A simple question on characteristic subgroups

Question

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 $2$?

Answer 1

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.

Answer 2

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 U ≤ V ≤ P are characteristic in P with [V : U] = 2. Then G has a normal subgroup of index 2.

Proof: Let N = N_G(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 = C_V(N) × [ V, N ], but since U is normal in N, [ V, N ] ≤ U < V is a proper subgroup of V, and so C_V(N) ≠ 1. Of course, then C_P(A) ≥ C_V(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 = U ≤ V = 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 C_V(N) ∩ [ V, N ] = 1. For example, P ≅ Q₈ has 1 = U ≤ V = Z(P) ≤ P characteristic, and N_G(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.

Answer 3

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.