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Topic: Groups with a maximal subgroup of order 3 (Read 1479 times) |
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ecoist
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Groups with a maximal subgroup of order 3
« on: Aug 25th, 2006, 6:56pm » |
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Some years ago I posed the puzzle of classifying all groups of order 3*2n with a maximal subgroup of order 3. It turns out that, with little change in the proof, one can:
Determine all non-isomorphic groups of order 3*pn with p a prime different from 3 and a Sylow 3-subgroup maximal.
(Aside: I don't know if the problem remains manageable if we drop the assumption that G is divisible by at most two distinct primes.)
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Deedlit
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Re: Groups with a maximal subgroup of order 3
« Reply #1 on: Aug 25th, 2006, 9:56pm » |
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on Aug 25th, 2006, 6:56pm, ecoist wrote:| Some years ago I posed the puzzle of classifying all groups of order 3*2n with a maximal subgroup of order 3. |
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You mean index 3. But such a subgroup is always maximal!
I wondered if you might have meant that there is a proper subgroup that contains all other subgroups, but of course that's not possible in a non-p-group.
So, I'm not sure what you mean.
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ecoist
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Re: Groups with a maximal subgroup of order 3
« Reply #2 on: Aug 25th, 2006, 10:38pm » |
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I meant exactly what I said, a maximal subgroup of order 3. The solution includes, but is not limited to, the symmetric group on three letters and the alternating group on four letters.
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Deedlit
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Re: Groups with a maximal subgroup of order 3
« Reply #3 on: Aug 25th, 2006, 11:18pm » |
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Whoops, my apologies! I read your second statement wrong, and pulled it back to the first.
Anyway, G (our group) cannot be the direct product of its Sylow subgroups, since every the p-Sylow group has proper subgroups for n > 1, and we can just take the product. For p > 3, the p-Sylow group is normal, so the 3-Sylow group must not be. The normallizer of the 3-Sylow group must contain itself, and can't be all of G, so it must be itself. So the 3-Sylow group has pn conjugates, and G consists of a group of order pn and pn groups of order 3.
Of course for n=1 the restriction is meaningless, so we get all cyclic groups of order 3p, and all metacyclic groups of order 3p for p = 1 mod 3.
Hmmm, more later...
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Eigenray
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Re: Groups with a maximal subgroup of order 3
« Reply #4 on: Aug 26th, 2006, 7:04pm » |
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As Deedlit points out, G is nonabelian unless n=1, and G is cyclic of order 3p. So assume G is nonabelian.
By Burnside's theorem, G is solvable. Let N be a nontrivial proper normal subgroup. Let C be a subgroup of order 3. Since CN contains C, either CN=C, in which case N=C, or CN=G.
Suppose N=C is normal, and let P be a subgroup of order p. Then C < CP implies CP=G, which means n=1. Moreover, since P acts nontrivially on C by conjugation, and |Aut(C)|=2, p=2 and G=S3.
Otherwise, suppose CN=G, so that |N|=pn. If K is characteristic in N, it is normal in G, so KC=C or G, i.e., K=1 or K=N. Since Z(N) is characteristic and nontrivial, we have Z(N)=N is abelian. K={x : xp=1 } is also characteristic and nontrivial, so K=N, and K is elementary abelian. Thus G = Zpn x| C, defined by some nontrivial (since G is nonabelian) homomorphism
phi : C -> Aut(Zpn) = GLn(Zp).
Let T=phi(a), where a generates C. Then T is an automorphism of the vector space Zpn of order 3.
Lemma: A subset W of V=Zpn is a T-invariant subspace iff H = WxC is a subgroup of G.
(=>) (v,Ti)(w,Tj) = (v+Tiw, Ti+j).
(<=) (v,I)(w,I)=(v+w,I), so W is a subspace. (0,T)(v,I) = (Tv,T), so W is T-invariant.
By the lemma, if W is a T-invariant subspace of V, then W x C = C or G, so W = 0 or V. In particular, span{v,Tv,T2v}=V for any non-zero v, so n<3.
If p=1 mod 3, x3-1 splits over Fp, so T has an eigenvector, which forces n=1. GL1(Zp)=Zp* is cyclic, so has a unique subgroup of order 3, hence G = Zp x| Z3 is uniquely determined.
If p=2 mod 3, then the two primitive cube roots of 1 lie in Fp^2 \ Fp. If 1 is an eigenvalue, then n=1. Otherwise, n=3 is impossible, so n=2, and T has characteristic polynomial x2+x+1.
Conversely, suppose p=2 mod 3. Then |Aut(Zp2)| = p(p-1)2(p+1) is divisible by 3, so there's some T of order 3. Then C=<T> is maximal in G = Zp2 x| C. For, suppose C < H < G, so |H|=3p. Note that if (v,Ti) is in H, then since (0,Tj) is in C, we have (v, Ti)(0,Tj) = (v, Ti+j) in H, for all j. Hence H = W x C for some set W of size p, and by the lemma it is a 1-dimensional T-invariant subspace of V. So T has an eigenvector, so its characteristic polynomial splits over Fp, but since p=2 mod 3, this forces T=I, a contradiction.
So the groups with n>1 are those of the form GT = Zp2 x| C, where p=2 mod 3, and T : C->GL2(Zp) has order 3.
Given T, S of order 3, they are both diagonalizable over Fp^2, so they have the same rational canonical form, hence they are similar over Fp. Pick F in GL2(F2) such that FT=SF. Then (v,Ti) -> (Fv, Si) shows that GT is isomorphic to GS.
To summarize: In addition to S3, we have, for each prime p != 3, one abelian group, of order 3p, and one nonabelian group, Zpr x| <T>, where r is the order of p mod 3, and {I,T,T2} is the unique conjugacy class of a subgroup of order 3 in GLr(Fp).
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Eigenray
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Re: Groups with a maximal subgroup of order 3
« Reply #5 on: Aug 26th, 2006, 7:18pm » |
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I thought I'd take this chance to remind people of this problem, which has gone unanswered. (The fact that I'm linking to it from here may be viewed as a hint  )
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Eigenray
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Re: Groups with a maximal subgroup of order 3
« Reply #6 on: Aug 27th, 2006, 3:59am » |
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The group of order 3p2 above has a geometric interpretation:
If H is a subgroup of GL(V), then V x| H is the group of affine maps v+Tx, v in V, T in H.
Let V = {(x,y,z) | x+y+z=0} in Fp3, and T(x,y,z) = (y,z,x). Then G = V x| <T> is the group of affine transformations of V generated by V and T.
In the case p=2, V is a tetrahedron, and G is its group of rotations in 3-space: the image of the vertex 0 is determined by v, and its neighbors are cycled by T. Thus, G=A4.
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ecoist
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Re: Groups with a maximal subgroup of order 3
« Reply #7 on: Aug 27th, 2006, 11:17am » |
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Eigenray, I couldn't tell if your list includes Z52 extended by an automorphism of order 3.
Here's my proof.
Let T be a subgroup of order 3 and let P be a sylow p-subgroup of G. If T is normal in G, then G has order 3 or 3p since T is maximal in G. Hence G is either cyclic or isomorphic to the symmetric group on 3 letters.
Now assume that T is not normal in G. Then G has (3-1)*pn elements of order 3 since T is maximal. That leaves pn elements for the Sylow p-subgroups. Hence P is normal in G. Since the set C of all elements in the center of P of order less or equal p form a characteristic subgroup of P, it follows that C is normal in G. By the maximality of T, we have C=P. Hence G is the semidirect product of Zpn and T.
Let x be a nonidentity element of P and let t be a generator of T. If t normalizes the subgroup generated by x, then, by the maximality of T and the fact that T is not normal in G, we have that G=<t,x> is a Frobenius group of order 3p, with p congruent 1 modulo 3. Otherwise, consider the product of the elements x, t-1xt, and t-2xt2. This product commutes with t. This product must be 1, for, otherwise, this product together with T generates a subgroup of G strictly between T and G, contradicting the maximality of T. Therefore, the product is 1 and the subgroup <x,t> has order 3*p2. By the maximality of T, this subgroup is G. Moreover, the matrix of t relative to the basis {x,t-1xt} of P=Zpn is
|0 -1|
|1 -1|.
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Eigenray
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Re: Groups with a maximal subgroup of order 3
« Reply #8 on: Aug 27th, 2006, 3:30pm » |
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The proofs are essentially the same... I was a bit redundant with the linear algebra though.
Suppose a nonabelian group G has order pnq, and a maximal q-subgroup. Let P, Q be Sylow subgroups. Since Q is maximal, either N(Q)=G or N(Q)=Q.
In the first case, n=1 and G = Q x| P. Then p|q-1, and there's a unique such G.
In the second case, G has (q-1)pn elements of order q, so P is normal. If K is characteristic in P, KQ=Q or KQ=G, so K is 1 or P, i.e., P is a characteristically simple p-group, so P=Zpn, and G = P x| Q is a nontrivial irreducible representation of Q -> GL(P).
Q = <T>, where T has order q. Since P has no T-invariant subspaces, T is similar to the companion matrix of some irreducible polynomial f(x), which is the minimal polynomial of T, and divides the q-th cyclotomic polynomial phiq(x). Since the latter has distinct roots in some extension of Fp, it follows that two such operators T, T' are conjugate (over Fp) iff they have the same eigenvalues.
Let r be the order of p mod q, so that Fp^r is the smallest field containing a q-th root of unity z. Then for 0<a<q, za, zap, zap^2, ..., zap^(r-1) all have the same minimal polynomial ga over Fp, so
phiq(x) = xq-1 + xq-2 + ... + x + 1 = ga_1(x) ga_2(x) ... ga_t(x)
is the factorization of phiq into degree-r irreducibles over Fp, where {a1,...,at} represent the cosets of <p> in Fq*.
Now, if T has eigenvalues {za, zap, ..., zap^(r-1) }, and T' has eigenvalues { za', za'p, ..., za'p^(r-1) }, it follows that Ta' ~ T'a, and therefore that the subgroups Q=<T> and Q'=<T'> are conjugate in GL(P), hence P x| Q and P x| Q' are isomorphic.
Thus G = P x| Q, where a generator of Q acts on P via the companion matrix of an irreducible factor of phiq(x) in Fp[x]. In particular, n=r is the order of p mod q.
Thus the groups are:
(1) Cyclic of order pq
(2) If p|q-1, Zq x| Zp
(3) Zpn x| Zq, where n is the order of p mod q.
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| « Last Edit: Aug 27th, 2006, 8:22pm by Eigenray » |
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ecoist
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Re: Groups with a maximal subgroup of order 3
« Reply #9 on: Aug 27th, 2006, 6:13pm » |
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Nice! (I think in (2) you mean q|(p-1) rather than p|(q-1).)
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Eigenray
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Re: Groups with a maximal subgroup of order 3
« Reply #10 on: Aug 29th, 2006, 7:54pm » |
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Thanks. Another way to state the result:
Aside from the nontrivial semidirect product Zq x| Zp when p|(q-1) (and I do mean that), all other groups can be constructed as follows:
Let T : Fp^q -> Fp^q be Frobenius, T(x)=xp, viewed as an Fp-linear operator (of course, any q-cycle of the coordinates will do, but, you know). Then V = Fp^q decomposes into irreducible subspaces
V = U + W1 + ... + Wk,
where T acts trivially on U = Fp, and each Wi has dimension r, the order of p mod q, and rk=(q-1). The Wi are non-isomorphic as Zq representations, but the semidirect products W x| T are isomorphic groups. Then G = U x| T (the cyclic case), or G = W x| T (the nonabelian case).
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Deedlit
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Re: Groups with a maximal subgroup of order 3
« Reply #11 on: Sep 6th, 2006, 4:42pm » |
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I hope you don't mind if I ask some silly questions.
on Aug 26th, 2006, 7:04pm, Eigenray wrote:
If p=1 mod 3, x3-1 splits over Fp,
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Does this follow from a general theorem?
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If p=2 mod 3, then the two primitive cube roots of 1 lie in Fp^2 \ Fp.
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Again, what's the general rule for this?
Quote: Otherwise, n=3 is impossible, so n=2, and T has characteristic polynomial x2+x+1.
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Does the fact that T^3 - I = 0 imply that the characteristic polynomial of T divides x^3 - 1?
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Note that if (v,Ti) is in H, then since (0,Tj) is in C, we have (v, Ti)(0,Tj) = (v, Ti+j) in H, for all j. Hence H = W x C for some set W of size p, and by the lemma it is a 1-dimensional T-invariant subspace of V.
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I would think we also need (0,Tj)(v, Ti) = (v, Ti+j), although that follows from p = 2 mod 3, so Aut(Z_p) = Z_(p-1), which doesn't how a subgroup of order 3.
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Given T, S of order 3, they are both diagonalizable over Fp^2, so they have the same rational canonical form, hence they are similar over Fp.
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Why is this?
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Eigenray
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Re: Groups with a maximal subgroup of order 3
« Reply #12 on: Nov 18th, 2006, 6:54pm » |
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Sorry, I meant to reply when I had time, but then completely forgot.
on Sep 6th, 2006, 4:42pm, Deedlit wrote:| Does this follow from a general theorem? / Again, what's the general rule for this? |
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The multiplicative group Fp* is cyclic of order p-1, so contains (two) elements of order 3 iff p = 1 mod 3. Hence if p=1 mod 3, x3-1 has 3 roots. Otherwise, x2+x+1 = (x3-1)/(x-1) is irreducible over Fp, so its roots have degree 2 over Fp, hence lie in (an isomorphic copy of) Fp^2.
As I mentioned before, we have more generally that if q is prime, and r is the order of p mod q, then (xq-1)/(x-1) factors into the product of (q-1)/r polynomials of degree r, each irreducible over Fp.
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Does the fact that T^3 - I = 0 imply that the characteristic polynomial of T divides x^3 - 1? |
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Since we're assuming V is irreducible as an Fp[T]-module, V ~= Fp[x]/(p(x)), where p is both the minimal polynomial and the characteristic polynomial of T. So, yes. But my original reasoning was:
If n>1, then T has no eigenvectors (over Fp), since we're assuming V is irreducible as an Fp[T]-module. Hence its eigenvalues can only be w and w2, where w in Fp^2 is a primitive cube root of 1. Since tr T, the sum of the eigenvalues, is an element of Fp, there is no way to have n=3. This forces n=2, and T has eigenvalues w and w2, each with multiplicity 1.
Quote:| I would think we also need (0,Tj)(v, Ti) = (v, Ti+j), although that follows from p = 2 mod 3, so Aut(Z_p) = Z_(p-1), which doesn't how a subgroup of order 3. |
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I don't know what you mean. I was showing that if (v, c) is in H for some c, then {v} x C is contained in H (so H = WxC for some set W).
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S,T are diagonalizable (to the same matrix) over Fp^2 since they have two distinct eigenvalues. In general, if two matrices with entries in some field F are similar over some larger field K, then they are similar over F. To see this, let S', T' denote the rational canonical forms of S,T, computed over F. Then S' ~ S ~ T ~ T' are similar over K, and since S', T' are both in rational canonical form, we have S'=T' by uniqueness of RCF. Then S ~ S' = T' ~ T are similar over F.
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