can’t. If it has something like then I can just swap p123qp456q, these two cycles, by doing

Where σ ¨ p123qp456qσ´1 “ p123qp456q σ “ p14qp25qp36q. The converse is established by the above lemma.

3.7. [?] Corollary. A5 is simple. The possible cycle types for A5 are

p1, 1, 1, 1, 1q of which there are 1 5! 5! p2, 2, 1q of which there are “ “ 15 2!22 8 5! 5! p3, 1, 1q of which there are “ “ 20 3 ¨ 2! 6 5! p5q of which there are “ 24 5

(By the above proposition, none of these split into two conjugacy classes in A5.)

3.7.1. No proper nontrivial subgroup can be a union of these conjugacy classes. Just arithmetic, what can divide the group order.

3.8. [?, p. 2] An is the only normal subgroup of Sn (n ě 5).

3.9. [?] Corollary. A map out of Sn n ě 5 can only have image size 1, 2, or be an isomorphism.

3.10. [?, p. 3] The next biggest subgroups of Sn are the Sn´1.(n ě 5).

3.10.1. Counterexample for n ă 5. In S4, we have some D4’s in there of index 3, bigger than the S3’s. These are the subgroups xp1234q, p12qp34qy, xp1234q, p13qp24qy, and so forth. I should think of these being weird in S4 because the V4 ă S4 is weirdly normal.

3.11. [?, p. 3] Corollary. When Sn acts on a set, any orbit is of size 1, 2, or at least n. (n ě 5q.

3.12. [?] An is simple for n ě 5. Note that An has n subgroups An´1 which are the stabilizers of 1, 2, . . . , n. They are all conjugate. SFSOC that there were N Ÿ An. Then since

IV. Solvable and Nilpotent Groups

4.1. K normalizes H iff rK,Hs ď H. 4.2. K centralizes H iff rK,Hs “ xey. 4.3. [?, p. 1] A group is solvable iff its commutator series reaches 1 in a finite number of steps.

11 4.3.1. Definition. A group is solvable if it has a subnormal series with successive abelian quotients. That is,

G ą G2 ą ¨ ¨ ¨ ą Gn “ 1 where the quotients are abelian.

Clearly if the commutator series terminates in finitely many stages, this is true: they’re all normal (characteristic), and the quotients are abelian by construction. And conversely, if we have a subnormal series

G ą G2 ą ¨ ¨ ¨ ą Gn “ 1,

p1q then G2 contains G “ rG, Gs because the quotient by it is abelian. Now G3 ą rG2,G2s ą rGp1q,Gp1qs “ Gp2q, and so forth. 4.3.2. Corollary [?, p. 2]. A subgroup of a solvable group is solvable. For if we examine the commutator series we have Hpnq ă Gpnq, so it will terminate. 4.3.3. Corollary [?, p. 2]. A quotient group of a solvable group is solvable. ϕ 4.3.4. Lemma. If G ÝÑ H, then ϕrGpnqs “ ϕrGspnq.

n n And now we can see that when we have a surjective map G  H, ϕpGp qq “ Hp q, and the subgroup series of H will terminate. 4.3.5. Corollary [?, p. 3]. A minimal normal subgroup of a finite solvable group is elementary abelian. Let N Ÿ G be the minimal normal subgroup. Then N p1q is char of normal so normal in G, and it is strictly contained in N. So N p1q “ e and N is abelian. Now to show it’s elementary abelian, we show that pN “ e for every p | #H. Again, pN is a char subgroup of N, so it’s normal. It can’t be all of N, because there are some elements of order p, and we pigeonhole. 4.4. If a group has some central series (is nilpotent), it has the standard ascending and descending central series. Much of what follows is taken from [?].

4.4.1. Definition A finite group is called nilpotent if it has a central series, i.e. where G ą G2 ą ... ą Gn “ 1 Gi{Gi`1 ď ZpG{Gi`1q.

4.4.2. Observation. For any series, Gi{Gi`1 ă ZpG{Gi`1q ðñ rGi{Gi`1,G{Gi`1s “ xey ðñ rG, Gis ă Gi`1. The first equivalence isn’t hard, and the third condition is just the lift of the second. Note the distinction from a solving series, where rGi,Gis ă Gi`1. The central condition is stronger.

4.4.3. Remark. Notice our conditions ensure the steps aren’t too spaced out

means Gi`1 doesn’t step rG, Gis ď Gi`1 down too far from Gi

means Gi doesn’t step up Gi{Gi`1 ď ZpG{Gi`1q too far from Gi`1

12 The standard ascending (descending) series takes the maximum possible step up (down) consistent with our requirement, and we can use this to show that if a central series exists, these two in particular exist.

So the standard 1 1 G2 ě rG, Gs “: G ùñ G3 ě rG, G2s ě rG, Gs series always takes bigger steps down

and ZpG{Gn´1q always has a smaller lift than ZpG{ZGq, so G {G ď ZpG{G q ùñ G {G ď ZpG{G q n´1 n n n´2 n´1 n´1 the standard series always takes bigger steps up.

4.5. A nilpotent group has no self-normalizing subgroups. SFSOC H is self-normalizing. Take the descending central series

G ą G2 ą G3 ą ... ą e. “ “ rG,Gs rG,G2s

Find where it “crosses” H: i.e. let Gi be the first term not inside H. So Gi`1 ă H.

rG, Gis “ Gi`1 ă H ă rH,Gis, so Gi normalizes H.

So the normalizer of H contains GiH ŋ H, K. 4.6. Corollary. Every maximal subgroup of a finite nilpotent group is normal. 4.7. Corollary. Finite p groups are nilpotent, and any finite nilpotent group is a product of p groups.

4.7.1. Finite p groups are nilpotent. They have nontrivial centers and P {ZpP q is a p group as well, so it has a nontrivial center, so... and the ascending central series will go all the way up. 4.7.2. Finite nilpotent groups are products of their Sylow subgroups. 4.7.3. Lemma. All Sylow subgroups in a finite nilpotent group are normal. For they have self-normalizing normalizers, which must then be the whole group.

Now we have a bunch of normal subgroups, disjoint, which, by order considerations, generate the group.

4.8. Corollary. Every maximal subgroup of a finite p group is normal, of index p. It must be of index p, or else P {M would have another group in there.

13 4.8.1. Remark. Just because we know P has a (normal) series with indices p, doesn’t imply the above result, because we don’t know that every subgroup can appear in such a chain. In a moment, however, we will show that every subgroup of a finite p group appears a (not necessarily normal) series of subgroups of orders p, p2, . . . , pk.

4.8.2. Example. In D4 of order 8, xsy isn’t normal, so it can’t appear in such a normal series.

4.9. Corollary. In a finite p group, I can choose a chain of groups of order p, p2, . . . , pk passing through any given subgroup. Given K ă G, K has such a seires, and the question is whether we can extend it until it gets to G. SFSOC there were a gap, i.e. K ă K1 ă G so that rK1 : Ks ą p and there were no intermediate groups. That contradicts that every maximal subgroup in K1 is of index p, K.

V. AutpZ{pnq is cyclic for odd primes

In any category the endomorphisms of a given element form a monoid in Set. The subgroup of units of this monoid is the group of automorphisms. Abelian groups are enriched over themselves, so the endomorphisms of abelian groups form a monoid in Ab, that is, a ring. The automorphisms of an object are the group of multiplicative units of this ring. In general, there’s no reason why the ring of endomorphisms should be commutative (why the auts should be an abelian group). But when A is a cyclic group, EndpAq will be a commutative ring, because everything is determined by where a generator is set, z ÞÑ zk, and then composition is done by multiplying these k. We then observe that AutpZ{mq – pZ{mqˆ.

Reminder Non-abelian groups aren’t enriched over themselves...they’re just enriched over pointed sets, in general.

Kevin [?] claims that any pointed category is enriched over pSet˚, ^, t˚, ˚uq. Further, if a category is enriched over pointed sets, and it has an initial and a final object, then the initial and final object coincide, and it is in fact pointed. So being a pointed category and being enriched over pointed sets are almost the same thing.

5.1. Aut distributes over cyclic terms of coprime order. This is the first thing that helps me when trying to compute automorphism groups of finite abelian groups.

AutpZ{m ‘ Z{nq “ AutpZ{mq ‘ AutpZ{nq when m and n are coprime

This is because if I map Z{m ‘ Z{n Ñ Z{m ‘ Z{n, I am essentially choosing four maps in a matrix, and when m and n are coprime, there are no maps between those factors. Therefore, I can reduce the problem of calculating aut groups to calculating

α α AutpZ{p 1 ‘ ¨ ¨ ¨ ‘ Z{p n q for particular p.

Unfortunately, when there’s more than one cyclic term here, things are difficult.

14 5.1.1. Example. When there are two cyclic factors of prime order. We have

AutpZ{p ‘ Z{pq “ GL2pFpq.

In case p “ 2, we have GL2pF2q – S3. The point of this section is to show that when there is one cyclic term, and p is odd, what we get out is cyclic. When p “ 2, the group pZ{2kqˆ is not cyclic in general, but it’s the next simplest thing.

5.1.2. The order of pZ{pnqˆ is pn´1pp ´ 1q. This isn’t too hard to see, thinking about which numbers aren’t divisible by p.

So we want to find an element of order pn´1pp ´ 1q, following Dummit p. 60-61. To do this, we find separately elements of orders p ´ 1 and pn´1, and use our exp magic.

5.1.3. Reminder In case the exponent on p is one, we already know pZ{pqˆ is cyclic, because it is a finite multiplicative subgroup of a field.

5.1.4. Reminder (The Euler function is multiplicative on coprime integers.) Let a and b be coprime.

Claim. ϕpabq “ ϕpaqϕpbq.

Proof.

Z{ab –Ring Z{a ˆ Z{b

ˆ ˆ ˆ ˆ pZ{abq – pZ{a ˆ Z{bq – pZ{aq ˆ pZ{bq

Therefore, to compute ϕpnq,

α1 αn n “ p1 ¨ ... ¨ pn α1 αn α1 αn ϕpnq “ ϕpp1 ¨ ... ¨ pn q“ ϕpp1 q ¨ ... ¨ ϕppn q

α1´1 αn´1 “ p1 pp1 ´ 1q ¨ ... ¨ pn ppn ´ 1q

5.2. pZ{pnqˆ has an element of order p ´ 1. It suffices to find an element of order divisible by p ´ 1, since then we can just raise it to the right power. Consider the ring map which reduces Z{pn by the ideal ppq Ă Z{ppnq, leaving the quotient Z{p. This is in particular a

n Z{p Ñ Z{p n pZ{p qˆ Ñ pZ{pqˆ z ÞÑ (generator) map of multiplicative monoids, and I can then restrict to the submonoid of units. It’s pretty clear that this map on units is surjective: an element in Z{pn is a unit iff it is coprime to

15 p. So since the original map is surjective, the map on units is surjective. The map on units says “quotient out by the multiplicative subgroup of pZ{pnqˆ consisting of elements with multiplicative order a pth power”. Now recalling that pZ{pqˆ is cyclic, consider an element z P pZ{pnqˆ that maps to a generator of pZ{pqˆ. It has the property that zp´1 P ker, since xp´1 “ 1 in the image. Recalling that the kernel of the map is those elements of multiplicative order a pth power, we’re saying that zp´1 is of p power order. Here’s our element.

5.3. 1 ` p P pZ{pnqˆ is of order pn´1. (Recall that we are assuming p odd here.) For this, we make use of a nifty lemma from number theory.

pk 5.3.1. ` is fairly divisible by p. Let νppnq be the exponent of p in n, that is, the largest j such that pj | n. ` ˘ pk Claim. ν ` “ k ´ νp`q. ` ˘ a 6 Remark. The relation ν b “ νpaq ´ νpbq doesn’t hold in general...ν2 3 “ νp20q “ 2. Proof. We look for ` ˘ ` ˘ pk! ν “ νppk!q ´ νp`!q ´ νpppk ´ `q!q `!ppk ´ `q! ˆ ˙ When we look at νpn!q for any integer n, we see (say for p “ 3)

1 ¨ 2 ¨ 3 ¨ 4 ¨ 5 ¨ 6 ¨ 7 ¨ 8 ¨ 9

that we want to count once for each k such that 3k ď n, and once more (not to double count) for each k such that k32 ď n, and so forth.

8 n νpn!q “ pj j“1 ÿ Z ^ Therefore in the above calculation we get

νppk!q ´ νp`!q ´ νpppk ´ `q!q

k pk k ` k pk ´ ` “ ´ ´ pj pj pj j“1 j“1 j“1 ÿ Z ^ ÿ Z ^ ÿ Z ^ The last term is

k pk ` k pk k ` ´ “ ´ pj pj pj pj j“1 j“1 j“1 ÿ Z ^ ÿ Z ^ ÿ R V k pk k ` “ ´ pj pj j“1 j“1 ÿ ÿ R V 16 and the first term cancels with the previous one in the sum.

k pk k ` k pk ´ ` ´ ´ pj pj pj j“1 j“1 j“1 ÿ Z ^ ÿ Z ^ ÿ Z ^ k pk k ` k pk ` “ ´ ´ ´ pj pj pj pj j“1 j“1 ˜j“1 ¸ ÿ Z ^ ÿ Z ^ ÿ Z ^ R V k ` ` ´ pj pj j“1 ÿ R V Z ^ which says count zero up the the point where pj no longer divides `, and then start counting ones from that point up until pk. So we get k ´ νp`q.

5.3.2. p1 ` pq P pZ{pnqˆ is of order dividing pn´1.

Reminder It’s always good to remind yourself when dealing with p powers that pxpqk “ xpxp ¨ ¨ ¨ xp “ k terms p ` p ` ¨ ¨ ¨ ` p looooomooooon x k terms “ xpk, whereas xpxk “ xp`k. loooooooomoooooooon Consider

p1 ` pqp1 ` pq ¨ ¨ ¨ p1 ` pq

pn´1 terms

pn´1 loooooooooooooomoooooooooooooonpn´1 pn´1 1 ` ¨ p ` ¨ p2 ` ¨ ¨ ¨ ` pk ` ¨ ¨ ¨ 1 2 k ˆ ˙ ˆ ˙ ˆ ˙ n and what we need in order for everything to be divisible by p is that νppkq ă k. This is always true.

n 2 Now, to show its order is exactly pn´1, we just need to show that p1`pqp ´ ı 1, because that is its last chance to have any other order.

5.3.3. p1 ` pq P pZ{pnqˆ is of order exactly pn´1. For this, we just need to show that n 2 p1 ` pqp ´ ı 1, because that is its last chance to have any other order.

p1 ` pqp1 ` pq ¨ ¨ ¨ p1 ` pq

pn´2 terms

pn´2 loooooooooooooomoooooooooooooonpn´2 pn´2 1 ` ¨ p ` ¨ p2 ` ¨ ¨ ¨ ` pk ` ¨ ¨ ¨ 1 2 k ˆ ˙ ˆ ˙ ˆ ˙ n´1 n I should get 1 ` p mod p , if I can guarantee that νppkq ă k ´ 2, which is true for all k when p ą 2.

17 5.4. pZ{2nqˆ is the next simplest thing it could be. We know that pZ{2nqˆ is of order 2n´1p2 ´ 1q “ 2n´1. We show that it’s Z{2 ‘ Z{2n´2. Here the trick is the look at powers of 5, in other words 1 ` 22.

5.4.1. p1 ` 22q P pZ{2nqˆ has order dividing 2n´2. For

n´2 n´2 n´2 2 2 p1 ` 22q2 “ 1 ` ¨ 22 ` ¨ 24 ` ¨ ¨ ¨ 1 2 ˆ ˙ ˆ ˙

and we need ν2pkq ď 2k ´ 2, which is true for all k.

5.4.2. p1 ` 22q P pZ{2nqˆ has order exactly 2n´2. For

n´3 n´3 n´3 2 2 p1 ` 22q2 “ 1 ` ¨ 22 ` ¨ 24 ` ¨ ¨ ¨ 1 2 ˆ ˙ ˆ ˙ ” 1 ` 2n´1

n´2 5.4.3. pZ{2nqˆ has two elements of order 2. These are p1 ` 22q2 and ´1. Therefore the group is not cyclic. And since it has an element of order 2n´2, the only thing it can be is Z{2n´2 ‘ Z{2.

References

[1] Dummit and Foote, Abstract Algebra.

[2] Kiyoshi Igusa online notes, Solvable Groups. http://people.brandeis.edu/~igusa/ Math101b/solv.pdf

[3] Kiyoshi Igusa online notes, Nilpotent Groups. http://people.brandeis.edu/~igusa/ Math101b/nilpotent

[4] Kiyoshi Igusa online notes, The Special Linear Group SLpn, F q. http://people. brandeis.edu/~igusa/Math101b/SL.pdf

[5] Keith Conrad, Transitive Group Actions. http://www.math.uconn.edu/~kconrad/ blurbs/grouptheory/transitive.pdf

[6] https://rohilprasad.wordpress.com/2015/12/17/constructing-a-polynomial-with-galois-group-s5/

[7] https://groupprops.subwiki.org/wiki/Borel_subgroup_is_self-normalizing_ in_general_linear_group

[8] “General Linear Group”, Gabe Cunningham, online notes. http://math.mit.edu/~dav/ genlin.pdf

18 [9] “Why is the group of unit upper triangular matrices solvable?”, Stack Exchange article. https://math.stackexchange.com/questions/1753556/ why-is-the-group-of-unit-upper-triangular-matrices-solvable

[10] Basic Algebra, Anthony Knapp.

[11] “Math 6310 Lecture Notes” based on lectures by Marcelo Aguiar.

[12] “Why is the group of unit upper triangular matrices solvable?”, Stack Ex- change Question. https://math.stackexchange.com/questions/1753556/ why-is-the-group-of-unit-upper-triangular-matrices-solvable

[13]“ An is Simple” (sic). Online note by Patrick J. Morandi. http://sierra.nmsu.edu/ morandi/notes/an-is-simple.pdf

[14] “The alternating group is generated by three-cycles”. Stack Ex- change question. https://math.stackexchange.com/questions/914338/ the-alternating-group-is-generated-by-three-cycles

[15] “Simplicity of An”. Expository paper by Keith Conrad. http://www.math.uconn.edu/ ~kconrad/blurbs/grouptheory/Ansimple.pdf [16] “Generating Sets”. Online notes by Keith Conrad. http://www.math.uconn.edu/ ~kconrad/blurbs/grouptheory/genset.pdf [17] “The Sum of an Irreducible Representation”. Stack Exchange article. https://math.stackexchange.com/questions/127002/ the-sum-of-an-irreducible-representation

19