Minimal Group Determinants For Dicyclic Groups
aa r X i v : . [ m a t h . N T ] F e b MINIMAL GROUP DETERMINANTS FOR DICYCLIC GROUPS
BISHNU PAUDEL AND CHRISTOPHER PINNER
Abstract.
We determine the minimal non-trivial integer group determinantfor the dicyclic group of order 4 n when n is odd. We also discuss the set of allinteger group determinants for the dicyclic groups of order 4 p . Introduction
For a finite group G = { g , . . . , g n } of order n we assign a variable x g for eachelement g ∈ G and define the group determinant D G ( x g , . . . , x g n ) to be the deter-minant of the n × n matrix whose ij th entry is x g i g − j . We are interested here inthe values that the group determinant can take when the variables are all integers S ( G ) = { D G ( x g , . . . , x g n ) : x g , . . . , x g n ∈ Z } . Notice that S ( G ) will be closed under multiplication:(1) D G ( a g , . . . , a g n ) D G ( b g , . . . , b g n ) = D G ( c g , . . . , c g n ) , c g = X uv = g a u b v . An old problem of Olga Taussky-Todd is to determine S ( Z n ), where the groupdeterminants are the n × n circulant determinants with integer entries. Here andthroughout we write Z n for the integers modulo n , and p will always denote a prime.Laquer [11] and Newman [16, 17] obtained divisibility conditions on the valuesof the group determinant for integer variables for cyclic groups and a completedescription of the values for certain cyclic groups. For example, Laquer [11] andNewman [16] showed that(2) S ( Z p ) = { p a m, ( m, p ) = 1 , a = 0 or a ≥ } , and Laquer [11] that for odd p (3) S ( Z p ) = { a p b m, ( m, p ) = 1 , a = 0 or a ≥ , b = 0 or b ≥ } . Newman [17] described S ( Z ) with upper and lower set inclusions for general Z p .For the general cyclic group Newman [16] showed that(4) { m ∈ Z : gcd( m, n ) = 1 } ⊂ S ( Z n ) , with a divisibility restriction for the values not coprime to the order:(5) p t k n, p | m ∈ S ( Z n ) ⇒ p t +1 | m. Date : February 10, 2021.2010
Mathematics Subject Classification.
Primary: 11R06, 15B36; Secondary: 11B83, 11C08,11C20, 11G50, 11R09, 11T22, 43A40.
Key words and phrases. group determinant, dicyclic group, Lind-Lehmer constant, Mahlermeasure.
For odd p the values for the Dihedral groups of order 2 p or 4 p were obtained in [2]: S ( D p ) = { a p b m : ( m, p ) = 1 , a = 0 or a ≥ , b = 0 or b ≥ } , S ( D p ) = { m ≡ p ∤ m or p | m }∪ { a p b m : ( m, p ) = 1 , a = 4 or a ≥ , b = 0 or b ≥ } , with a counterpart to (4)(6) { m ∈ Z : gcd( m, n ) = 1 } ⊂ S ( D n ) , for n odd, but only those gcd( m, n ) = 1 with m ≡ n is even, andthe divisibility condition (5)(7) p t k n, p | m ∈ S ( D n ) ⇒ p t +1 | m, for odd p , with 2 , or 2 t +4 | m when p = 2 and t = 0 , t ≥ S in [21]. For example for the two dicyclic groups of order less than 14:(8) S ( Q ) = { m + 1 , (8 m − p , and 2 m : m ∈ Z , p ≡ } and S ( Q ) = { a b m : a = 0 , a ≥ , b = 0 or b ≥ , gcd( m,
6) = 1 } (9) ∪ { b m : b = 4 or b ≥ , gcd( m,
6) = 1 }∪ { b mp : b = 0 , , gcd( m,
6) = 1 , p ≡ }∪ { b mp : b = 0 , , gcd( m,
6) = 1 , p ≡ } . 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 integerdeterminant(10) λ ( G ) := min {| D G ( x g , . . . , x g n ) | ≥ x g i ∈ Z } can be difficult. For a group of order n taking x e = 0 and x g = 1 for g = e alwaysgives determinant ( − n − ( n − λ ( G ) ≤ | G | − | G | ≥
3, with λ ( { e } ) = 2, λ ( Z ) = 3.Kaiblinger [9] obtained λ ( Z n ) when 420 ∤ n , with this extended to 2 · · · · · · · · ∤ n in [18]. Values of λ ( G ) for non-cyclic abelian G are consideredin [6, 7, 20, 3, 15]. In [2] the value of λ ( D n ) was obtained for any dihedral groupof order 2 n with 2 · · · · · · . . . · · · ∤ n . Our goal here is todetermine similar results for Q n , the dicyclic group of order 4 n , when n is odd.2. Lind Mahler Measure
For a polynomial F ∈ Z [ x, x − ] one defines the traditional logarithmic Mahlermeasure by(12) m ( F ) = Z log | F ( e πiθ ) | dθ. INIMAL GROUP DETERMINANTS FOR DICYCLIC GROUPS 3
Lind [14] regarded this as a measure on the group R / Z and extended the concept toa compact abelian group with a Haar measure. For example for an F ∈ Z [ x, x − ]and cyclic group Z n we can define a Z n -logarithmic measure m Z n ( F ) = 1 n X z n =1 log | F ( z ) | That is m Z n ( F ) = n log | M Z n ( F ) | where M Z n ( F ) := n − Y j =0 F ( w jn ) , w n := e πi/n . More generally for a finite abelian group(13) G = Z n × · · · × Z n k we can define a logarithmic G -measure on Z [ x , . . . , x k ] by m G ( F ) = 1 | G | log | M G ( F ) | , M G ( F ) = n − Y j =0 · · · n k − Y j k =0 F (cid:0) w j n , · · · , w j k n k (cid:1) . As by observed by Dedekind the group determinant for a finite abelian group canbe factored into linear factors using the group characters ˆ G (14) D G ( x g , . . . , x g n ) = Y χ ∈ ˆ G ( χ ( g ) x g + · · · + χ ( g n ) x g n ) , 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 Z n ( a , a , . . . , a n − ) = M Z n ( a + a x + · · · + a n − x n − ) , and in the general finite abelian case (13) D G ( a g , . . . , a g n ) = M G X g =( j ,...,j k ) ∈ G a g x j · · · x j k k . For a finite non-abelian group the group determinant will not factor into linearfactors but can still be factored using the group representations ˆ G D G ( x g , . . . , x g n ) = Y ρ ∈ ˆ G det X g ∈ G x g ρ ( g ) deg( ρ ) as discovered by Frobenius, see for example [8, 4]. In [2] it was shown that thegroup determinants for the dihedral group of order 2 n , D n = h x, y : x n = 1 , y = 1 , xy = yx − i = { , x, · · · , x n − , y, yx, . . . , yx n − } , can be written as a Z n -measure(15) D D n ( a , . . . , a n − , b , . . . , b n − ) = M Z n (cid:0) f ( x ) f ( x − ) − g ( x ) g ( x − ) (cid:1) , where(16) f ( x ) = a + · · · + a n − x n − , g ( x ) = b + · · · + b n − x n − . Similarly for the dicyclic group of order 4 n , Q n = h x, y : x n = 1 , y = x n , xy = yx − i = { , x, · · · , x n − , y, yx, . . . , yx n − } , B. PAUDEL AND C. PINNER it was shown in [19] that the group representations give(17) D Q n ( a , . . . , a n − , b , . . . , b n − ) = M Z n (cid:0) f ( x ) f ( x − ) − x n g ( x ) g ( x − ) (cid:1) , where(18) f ( x ) = a + · · · + a n − x n − , g ( x ) = b + · · · + b n − x n − . Notice that we can conversely use the group determinant to define a Lind stylepolynomial measure for non-abelian finite groups. For example we can define D n and Q n measures on Z [ x, y ] by M D n ( f ( x ) + yg ( x )) = M Z n (cid:0) f ( x ) f ( x − ) − g ( x ) g ( x − ) (cid:1) ,M Q n ( f ( x ) + yg ( x )) = M Z n (cid:0) f ( x ) f ( x − ) − x n g ( x ) g ( x − ) (cid:1) , (19)although here the polynomial ring is no longer commutative, the monomials havingto satisfy the group relations y = 1, xy = yx − etc., the relations allowing us toreduce 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 ) > F ∈ Z [ x ] } .Given the correspondence between the Lind measures and group determinants in theabelian case we can regard determining λ ( G ) for a finite group as the Lind-Lehmerproblem for that group. An alternative way of extending the Mahler measure togroups can be found in [5].3. Minimal determinants for odd n For the dicyclic groups G = Q n we have some extra properties when n is odd.For example, since(20) M G ( f ( x ) + yg ( x )) = ( − n M G ( g ( x ) + yf ( x )) , if n is odd we have − m ∈ S ( G ) whenever m ∈ S ( G ). This is certainly not truewhen n is even as we saw for Q . When n is odd we also have M G (cid:16) x n + 1)( x + · · · + x ( n − / ) + y ( x n + 1)( x + · · · + x ( n − / ) (cid:17) = 2 n − , always improving on the trivial bound (11), and(21) M G ( x + 1) = 16 , giving us an absolute bound λ ( Q n ) ≤
16 for n odd. In the next section we will seethat an analog to (4) and (6) holds for n odd:(22) { m ∈ Z : gcd( m, n ) = 1 } ⊂ S ( Q n ) , and, corresponding to the divisibility conditions (5) and (7),(23) 2 | m ∈ S ( Q n ) ⇒ | m, p t k n, p | m ∈ S ( Q n ) ⇒ p t +1 | m. Properties (21),(22),(23), are enough for us to completely determine λ ( Q n ): Theorem 3.1. If n is odd then λ ( Q n ) = min { , p } INIMAL GROUP DETERMINANTS FOR DICYCLIC GROUPS 5 where p is the smallest prime not dividing n . That is, λ ( Q n ) = if ∤ n , if | n, ∤ n , if · | n, ∤ n , if · · | n, ∤ n , if · · · | n, ∤ n , if · · · · | n . A complete description of the determinants for D p and D p was given in [2].As we saw for Q in (9) the determinants for Q p must depend more subtly on p ,or at least those determinants M with 2 k M. We can be precise about the othervalues.
Theorem 3.2.
Suppose that p is an odd prime. The determinants for Q p willtake the form k p ℓ m , gcd( m, p ) = 1 , with k = 0 or k ≥ and ℓ = 0 or ℓ ≥ .We can achieve all such values with k = 0 , k = 4 or k ≥ , and all with k = 5 and ℓ = 4 or ℓ ≥ .This just leaves m , p m , p m , gcd( m, p ) = 1 . Not all m are possible.The smallest determinant of the form | m | , gcd( m, p ) = 1 has m = ( p + 1) .If p ≡ mod 4 the smallest p | m | , p | m | , gcd( m, p ) = 1 have m = ( p + 1) .If p ≡ mod 4 then all the multiples of p are determinants. For p = 5 allmultiples of p are determinants. For the p ≡ p > p m , gcd( m, p ) = 1, with | m | smaller than m = ( p + 1).4. The case of even n When G = Q n with n even it is not at all obvious which values coprime to 2 n are determinants; (22) is far from true, the odd determinants must be 1 mod 4 withonly some of those obtainable. The observation that when g = 0 we have(24) M Q n ( f ( x )) = M Z n ( f ( x )) , does give us { m : gcd( m, n ) = 1 } ⊂ { m : m ∈ S ( Z n ) } ⊂ S ( Q n ) , where, writing Φ ℓ ( x ) for the ℓ th cyclotomic polynomial,gcd( m, n ) = 1 ⇒ M G Y p α k m φ p ( x ) α = m . As a counterpart to (21) we do have(25) 2 t k n ⇒ M G ( x t +1 + 1) = 2 t +2 . In particular we always have(26) λ ( Q n ) ≤ min n t +2 , p o , where p is the smallest prime not dividing 2 n . With our divisibility conditionsand Lemma 5.4 we can certainly come up with cases of equality in (26), though notalways, for example λ ( Q ) = 7. B. PAUDEL AND C. PINNER
In a future paper we hope to consider the case of Q n , 2 k n . The general caseof even n seems far out of reach; for t = 1 we know that (25) does give the smallesteven determinant, but for t ≥ Divisibility restrictions and values achieved
We work with the dicyclic measures of polynomials F = f ( x ) + yg ( x ), M G ( F ) = M Z n (cid:0) f ( x ) f ( x − ) − x n g ( x ) g ( x − ) (cid:1) , f, g ∈ Z [ x ] , where if the degree of f or g exceeds 2 n − x n − Lemma 5.1.
Suppose that p α k n then (27) p | M Z n ( F ( x )) ⇒ p α +1 | M Z n ( F ( x )) . Since M Z n (cid:18) x − x n − x − (cid:19) = n , and for odd p p α +1 k M Z n ( p + ( x − , this is sharp for α ≥ when p is odd and for α = 1 when p = 2 .For p = 2 and α ≥ we have (28) 2 | M Z n ( F ( x )) ⇒ α +2 | M Z n ( F ( x )) . Since (29) 2 α +2 k M Z n (4 + ( x − , this exponent is again sharp. Although the exponent is sharp we do not necessarily get that prime power itself(let alone all multiples); for example Newman [17] showed that p is in S ( Z p ) for p = 3 but not for any p ≥ Lemma 5.2.
Suppose that G = Q n .(i) For odd p, if p α k n and p | M G ( F ) then p α +1 | M G ( F ) .This is best possible, for example p α +1 k M G (cid:18) − (1 + x n )(1 − x ) + y (cid:18) p − (cid:19) (1 + x n ) (cid:19) . (ii) Suppose that α || n and | M G ( F ) .(a) If α = 0 then | M G ( F ) .(b) If α ≥ then α +6 | M G ( F ) .Since M G (cid:18) x α +1 + 1 + m x n − x − ym x n − x − (cid:19) = 2 α +2 (1 + 2 mn ) , INIMAL GROUP DETERMINANTS FOR DICYCLIC GROUPS 7 and (30) 2 α +6 k M Q n (4 + ( x − the exponents in (a) and (b) are optimal. For Q p , p odd, we achieve all odd multiples of p , and for Q all multiples of2 , but in general it is not clear whether we can achieve the prime power p α or2 α +6 itself. Property (30) is just (29) and (24).When n is odd the next lemma, the counterpart to [2, Lemma 4.2], shows thatwe can achieve any integer coprime to 2 n . 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 with δ = ± . Set t = ( p − δ ) / , and f = δ + ( x n + 1) H ( x ) , g = ( x n + 1) H ( x ) , with H ( x ) = (cid:18) x m + 1 x + 1 (cid:19) ( x a + · · · + x a t ) , where m is odd with pm ≡ mod n , and pa , . . . , pa t ≡ , , . . . , ( p − / mod n if δ = 1 , and , , . . . , ( p − / mod n if δ = − , then M Q n ( f ( x ) + yg ( x )) = δp. When 2 | n we have additional restrictions on the odd determinants, showingthat we can no longer achieve all integers coprime to 2 n : Lemma 5.4.
Suppose that G = Q n with | n .If ∤ M G ( F ) then M G ( F ) ≡ or − mod .If k n and M G ( F ) ≡ − mod 8 then M G ( F ) = (8 m − k for some integer m and positive integer k ≡ mod . Further we can assume that gcd( k, n ) = 1 or M G ( F ) = (8 m − p with p | n . In either case if q | gcd( n, (8 m − is prime then q | (8 m − .If k n and M G ( F ) ≡ − mod is of the form ± q β with q α k n , α ≥ , then β ≥ α + 3 . Proofs
We shall need to know the resultant of two cyclotomic polynomials, see [1] or[13]; if m > n then | Res(Φ n , Φ m ) | = ( p φ ( n ) if m = np t ,1 else.It will be useful to split the product over the 2 n th roots of unity in (19) into theprimitive d th roots of unity with d | n : M G ( F ) = Y d | n M d , where M d := d Y j =1 ( j,d )=1 f ( w jd ) f ( w − jd ) − w njd g ( w jd ) g ( w − jd ) , w d := e πi/d . B. PAUDEL AND C. PINNER
Since we run through complete sets of conjugates the M d are integers. Moreover,since f ( x ) f ( x − ) − x n g ( x ) g ( x − ) is fixed by x x − , x n = 1, when d = 1 , M d will actually be the squareof an integer for d ≥ Proof of Lemma 5.1.
Suppose that G = Z n and write M G ( f ) = Y d | n U d ( f ) , U d ( f ) = Res(Φ d , f ) ∈ Z . Suppose p | M G ( f ) then p | U mp j ( f ) some p ∤ m , 0 ≤ j ≤ α , and since (1 − w p j ) | p we have U mp j ( f ) = m Y r =1 gcd( r,m )=1 p j Y s =1 gcd( s,p )=1 f ( w rm w sp j ) ≡ U m ( f ) φ ( p j ) mod p and p | U mp j ( f ) all j = 0 , . . . , α, and p α +1 | M G ( f ).Observe that F ( x ) = Q m r =1 ( r,m )=1 f ( w rm x ) is in Z [ x ] (since, for example, its coeffi-cients are fixed by the automorphisms of Q ( w m )). Hence when p = 2 and α ≥ U m ( f ) U m ( f ) U m ( f ) = U ( F ) U ( F ) U ( F ) = M Z ( F ) . From [10] we have S ( Z ) = { a c : gcd( c,
2) = 1 , a = 0 or a ≥ } . Hence we have 2 | U m ( f ) U m ( f ) U m ( f ) and 2 | U mp j any j = 3 , . . . , α , and2 α +2 | M G ( F ) . For the examples observe that U d ( p + ( x − ≡ U d ( x −
1) = Φ d (1) p unless d is a power of p , while U ( p + x −
1) = p , for the d = p j , j = 1 , . . . , α and p ≥ x a primitive p j th root of unity we can write p + ( x −
1) = ( x − v , v ≡ − w p j ) and U p j ( p + ( x − U p j ( x − tp ) = p (1 + tp ) and p k U p j ( p +( x − p α +1 k M G ( p +( x − p = 2 and M G (4+( x − k U (4 + ( x − (cid:3) Proof of Lemma 5.2.
Observe that if d = mp j with gcd( m, p ) = 1 then the primi-tive dth roots of unity can be written in the form w rm w sp j , r = 1 , ..., m , gcd( r, m ) = 1and s = 1 , ..., p j , gcd( p, s ) = 1. Notice that w rm w sp j ≡ w rm mod (1 − w p j ) where | − w p j | p = p − /φ ( p j ) . Hence we have a mod (1 − w p j ) congruence relating M mp j and M m and, since we are dealing with integers, actually a mod p congruence:(31) M mp j ≡ M φ ( p j ) m mod p. Suppose that p α k n and p | M G ( F ). Then p | M mp j for some mp j | n ,gcd( m, p ) = 1 and 0 ≤ j ≤ α for p ≥ ≤ j ≤ α + 1 for p = 2. By (31) weget that p | M mp j for all these j and hence p | M mp j for all the j if m > j ≥ m = 1 or 2 and p ≥ j ≥ m = 1 and p = 2.Hence for p odd and α ≥ p | M m , p | M mp , . . . , M mp α and p α | M G ,improving to p α | M G except when m = 1 or 2.Suppose that p = 2 and write n = 2 α N . INIMAL GROUP DETERMINANTS FOR DICYCLIC GROUPS 9
Suppose first that α = 0. If m > | M m , M m and 2 | M G ( F ). If m = 1 then M = f (1) − g (1) , M = f ( − + g ( − where f (1) , g (1) , f ( −
1) and g ( −
1) must have the same parity. If both are odd then2 | M and 2 k M , while if both are even 2 | M , M . Hence in either case2 | M G ( F ).Suppose that α ≥
1. We write M G ( F ) = AB, A = Y d | n M d , B = Y d | N M d α +1 where, since M m is in A and M m α +1 is in B both are even, with 2 β k B since the M d α +1 are squares. Now A = M Z n (cid:0) f ( x ) f ( x − ) − g ( x ) g ( x − ) (cid:1) = M D n ( F ) , and it was shown in [2, Lemma 4.4] that even M D n ( F ) had 2 k A or 2 | A if α = 1 and 2 α +4 | A if α ≥
2, giving us 2 k AB or 2 | AB when α = 1 and (b)when α ≥
2. It remains to show that we do not have 2 k M G ( F ) when α = 1.If m = 1 then 2 k M M M = M Q ( F ), but from (8) this can not occur. Sosuppose that for some odd m ≥ k M m , M m , M m . Write: H ( x ) = ( m − / Y j =1 gcd( j,m )=1 (cid:0) f ( w jm x ) f ( w − jm x − ) − x n g ( w jm x ) g ( w − jm x − ) (cid:1) × (cid:0) f ( w − jm x ) f ( w jm x − ) − x − n g ( w − jm x ) g ( w jm x − ) (cid:1) . and observe that M m = H (1) , M m = H ( − , M m = H ( i ) . Observe that H ( x − ) = H ( x ), so H ( x ) is a sum of terms a i ( x i + x − i ) and hence H ( x ) = A + N X j =1 A j ( x + x − ) j , A j ∈ Z . So M m ≡ A + 2 A + 4 A mod 8 , M m ≡ A − A + 4 A mod 8 , and if 2 k M m , M m A ≡ M m + M m ≡ . Hence A ≡ | M m and 2 | M G ( F ).Suppose that p is odd and F = f + yg with f ( x ) = 1 − (1 + x n )(1 − x ) , g ( x ) = (cid:18) p − (cid:19) (1 + x n ) , then for x n = − x n = 1 we have f ( x ) f ( x − ) − x n g ( x ) g ( x − ) = 1 or − x − ( x − + 2 p − p , and M G ( F ) = Y d | n M d , M d = Res( − x − ( x − + 2 p − p , Φ d ) . Now M d ≡ φ ( d ) Res(Φ , Φ d ) p unless d = 1 , p, . . . , p α . Plainly p k M = p (2 − p ). Since p = p j Y u =1 gcd( u,p )=1 (1 − w up j ) = (1 − w p j ) φ ( p j ) A ( w p j ) , for φ ( p j ) > x = w up j we have − x − ( x − + 2 p − p = ( x − ℓ ( x ) , ℓ ( x ) ≡ − − ω p j , and M p j = Res(1 − x, Φ p j ) L = p L, where L = p j Y u =1 gcd( u,p )=1 ℓ ( ω up j ) ≡ ( − φ ( p j ) ≡ − ω p j . Since it is an integer, L ≡ p . When φ ( p j ) = 2, that is p = 3, j = 1, one has M = 3 . Hence p k M p j , j = 1 , . . . , p α and p α k M G . (cid:3) Proof of Lemma 5.3.
We set H ( x ) = (cid:16) x m +1 x +1 (cid:17) ( x a + · · · + x a t ) and B ( x ) = f ( x ) f ( x − ) − x n g ( x ) g ( x − ) . For the values with x n = − B ( x ) = δ = 1 and when x n = 1 B ( x ) = ( δ + 2 H ( x ))( δ + 2 H ( x − ) − H ( x ) H ( x − ) = 1 + 2 δ ( H ( x ) + H ( x − )) . Notice that if x = 1 then B ( x ) = 1 + 4 δH (1) = 1 + 4 δt = δp , and since 2 ∤ n Y x n =1 ,x =1 ( x + 1) = Y d | n,d =1 Res(Φ d ( x ) , Φ ( x )) = 1 , so we have M G ( f + yg ) = M Z n ( B ( x )) = M Z n ( B ( x )) = ( δp ) M ′ , where M ′ = Y x n =1 ,x =1 ( x + 1)(1 + 2 δ ( H ( x ) + H ( x − ))= Y x n =1 ,x =1 (cid:0) x + 1 + 2 δ ( x m + 1)( x a + · · · x a t ) + 2 δ ( x − m + 1)( x − a + · · · + x − a t ) (cid:1) . As p ∤ n the values of x p run through the n th roots of unity as x does and M ′ = Y x n =1 ,x =1 (cid:0) x p + 1 + 2 δ ( x mp + 1)( x pa + · · · x pa t ) + 2 δ ( x − pm + 1)( x p − pa + · · · + x p − pa t ) (cid:1) . Taking mp = 1 mod nM ′ = Y x n =1 ,x =1 ( x + 1) (cid:18) x p + 1 x + 1 + 2 δ (cid:0) x pa + x p − − pa + · · · + x pa t + x p − − pa t (cid:1)(cid:19) INIMAL GROUP DETERMINANTS FOR DICYCLIC GROUPS 11
When δ = 1 taking pa , . . . , pa t ≡ , , . . . , ( p − / n gives x p + 1 x + 1 +2 δ (cid:0) x pa + x p − − pa + · · · + x pa t + x p − − pa t (cid:1) = 1 + x + · · · + x p − = Φ p ( x ) , and when δ = − pa , . . . , pa t ≡ , , . . . , ( p − / n gives − Φ p ( x ).Since p ∤ n we have Y x n =1 ,x =1 Φ p ( x ) = Y d | n,d =1 Res (Φ d ( x ) , Φ p ( x )) = 1and M ′ = 1. (cid:3) Proof of Lemma 5.4.
Suppose that M G ( F ) is odd. We write M G ( F ) = Q d | n M d .Then, since the M d are odd squares for d >
2, and so 1 mod 8, we have M G ( F ) ≡ M M mod 8 where M = f (1) − g (1) , M = f ( − − g ( − . Since M isodd 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 M , M ≡ − − | g (1) , g ( −
1) then M , M ≡ M G ( F ) ≡ | g (1) and 2 k g ( −
1) (or vice versa) then M M ≡ − M G ( F ) ≡ − k n and M G ( F ) ≡ − M G ( F ) =(8 m − M where M = k , k = | f ( i ) | + | g ( i ) | , where from above we can assumethat f (1) is odd, 4 | g (1), 2 k g ( −
1) (or vice versa). Now | f ( i ) | ≡ f (1) mod 2 isodd and of the form a + b so must be 1 mod 4. Separating the monomials intothe exponents mod 4 we have g (1) = a + a + a + a , g ( −
1) = a − a + a − a , | g ( i ) | = ( a − a ) + ( a − a ) . Since 4 | g (1), 2 k g ( −
1) (or vice versa) we have a + a = ( g (1) + g ( − a + a = ( g (1) − g ( − a − a and a − a are both odd and | g ( i ) | ≡ k ≡ p | k and p | n then p | M , M p and so p | M M p . In either case if q | (8 m −
3) and n , then either q | M or M and q | M M q or M M q or the extra q came from a square M d with d > q ’s.Suppose M G ( F ) = ± q β ≡ − q α k n , α ≥
1. Since β is odd wemust have q | M M and so q α | M M q · · · M q α or M M q · · · M q α in additionto the q α | M M q · · · M q α . (cid:3) Proof of Theorem 3.1.
Suppose that n is odd. From Lemma 5.2 we can achieve16 and from Lemma 5.3 achieve the smallest odd prime p ∤ n . The minimum ofthese is the value claimed for λ ( G ). By Lemma 5.2 an even determinant must be amultiple of 16 and a value containing a prime p | n must be divisible by p (and sobe at least 27). Hence we can’t beat 16 or the smallest odd prime p ∤ n . (cid:3) Proof of Theorem 3.2.
From Lemmas 5.2 we know that the determinants must beof the form 2 k p ℓ m , gcd( m, p ) = 1, with k = 0 or k ≥ ℓ = 0 or ℓ ≥
3. ByLemma 5.3 we can obtain all the m with gcd( m, p ) = 1, so by multiplication itwill be enough to achieve the appropriate 2 k p ℓ .We get the even powers 2 k , k ≥
4, from g ( x ) = 0 and f ( x ) = x + 1 ⇒ M = 2 , f ( x ) = x + 1 + ( x p + 1) x ⇒ M = 2 , and the odd powers k ≥ g ( x ) = ( x p + 1) and f ( x ) = x + 1 + ( x p + 1)( x + x ) ⇒ M = 2 ,f ( x ) = ( x + 1)( x + 1) + x ( x p + 1) ⇒ M = 2 , where to see that the p th roots give 1 it may be useful to note that (cid:0) x + 1 + 2( x + x ) (cid:1) (cid:0) x − + 1 + 2( x − + x − ) (cid:1) − x − ( x + 1) ( x + 1) , (cid:0) ( x + 1)( x + 1) + 2 x (cid:1) (cid:0) ( x − + 1)( x − + 1) + 2 x − (cid:1) − x − ( x + 1)( x + 1) . For the powers of p we write p = 4 b + δ , δ = ± a = 2 b + δ . Then f ( x ) = ( x a − x −
1) + mh ( x ) , g ( x ) = ( x p + 1) ( x b − x −
1) + mh ( x ) ⇒ M = δp (1 + 4 m ) , where as usual h ( x ) = ( x p − / ( x − ± p ℓ for all the ℓ ≥ m . To see that the p th roots give p observe that p − a = 2 b and( x p − a − x − ( p − a ) − − x b − x − b −
1) = − ( x b − ( x − b − . We get the 2 p ℓ with ℓ = 4 or ℓ ≥ p and M = − p t +4 , t ≥ , from f ( x ) = 1 − x + 2Φ p ( x ) t +1 − p t h ( x ) , g ( x ) = ( x p + 1) + 2Φ p ( x ) t +1 − p t h ( x ) . Finally, suppose that we have a determinant M = 2 m , or when p = 3 mod 4an M = 2 p m or 2 p m , with gcd( m, p ) = 1 and 1 ≤ | m | < ( p + 1).We write M = M M M p M p where M = f (1) − g (1) , M = f ( − + g ( − ,M p = p − Y j =1 | f ( ω j ) | − | g ( ω j ) | , M p = p − Y j =1 | f ( − ω j ) | + | g ( − ω j ) | , ω := e πi/p . Since M p , M p are squares we must have M M even. Thus f (1), g (1) have thesame parity and 2 | M M and M p , M p are odd. Likewise when p ≡ p and sothe multiples of p and p must have p | M , p | M p and p ∤ M M p . Now M p ≡ M p − ≡ p and so M p = 1, else m is divisible by the square of an odd integer ≡ ± p and | m | ≥ (2 p − . But M p ≥ Q p − j =1 | f ( − ω j ) | + Q p − j =1 | g ( − ω j ) | ,so one of these integers must be zero, say g ( − ω ) = 0. Hence g ( x ) = Φ p ( − x ) g ( x ).This gives g ( −
1) = pg ( −
1) and hence g ( −
1) = 0, otherwise m has a factor of sizeat least ( p + 1). Hence M = f ( − is divisible by an even power 2 t , t ≥ g (1), f (1) both even forces 2 k M or 2 | M , contradicting 2 k M M .We can though get determinants of this form with m = ( p + 1): f ( x ) = 1 + x , g ( x ) = ( x − p ( x ) ⇒ M = 12 ( p + 1) 2 ,f ( x ) = − µh ( x ) , g ( x ) = Φ p ( − x ) + µh ( x ) ⇒ M = −
12 ( p + 1) 2 p µ, on observing that 1 − Φ p ( − ω )Φ p ( − ω − ) = 1 − ω )(1+ ω − ) = (1 − ω ) (1+ ω ) . When p = 1 mod 4 we can write 2 p = A + B and f ( x ) = (1 + x ) + A ( x p − p ( x ) , g ( x ) = B ( x p − p ( x ) ⇒ M = 2 p . For p = 5 we also get the missing values 2 p . f ( x ) = 1 − x + x + (1 + x p ) x, g ( x ) = 1 + (1 + x p )( x + x ) ⇒ M = − p . (cid:3) INIMAL GROUP DETERMINANTS FOR DICYCLIC GROUPS 13
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