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Suppose $P$ is a finite abelian $2$-group and $U$,$V$ are characteristic subgroups of $P$ such that $|V:U|=2$. Does it follow that $P$ has a characteristic subgroup of order $2$?

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3 Answers 3

up vote 2 down vote accepted

To answer a (deleted now, I guess) question of Stefanos in the comments: show a finite simple group with a Sylow 2-subgroup satisfying the hypothesis on P has order 2. One uses transfer and coprime action.

Proposition: Suppose G is a finite group with abelian Sylow 2-subgroup P such that UVP are characteristic in P with [V : U] = 2. Then G has a normal subgroup of index 2.

Proof: Let N = NG(P) and since P is abelian, let A = N/P be the induced group of automorphisms. Since the action of A on P is coprime, V = CV(N) × [ V, N ], but since U is normal in N, [ V, N ] ≤ U < V is a proper subgroup of V, and so CV(N) ≠ 1. Of course, then CP(A) ≥ CV(N) ≠ 1 and [ P, N ] is a proper subgroup of P. [ P, N ] is known as the focal subgroup of P in G (because P is abelian), and G has a quotient isomorphic to the (non-identity) 2-group P/[ P, N ], and so G has a normal subgroup of index 2 corresponding to a maximal subgroup of P under the transfer map. $\square$.

The key part of the argument is that [ V, N ] ≠ V. In more detail: let v be an element of V not contained in U. Since U is normal in N, no conjugate of v can be in U, and so N permutes the non-identity elements of V/U by conjugation. In particular, (vU)n must be a non-identity coset of V/U, but there is only one: vU. Hence [ v, n ] = u, and [ V, N ] ≤ U.

I use that P is abelian, that U, V are normal in N, and that [ V : U ] = 2, not just that it is prime.

Indeed, G non-abelian of order 6 satisfies similar hypotheses when p=3, with 1 = UV = P, but G has no normal subgroup of index 3 (the corresponding conclusion). It is also necessary to assume P abelian in order to get CV(N) ∩ [ V, N ] = 1. For example, P ≅ Q8 has 1 = UV = Z(P) ≤ P characteristic, and NG(P) always controls G-fusion (and so transfer), but G = SL(2, q) for q ≡ ±3 mod 8 has P as a Sylow 2-subgroup and no normal subgroup of index 2.

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Thank you for your answer Jack. I deleted my question because I think I managed to solve it: The main result I am assuming is that if $G$ is finite and $P$ is an abelian Sylow $p$-subgroup of $G$, then $G' \cap P \cap Z(N_G(P))=1$. Now assume $P$ is a proper subgroup of $G$. Then $G$ is not abelian (otherwise $P$ would not be proper). Since $G$ is simple, this forces $G'=G$ and hence $P \cap Z(N_G(P))=1$. –  the_fox Oct 8 '11 at 15:59
    
Both $U$ and $V$ are characteristic in $P$ which is normal in $N_G(P)$. Thus both $U$ and $V$ are normal in $N_G(P)$. Then $V/U$ is normal in $N_G(P)/U$. Since $|V/U|=2$ the involution, say $vU$, which generates it is in $Z(N_G(P)/U)$ which equals $Z(N_G(P))/U$. Therefore $vU=gU$ for some $g \in Z(N_G(P))$. Hence $v^-1g \in U \leq V \leq P$, which implies that $g \in P$. So it must be the case that $g=1$ and hence $v \in U$, a contradiction. Thus $P$ is not proper in $G$ which means that (since G is simple) $G=C_2$. What do you think of this argument? –  the_fox Oct 8 '11 at 15:59
    
@StefanosA: I think that sounds good. $P\cap Z(N) = C_P(N)$ and so our proofs are pretty similar. I didn't check: why is Z(N/U) = Z(N)/U? I think it is true, just because P is abelian and N is coprime. Everything else seems very clear. –  Jack Schmidt Oct 8 '11 at 21:52
    
You are right to question the equality between the centers. As far as I can see only one containment holds (vacuously): $Z(N)/U \leq Z(N/U)$ and it's not even the useful one. –  the_fox Oct 9 '11 at 1:58

Let $G$ be the abelian group $C_4 \times C_4 \times C_2$, and let the three direct factors have generators $x,y,z$.

Let $V = \{ x \in G \mid x^2 = 1 \}$. Then $|V|=8$, $V$ is characteristic in $G$ and $V$ is generated by $x^2,y^2,z$.

Let $U = \{ x^2 \mid x \in G \}$. Then $|U|=4$, $U$ is characteristic in $G$ and is generated by $x^2,y^2$. So $U \le V$ with $|V:U|=2$.

But $G$ has no characteristic subgroup of order 2. To see this, note that there are 7 elements of order 2, $x^2, y^2, (xy)^2, z, zx^2, zy^2, z(xy)^2$. The automorphism induced by $(x,y,z) \rightarrow (y,xy,z)$ moves the three elements $x^2,y^2,(xy)^2$, whereas the automorphisms $(x,y,z) \rightarrow (x,y,zt)$ with $t= x^2,y^2$ or $(xy)^2$ move the other elements of order 2.

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Note: As pointed out in the comments below, this answer is not correct, as I was wrong in assuming that characteristic subgroups stay characteristic in quotients by characteristic subgroups (it is the other way around).

Yes, this follows because characteristic subgroups of characteristic subgroups are themselves characteristic. This means that we just need to show that $V$ has a characteristic subgroup of order 2.

Now, if $U$ is a characteristic subgroup of index 2 in $V$, then it corresponds to a characteristic subgroup of $V/\Phi(V)$ of index 2, but since this quotient is characteristically simple (it is elementary abelian), we must have that $U = \Phi(V)$ and thus $V$ is cyclic, which indeed means that $V$ has a characteristic subgroup of order 2.

Note that I have not used that $P$ was assumed abelian, nor that it was assumed to be a 2-group (any prime will do).

Edit: As pointed out below, I am still assuming that $P$ is a $p$-group for some prime $p$, just not necessarily 2.

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Yes, I am assuming that P is a p-group for some prime p (just not necessarily 2) –  Tobias Kildetoft Oct 7 '11 at 9:16
    
Tobias: The dihedral group of order 8 has a characteristic subgroup of index 2, but it is not cyclic. Characteristic subgroups may not remain characteristic in quotients (some automorphisms do not extend to extensions, they have to stabilize the cohomology class). At any rate, there are some groups of order 32 with characteristic subgroups of index 2 and no characteristic subgroups of size 2. –  Jack Schmidt Oct 7 '11 at 13:35
    
I don't think I buy this argument, since it can happen that all the automorphisms of $V$ descend to the trivial automorphism of $V/\Phi(V)$ (or e.g. to automorphisms fixing a particular line). In particular, your conclusion that $V$ is necessarily cyclic is false. A counterexample is given e.g. by $G$=SmallGroup(32,9) in MAGMA notation. The centre of this group is $C_2\times C_2$ (and, being the centre, is characteristic), but it contains a characteristic subgroup of order 2. Namely, $G$ is a semidirect product of $C_4$ by $D_8$ and the centre of $D_8$ is (necessarily) characteristic in $G$. –  Alex B. Oct 7 '11 at 13:49
    
I didn't see Jack's comment before posting mine, but I guess I might as well leave it up there. –  Alex B. Oct 7 '11 at 13:51

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