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For odd p the values for the Dihedral groups of order 2p or 4p were obtained in [2]:

a b S(D2p)= {2 p m : (m, 2p)=1, a =0 or a ≥ 2, b =0 or b ≥ 3}, 3 S(D4p)= {m ≡ 1 mod 4 : p ∤ m or p | m} ∪{2apbm : (m, 2p)=1, a =4 or a ≥ 6, b =0 or b ≥ 3}, with a counterpart to (4)

(6) {m ∈ Z : gcd(m, 2n)=1} ⊂ S(D2n), for n odd, but only those gcd(m, 2n) = 1 with m ≡ 1 mod 4 when n is even, and the divisibility condition (5)

t 2t+1 (7) p k n, p | m ∈ S(D2n) ⇒ p | m, for odd p, with 22, 24 or 22t+4 | m when p = 2 and t =0, 1 or t ≥ 2 respectively. A complete description for all groups of order at most 14 was found in [19] and for S4 in [21]. For example for the two dicyclic groups of order less than 14: 2 8 (8) S(Q8)= {8m +1, (8m − 3)p , and 2 m : m ∈ Z, p ≡ 3 mod 4} and

a b (9) S(Q12)= {2 3 m : a =0, 4 or a ≥ 6,b =0 or b ≥ 3, gcd(m, 6)=1} ∪{253bm : b =4 or b ≥ 6, gcd(m, 6)=1} ∪{253bmp : b =0, 3 or5, gcd(m, 6)=1,p ≡ 5 mod 12} ∪{253bmp2 : b =0, 3 or 5, gcd(m, 6)=1, p ≡ 5 mod 6}. The complexity encountered even for small groups [19] makes it clear that obtaining S(G) is not in general feasible. Indeed simply finding the smallest non-trivial integer determinant D (10) λ(G) := min{| G(xg1 ,...,xgn )|≥ 2 : xgi ∈ Z} can be difficult. For a group of order n taking xe = 0 and xg = 1 for g 6= e always gives determinant (−1)n−1(n − 1), so we have as our trivial bound (11) λ(G) ≤ |G|− 1 for |G|≥ 3, with λ({e}) = 2, λ(Z2) = 3. 3 Kaiblinger [9] obtained λ(Zn) when 420 ∤ n, with this extended to 2 · 3 · 5 · 7 · 11 · 13 · 17 · 19 · 23 ∤ n in [18]. Values of λ(G) for non-cyclic abelian G are considered in [6, 7, 20, 3, 15]. In [2] the value of λ(D2n) was obtained for any of order 2n with 22 · 32 · 5 · 7 · 11 · 13 · ... · 107 · 109 · 113 ∤ n. Our goal here is to determine similar results for Q4n, the dicyclic group of order 4n, when n is odd.

2. Lind Mahler Measure For a polynomial F ∈ Z[x, x−1] one defines the traditional logarithmic Mahler measure by 1 (12) m(F )= log |F (e2πiθ)|dθ. Z0 MINIMAL GROUP DETERMINANTS FOR DICYCLIC GROUPS 3

Lind [14] regarded this as a measure on the group R/Z and extended the concept to a compact with a Haar measure. For example for an F ∈ Z[x, x−1] and Zn we can define a Zn-logarithmic measure 1 mZ (F )= log |F (z)| n n zn=1 X 1 Z Z That is m n (F )= n log |M n (F )| where n−1 j 2πi/n MZn (F ) := F (wn), wn := e . j=0 Y More generally for a finite abelian group

(13) G = Zn1 ×···× Znk we can define a logarithmic G-measure on Z[x1,...,xk] by n −1 n −1 1 1 k m (F )= log |M (F )|,M (F )= ··· F wj1 , ··· , wjk . G |G| G G n1 nk j1=0 j =0 Y Yk  As by observed by Dedekind the group determinant for a finite abelian group can be factored into linear factors using the group characters Gˆ D (14) G(xg1 ,...,xgn )= (χ(g1)xg1 + ··· + χ(gn)xgn ) , χY∈Gˆ and can be related directly to a Lind Mahler measure for the group, see for example [20]. For example in the cyclic case, see [10] D n−1 Zn (a0,a1,...,an−1)= MZn (a0 + a1x + ··· + an−1x ), and in the general finite abelian case (13)

D (a ,...,a )= M a xj1 ··· xjk . G g1 gn G  g 1 k  g=(j1X,...,jk)∈G For a finite non-abelian group the group determinant will not factor into linear factors but can still be factored using the group representations Gˆ deg(ρ) D (x ,...,x )= det x ρ(g) G g1 gn  g  ρY∈Gˆ gX∈G as discovered by Frobenius, see for example [8, 4]. In [2] it was shown that the group determinants for the dihedral group of order 2n, n 2 −1 n−1 n−1 D2n = hx, y : x =1,y =1, xy = yx i = {1, x, ··· , x ,y,yx,...,yx }, can be written as a Zn-measure D −1 −1 (15) D2n (a0,...,an−1,b0,...,bn−1)= MZn f(x)f(x ) − g(x)g(x ) , where  n−1 n−1 (16) f(x)= a0 + ··· + an−1x , g(x)= b0 + ··· + bn−1x . Similarly for the dicyclic group of order 4n, 2n 2 n −1 2n−1 2n−1 Q4n = hx, y : x =1,y = x , xy = yx i = {1, x, ··· , x ,y,yx,...,yx }, 4 B. PAUDEL AND C. PINNER it was shown in [19] that the group representations give D −1 n −1 (17) Q4n (a0,...,a2n−1,b0,...,b2n−1)= MZ2n f(x)f(x ) − x g(x)g(x ) , where 

2n−1 2n−1 (18) f(x)= a0 + ··· + a2n−1x , g(x)= b0 + ··· + b2n−1x . Notice that we can conversely use the group determinant to define a Lind style polynomial measure for non-abelian finite groups. For example we can define D2n and Q4n measures on Z[x, y] by −1 −1 MD2n (f(x)+ yg(x)) = MZn f(x)f(x ) − g(x)g(x ) , −1 n −1 Z (19) MQ4n (f(x)+ yg(x)) = M 2n f(x)f(x ) − x g(x)g(x ) , although here the polynomial ring is no longer commutative, the monomials having to satisfy the group relations y2 = 1, xy = yx−1 etc., the relations allowing us to reduce any F (x, y) to the form f(x)+ yg(x), with f and g of the form (16) or (18) if we want, and to multiply and reduce two polynomials. The classical Lehmer problem [12] is to determine inf{m(F ) > 0 : F ∈ Z[x]}. Given the correspondence between the Lind measures and group determinants in the abelian case we can regard determining λ(G) for a finite group as the Lind-Lehmer problem for that group. An alternative way of extending the Mahler measure to groups can be found in [5].

3. Minimal determinants for odd n

For the dicyclic groups G = Q4n we have some extra properties when n is odd. For example, since

n (20) MG(f(x)+ yg(x)) = (−1) MG(g(x)+ yf(x)), if n is odd we have −m ∈ S(G) whenever m ∈ S(G). This is certainly not true when n is even as we saw for Q8. When n is odd we also have

n (n−1)/2 n (n−1)/2 MG 1 + (x + 1)(x + ··· + x )+ y(x + 1)(x + ··· + x ) =2n − 1,   always improving on the trivial bound (11), and

2 (21) MG(x +1)= 16, giving us an absolute bound λ(Q4n) ≤ 16 for n odd. In the next section we will see that an analog to (4) and (6) holds for n odd:

(22) {m ∈ Z : gcd(m, 2n)=1} ⊂ S(Q4n), and, corresponding to the divisibility conditions (5) and (7),

t 2t+1 (23) 2 | m ∈ S(Q4n) ⇒ 16 | m, p k n, p | m ∈ S(Q4n) ⇒ p | m.

Properties (21),(22),(23), are enough for us to completely determine λ(Q4n): Theorem 3.1. If n is odd then

λ(Q4n) = min{16,p0} MINIMAL GROUP DETERMINANTS FOR DICYCLIC GROUPS 5 where p0 is the smallest prime not dividing 2n. That is, 3 if 3 ∤ n, 5 if 3 | n, 5 ∤ n,  7 if 3 · 5 | n, 7 ∤ n, λ(Q4n)=  11 if 3 · 5 · 7 | n, 11 ∤ n,  13 if 3 · 5 · 7 · 11 | n, 13 ∤ n,  16 if 3 · 5 · 7 · 11 · 13 | n.   A complete description of the determinants for D2p and D4p was given in [2]. As we saw for Q12 in (9) the determinants for Q4p must depend more subtly on p, or at least those determinants M with 25 k M. We can be precise about the other values.

Theorem 3.2. Suppose that p is an odd prime. The determinants for Q4p will take the form 2kpℓm, gcd(m, 2p)=1, with k =0 or k ≥ 4 and ℓ =0 or ℓ ≥ 3. We can achieve all such values with k = 0, k = 4 or k ≥ 6, and all with k = 5 and ℓ =4 or ℓ ≥ 6. This just leaves 25m, 25p3m, 25p5m, gcd(m, 2p)=1. Not all m are possible. 5 1 2 The smallest determinant of the form 2 |m|, gcd(m, 2p)=1 has m = 2 (p + 1). 5 3 5 5 1 2 If p ≡ 3 mod 4 the smallest 2 p |m|, 2 p |m|, gcd(m, 2p)=1 have m = 2 (p + 1). If p ≡ 1 mod 4 then all the multiples of 25p5 are determinants. For p = 5 all multiples of 25p3 are determinants. For the p ≡ 1 mod 4 with p > 5 it remains unclear whether we achieve any 5 3 1 2 2 p m, gcd(m, 2p) = 1, with |m| smaller than m = 2 (p + 1). 4. The case of even n

When G = Q4n with n even it is not at all obvious which values coprime to 2n are determinants; (22) is far from true, the odd determinants must be 1 mod 4 with only some of those obtainable. The observation that when g = 0 we have 2 (24) MQ4n (f(x)) = MZ2n (f(x)) , does give us 2 2 {m : gcd(m, 2n)=1}⊂{m : m ∈ S(Z2n)} ⊂ S(Q4n), where, writing Φℓ(x) for the ℓth cyclotomic polynomial,

α 2 gcd(m, 2n)=1 ⇒ MG φp(x) = m .  α  pYkm As a counterpart to (21) we do have   t 2t+1 2t+2 (25) 2 k n ⇒ MG(x +1)=2 . In particular we always have

2t+2 2 (26) λ(Q4n) ≤ min 2 , p0 , n o where p0 is the smallest prime not dividing 2n. With our divisibility conditions and Lemma 5.4 we can certainly come up with cases of equality in (26), though not always, for example λ(Q8) = 7. 6 B. PAUDEL AND C. PINNER

In a future paper we hope to consider the case of Q4n, 2 k n. The general case of even n seems far out of reach; for t = 1 we know that (25) does give the smallest even determinant, but for t ≥ 2 this is not at all clear, indeed the counterpart for cyclic groups remains unresolved.

5. Divisibility restrictions and values achieved We work with the dicyclic measures of polynomials F = f(x)+ yg(x), −1 n −1 MG(F )= MZ2n f(x)f(x ) − x g(x)g(x ) , f,g ∈ Z[x], where if the degree of f or gexceeds 2n−1 we can still recover a group determinant by reducing the polynomial mod x2n − 1 to the form (18). For the dicyclic determinants we obtain a divisibility Lemma very much like that obtained for the cyclic groups and the dihedral groups [2, Lemma 4.4]. We begin by observing that the cyclic results [16, Theorem 2] and [10, Theorem 5.8] are in fact best possible: Lemma 5.1. Suppose that pα k n then α+1 (27) p | MZn (F (x)) ⇒ p | MZn (F (x)). Since n x − 1 2 MZ x − 1+ = n , n x − 1   and for odd p α+1 p k MZn (p + (x − 1)) , this is sharp for α ≥ 1 when p is odd and for α =1 when p =2. For p =2 and α ≥ 2 we have α+2 (28) 2 | MZn (F (x)) ⇒ 2 | MZn (F (x)). Since α+2 (29) 2 k MZn (4 + (x − 1)) , this exponent is again sharp. Although the exponent is sharp we do not necessarily get that prime power itself 3 (let alone all multiples); for example Newman [17] showed that p is in S(Zp2 ) for p = 3 but not for any p ≥ 5. For the dicyclic groups we have:

Lemma 5.2. Suppose that G = Q4n. α 2α+1 (i) For odd p, if p k n and p | MG(F ) then p | MG(F ). This is best possible, for example p − 1 p2α+1 k M 1 − (1 + xn)(1 − x)+ y (1 + xn) . G 2     α (ii) Suppose that 2 ||n and 2 | MG(F ). 4 (a) If α =0 then 2 | MG(F ). 2α+6 (b) If α ≥ 1 then 2 | MG(F ). Since 2n 2n α+1 x − 1 x − 1 α+2 M x2 +1+ m + ym =22 (1+2mn), G x − 1 x − 1   MINIMAL GROUP DETERMINANTS FOR DICYCLIC GROUPS 7 and 2α+6 (30) 2 k MQ4n (4+(x − 1)) the exponents in (a) and (b) are optimal. 3 For Q4p, p odd, we achieve all odd multiples of p , and for Q8 all multiples of 28, but in general it is not clear whether we can achieve the prime power p1+2α or 22α+6 itself. Property (30) is just (29) and (24). When n is odd the next lemma, the counterpart to [2, Lemma 4.2], shows that we can achieve any integer coprime to 2n. By (1) and (20) it is enough to achieve p or −p for any p ∤ n. Lemma 5.3. Suppose that n is odd and p ∤ n is an odd prime, where p ≡ δ mod 4 with δ = ±1. Set t = (p − δ)/4, and f = δ + (xn + 1)H(x), g = (xn + 1)H(x), with xm +1 H(x)= (xa1 + ··· + xat ) , x +1   where m is odd with pm ≡ 1 mod n, and pa1,...,pat ≡ 1, 3,..., (p − 3)/2 mod n if δ =1, and 0, 2,..., (p − 3)/2 mod n if δ = −1, then

MQ4n (f(x)+ yg(x)) = δp. When 2 | n we have additional restrictions on the odd determinants, showing that we can no longer achieve all integers coprime to 2n :

Lemma 5.4. Suppose that G = Q4n with 2 | n. If 2 ∤ MG(F ) then MG(F ) ≡ 1 or −3 mod 8. 2 If 2 k n and MG(F ) ≡ −3 mod 8 then MG(F ) = (8m − 3)k for some integer m and positive integer k ≡ 3 mod 4. Further we can assume that gcd(k,n)=1 or 4 MG(F )=(8m − 3)p with p | n. In either case if q | gcd(n, (8m − 3)) is prime then q2 | (8m − 3). β α If 2 k n and MG(F ) ≡ −3 mod 8 is of the form ±q with q k n, α ≥ 1, then β ≥ 4α +3.

6. Proofs We shall need to know the resultant of two cyclotomic polynomials, see [1] or [13]; if m>n then pφ(n) if m = npt, |Res(Φn, Φm)| = (1 else. It will be useful to split the product over the 2nth roots of unity in (19) into the primitive dth roots of unity with d | 2n:

MG(F )= Md, dY|2n where d j −j nj j −j 2πi/d Md := f(wd)f(wd ) − wd g(wd)g(wd ), wd := e . j=1 (j,dY)=1 8 B. PAUDEL AND C. PINNER

Since we run through complete sets of conjugates the Md are integers. Moreover, since f(x)f(x−1) − xng(x)g(x−1) is fixed by x 7→ x−1, x2n = 1, when d 6= 1, 2 we run through a complete set of conjugates twice and Md will actually be the square of an integer for d ≥ 3.

Proof of Lemma 5.1. Suppose that G = Zn and write

MG(f)= Ud(f),Ud(f) = Res(Φd,f) ∈ Z. Yd|n Suppose p | MG(f) then p | Umpj (f) some p ∤ m, 0 ≤ j ≤ α, and since (1 − wpj ) | p we have

m pj r s φ(pj ) Umpj (f)= f(wmwpj ) ≡ Um(f) mod p r=1 s=1 gcd(Yr,m)=1 gcd(Ys,p)=1

α+1 and p | Umpj (f) all j =0, . . . , α, and p | MG(f). m r Observe that F (x) = r=1 f(wmx) is in Z[x] (since, for example, its coeffi- (r,m)=1 cients are fixed by the automorphismsQ of Q(wm)). Hence when p = 2 and α ≥ 2 we can write

Um(f)U2m(f)U4m(f)= U1(F )U2(F )U4(F )= MZ4 (F ). From [10] we have a S(Z4)= {2 c : gcd(c, 2)=1, a =0 or a ≥ 4}. 4 Hence we have 2 | Um(f)U2m(f)U4m(f) and 2 | Umpj any j = 3,...,α, and α+2 2 | MG(F ). For the examples observe that Ud(p + (x − 1)) ≡ Ud(x − 1)=Φd(1) 6≡ 0 mod p j unless d is a power of p, while U1(p + x − 1) = p, for the d = p , j =1,...,α and p ≥ 3 and x a primitive pj th root of unity we can write p + (x − 1) = (x − 1)v, v ≡ 1 mod (1 − wpj ) and Upj (p + (x − 1)) = Upj (x − 1)(1 + tp) = p(1 + tp) and α+1 p k Upj (p+(x−1)) and p k MG(p+(x−1)). The case p = 2 and MG(4+(x−1)) 2 is similar, except that 2 k U1(4 + (x − 1)). 

Proof of Lemma 5.2. Observe that if d = mpj with gcd(m,p) = 1 then the primi- r s tive dth roots of unity can be written in the form wmwpj , r =1, ..., m, gcd(r, m)=1 j r s r and s = 1, ..., p , gcd(p,s) = 1. Notice that wmwpj ≡ wm mod (1 − wpj ) where −1/φ(pj ) |1 − wpj |p = p . Hence we have a mod (1 − wpj ) congruence relating Mmpj and Mm and, since we are dealing with integers, actually a mod p congruence:

φ(pj ) (31) Mmpj ≡ Mm mod p. α j Suppose that p k n and p | MG(F ). Then p | Mmpj for some mp | 2n, gcd(m,p)=1 and 0 ≤ j ≤ α for p ≥ 3 and 0 ≤ j ≤ α +1 for p = 2. By (31) we 2 get that p | Mmpj for all these j and hence p | Mmpj for all the j if m> 2 and for j ≥ 1 if m =1 or 2 and p ≥ 3 and j ≥ 2 if m = 1 and p = 2. 2 1+2α Hence for p odd and α ≥ 1 we get p | Mm,p | Mmp,...,Mmpα and p | MG, 2+2α improving to p | MG except when m = 1 or 2. Suppose that p = 2 and write n =2αN. MINIMAL GROUP DETERMINANTS FOR DICYCLIC GROUPS 9

2 4 Suppose first that α = 0. If m > 1 then 2 | Mm,M2m and 2 | MG(F ). If m = 1 then 2 2 2 2 M1 = f(1) − g(1) ,M2 = f(−1) + g(−1) where f(1),g(1),f(−1) and g(−1) must have the same parity. If both are odd then 3 2 2 | M1 and 2 k M2, while if both are even 2 | M1,M2. Hence in either case 4 2 | MG(F ). Suppose that α ≥ 1. We write

MG(F )= AB, A = Md, B = Md2α+1 Yd|n dY|N 2β where, since Mm is in A and Mm2α+1 is in B both are even, with 2 k B since the Md2α+1 are squares. Now −1 −1 A = MZn f(x)f(x ) − g(x)g(x ) = MD2n (F ), 4 6 and it was shown in [2, Lemma 4.4] that even MD2n (F ) had 2 k A or 2 | A if α =1 and 22α+4 | A if α ≥ 2, giving us 26 k AB or 28 | AB when α = 1 and (b) 6 when α ≥ 2. It remains to show that we do not have 2 k MG(F ) when α = 1. 6 If m = 1 then 2 k M1M2M4 = MQ8 (F ), but from (8) this can not occur. So 2 suppose that for some odd m ≥ 3 we have 2 k Mm,M2m,M4m. Write:

(m−1)/2 j −j −1 n j −j −1 H(x)= f(wmx)f(wm x ) − x g(wmx)g(wm x ) j=1 gcd(Yj,m)=1  −j j −1 −n −j j −1 × f(wm x)f(wmx ) − x g(wm x)g(wmx ) . 2 −1 and observe that Mm = H(1),M2m = H(−1),M4m = H(i) . Observe thatH(x )= i −i H(x), so H(x) is a sum of terms ai(x + x ) and hence N −1 j H(x)= A0 + Aj (x + x ) , Aj ∈ Z. j=1 X So

Mm ≡ A0 +2A1 +4A2 mod 8,M2m ≡ A0 − 2A1 +4A2 mod 8, 2 and if 2 k Mm,M2m 2A0 ≡ Mm + M2m ≡ 0 mod 8. 2 8 Hence A0 ≡ 0 mod 4 and 4 | M4m and 2 | MG(F ). Suppose that p is odd and F = f + yg with p − 1 f(x)=1 − (1 + xn)(1 − x), g(x)= (1 + xn), 2   then for xn = −1 or xn = 1 we have f(x)f(x−1) − xng(x)g(x−1)=1or − 2x−1(x − 1)2 +2p − p2, and −1 2 2 MG(F )= Md,Md = Res(−2x (x − 1) +2p − p , Φd). Yd|n φ(d) 2 α Now Md ≡ 2 Res(Φ1, Φd) 6≡ 0 mod p unless d =1,p,...,p . 10 B. PAUDEL AND C. PINNER

Plainly p k M1 = p(2 − p). Since

pj u φ(pj ) p = (1 − wpj )=(1 − wpj ) A(wpj ), u=1 gcd(Yu,p)=1

j u for φ(p ) > 2 and x = wpj we have −1 2 2 2 −2x (x − 1) +2p − p = (x − 1) ℓ(x), ℓ(x) ≡−2 mod 1 − ωpj , and 2 2 Mpj = Res(1 − x, Φpj ) L = p L, where pj u φ(pj ) L = ℓ(ωpj ) ≡ (−2) ≡ 1 mod 1 − ωpj . u=1 gcd(Yu,p)=1 Since it is an integer, L ≡ 1 mod p. When φ(pj ) = 2, that is p = 3, j = 1, one has 2 2 α 1+2α M3 =3 . Hence p k Mpj , j =1,...,p and p k MG. 

m x +1 a1 at Proof of Lemma 5.3. We set H(x)= x+1 (x + ··· + x ) and   B(x)= f(x)f(x−1) − xng(x)g(x−1). For the values with xn = −1 we plainly have B(x)= δ2 = 1 and when xn =1 B(x) = (δ +2H(x))(δ +2H(x−1) − 4H(x)H(x−1)=1+2δ(H(x)+ H(x−1)). Notice that if x = 1 then B(x)=1+4δH(1)=1+4δt = δp, and since 2 ∤ n

(x +1)= Res(Φd(x), Φ2(x))=1, n x =1Y,x6=1 d|n,dY6=1 so we have ′ MG(f + yg)= MZ2n (B(x)) = MZn (B(x)) = (δp)M , where M ′ = (x +1)(1+2δ(H(x)+ H(x−1)) n x =1Y,x6=1 = x +1+2δ(xm + 1)(xa1 + ··· xat )+2δ(x−m + 1)(x1−a1 + ··· + x1−at ) . xn=1,x6=1 Y  As p ∤ n the values of xp run through the nth roots of unity as x does and

M ′ = xp +1+2δ(xmp + 1)(xpa1 + ··· xpat )+2δ(x−pm + 1)(xp−pa1 + ··· + xp−pat ) . xn=1,x6=1 Y  Taking mp = 1 mod n xp +1 M ′ = (x + 1) +2δ xpa1 + xp−1−pa1 + ··· + xpat + xp−1−pat x +1 xn=1,x6=1   Y  MINIMAL GROUP DETERMINANTS FOR DICYCLIC GROUPS 11

When δ = 1 taking pa1,...,pat ≡ 1, 3,..., (p − 3)/2 mod n gives xp +1 +2δ xpa1 + xp−1−pa1 + ··· + xpat + xp−1−pat x +1 p−1 =1+ x + ··· + x =Φp(x),  and when δ = −1 taking pa1,...,pat ≡ 0, 2,..., (p − 3)/2 mod n gives −Φp(x). Since p ∤ n we have

Φp(x)= Res (Φd(x), Φp(x)) =1 n x =1Y,x6=1 d|n,dY6=1 and M ′ = 1. 

Proof of Lemma 5.4. Suppose that MG(F ) is odd. We write MG(F )= d|2n Md. Then, since the Md are odd squares for d> 2, and so 1 mod 8, we have MG(F ) ≡ 2 2 2 2 Q M1M2 mod 8 where M1 = f(1) − g(1) , M2 = f(−1) − g(−1) . Since M1 is odd the f(1), g(1) have opposite parity. Suppose that f(1) is odd and g(1) even (else switch f and g). If 2 k g(1),g(−1) then M1,M2 ≡ 1 − 4 = −3 mod 8 and if 4 | g(1),g(−1) then M1,M2 ≡ 1 mod 8, and in both cases MG(F ) ≡ 1 mod 8. If 4 | g(1) and 2 k g(−1) (or vice versa) then M1M2 ≡ −3 mod 8 and MG(F ) ≡ −3 mod 8. Suppose that 2 k n and MG(F ) ≡ −3 mod 8 then we can write MG(F ) = 2 2 2 (8m − 3)M4 where M4 = k , k = |f(i)| + |g(i)| , where from above we can assume that f(1) is odd, 4 | g(1), 2 k g(−1) (or vice versa). Now |f(i)|2 ≡ f(1)2 mod 2 is odd and of the form a2 + b2 so must be 1 mod 4. Separating the monomials into the exponents mod 4 we have g(1) = a0 + a1 + a2 + a3, g(−1) = a0 − a1 + a2 − a3, 2 2 2 |g(i)| = (a0 − a2) + (a1 − a3) . Since 4 | g(1), 2 k g(−1) (or vice versa) we have 1 1 a0 + a2 = 2 (g(1) + g(−1)), a1 + a3 = 2 (g(1) − g(−1)) both odd. So a0 − a2 and 2 a1 − a3 are both odd and |g(i)| ≡ 2 mod 8 and k ≡ 3 mod 4. 2 4 Now if p | k and p | n then p | M4,M4p and so p | M4M4p. In either case if 3 q | (8m − 3) and n, then either q | M1 or M2 and q | M1Mq or M2M2q or the extra q came from a square Md with d> 2 so we must have at least two extra q’s. β α Suppose MG(F ) = ±q ≡ −3 mod 8, with q k n, α ≥ 1. Since β is odd we 1+2α must have q | M1M2 and so q | M1Mq ··· Mqα or M2M2q ··· M2qα in addition 2+2α to the q | M4M4q ··· M4qα . 

Proof of Theorem 3.1. Suppose that n is odd. From Lemma 5.2 we can achieve 16 and from Lemma 5.3 achieve the smallest odd prime p ∤ n. The minimum of these is the value claimed for λ(G). By Lemma 5.2 an even determinant must be a multiple of 16 and a value containing a prime p | n must be divisible by p3 (and so be at least 27). Hence we can’t beat 16 or the smallest odd prime p ∤ n. 

Proof of Theorem 3.2. From Lemmas 5.2 we know that the determinants must be of the form 2kpℓm, gcd(m, 2p) = 1, with k = 0 or k ≥ 4 and ℓ = 0 or ℓ ≥ 3. By Lemma 5.3 we can obtain all the m with gcd(m, 2p) = 1, so by multiplication it will be enough to achieve the appropriate 2kpℓ. We get the even powers 2k, k ≥ 4, from g(x) = 0 and

f(x)= x2 +1 ⇒ M =24, f(x)= x2 +1+(xp + 1)x ⇒ M =26, 12 B. PAUDEL AND C. PINNER and the odd powers k ≥ 7 from g(x) = (xp + 1) and f(x)= x4 +1+(xp + 1)(x2 + x) ⇒ M =27, f(x) = (x4 + 1)(x2 +1)+ x2(xp + 1) ⇒ M =29, where to see that the pth roots give 1 it may be useful to note that x4 +1+2(x2 + x) x−4 +1+2(x−2 + x−1) − 4= x−4(x + 1)2(x2 + 1)3, 4 2 2 −4 −2 −2 −6 4 2 4 (x + 1)(x +1)+2x (x + 1)(x +1)+2x  − 4= x (x + 1)(x + 1) . For the powers of p we write p =4b + δ, δ = ±1, a =2b + δ. Then (xa − 1) (xb − 1) f(x)= + mh(x), g(x) = (xp + 1) + mh(x) ⇒ M = δp3(1+4m), (x − 1) (x − 1) where as usual h(x) = (x2p − 1)/(x − 1), giving ±pℓ for all the ℓ ≥ 3 with suitable choices of m. To see that the pth roots give p2 observe that p − a =2b and (xp−a − 1)(x−(p−a) − 1) − 4(xb − 1)(x−b − 1) = −(xb − 1)2(x−b − 1)2. We get the 25pℓ with ℓ =4 or ℓ ≥ 6 using p3 and M = −25p2t+4, t ≥ 0, from 2 2 t+1 t p 2 t+1 t f(x)=1 − x + 2Φp(x ) − p h(x), g(x) = (x +1)+2Φp(x ) − p h(x). Finally, suppose that we have a determinant M = 25m, or when p = 3 mod 4 5 3 5 5 1 2 an M =2 p m or 2 p m, with gcd(m, 2p)=1 and1 ≤ |m| < 2 (p + 1). We write M = M1M2MpM2p where 2 2 2 2 M1 = f(1) − g(1) ,M2 = f(−1) + g(−1) , p−1 p−1 j 2 j 2 j 2 j 2 2πi/p Mp = |f(ω )| − |g(ω )| ,M2p = |f(−ω )| + |g(−ω )| , ω := e . j=1 j=1 Y Y Since Mp, M2p are squares we must have M1M2 even. Thus f(1), g(1) have the 4 same parity and 2 | M1M2 and Mp, M2p are odd. Likewise when p ≡ 3 mod 4 we know that a sum of two squares must be divisible by an even power of p and so 3 5 2 the multiples of p and p must have p | M1, p | Mp and p ∤ M2M2p. Now M2p ≡ p−1 M2 ≡ 1 mod p and so M2p = 1, else m is divisible by the square of an odd integer 2 p−1 j 2 p−1 j 2 ≡ ±1 mod p and |m| ≥ (2p − 1) . But M2p ≥ j=1 |f(−ω )| + j=1 |g(−ω )| , so one of these integers must be zero, say g(−ω) = 0. Hence g(x)=Φ (−x)g (x). Q Q p 1 This gives g(−1) = pg1(−1) and hence g1(−1) = 0, otherwise m has a factor of size 1 2 2 2t at least 2 (p + 1). Hence M2 = f(−1) is divisible by an even power 2 , t ≥ 1. 2 4 5 But g(1), f(1) both even forces 2 k M1 or 2 | M1, contradicting 2 k M1M2. 1 2 We can though get determinants of this form with m = 2 (p + 1): 1 f(x)=1+ x2, g(x) = (x − 1)Φ (x2) ⇒ M = (p2 +1)25, p 2 1 f(x)= −1+ µh(x), g(x)=Φ (−x)+ µh(x) ⇒ M = − (p2 +1)24p3µ, p 2 2 −1 4 (1−ω) on observing that 1 − Φp(−ω)Φp(−ω )=1 − (1+ω)(1+ω−1) = (1+ω)2 . When p =1 mod 4 we can write 2p = A2 + B2 and p 2 p 2 5 5 f(x)=(1+ x)+ A(x − 1)Φp(x ), g(x)= B(x − 1)Φp(x ) ⇒ M =2 p . For p = 5 we also get the missing values 25p3. f(x)=1 − x + x2 +(1+ xp)x, g(x)=1+(1+ xp)(x + x2) ⇒ M = −25p3.  MINIMAL GROUP DETERMINANTS FOR DICYCLIC GROUPS 13

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Department of Mathematics, Kansas State University, Manhattan, KS 66506, USA Email address: [email protected], [email protected]