Group with given characteristic subgroup

If $N$ and $H$ are finite groups, then there exist group $G$ s.t $N$ is normal in $G$ and $G/N\cong H$ (ex. $N\times H, N\rtimes H$ etc.)

Does there exist a group $G$ s.t. $N$ is characteristic subgroup of $G$ and $G/N\cong H$?

• I deleted my answer because I misread the question. – Geoff Robinson Aug 27 '11 at 8:59
• @Geoff: It seemed to work to me, what was wrong? – Zev Chonoles Aug 27 '11 at 9:02
• The PO seems to want to fix both $H$ and $N.$ I thought $N$ was allowed to be chosen. – Geoff Robinson Aug 27 '11 at 9:30
• My gut feeling is that the answer is "no"; here's where I would look for a counterexample: there are $2$-groups $M$ and $K$ such that the "base" subgroup of the wreath product $M\wr K$ is not characteristic, and I would try $N=M$ and $H=K$. But I don't remember off-hand the structures of the groups $M$ and $K$, so I can't check. – Arturo Magidin Aug 28 '11 at 3:27

Try taking $N = H =$ non-abelian finite simple group. It seems rather unlikely that $G$ could be anything other than $N \times N$, and in that case $N$ is not characteristic since there is an automorphism that exchanges the two copies of $N$.

• there doesn't seem to be a lot of 'proving' going on here. – jspecter Aug 27 '11 at 14:19
• As he said, "try taking...". So, take your simplest, easiest group to work with. Say, $C_2$...Done. – user1729 Aug 27 '11 at 14:49
• @Swlabr: $C_2$ doesn't work because $G$ could be $C_4$ which has a characteristic copy of $C_2$. Abelian groups can combine as composition factors in many different ways; that's why I suggested trying a non-abelian simple group. – Ted Aug 27 '11 at 16:51
• @jspecter: I know I haven't proven my answer. I couldn't figure out how to prove it so I just offered it as a plausible suggestion. – Ted Aug 27 '11 at 16:53
• @Ted: Eeverything you say in your answer is accurate, or, at least, can be made so. As Jack Schmidt does in a particular case, the key point to note is that if $N$ is a finite simple (non-Abelian) group, then $N$ does not embed in its outer automorphism group ${\rm Out}(N).$ However, for the moment I do not see an elementary proof of this fact, apart from the fact that the outer automorphism group of a finite simple group is solvable, which requires the classification of finite simple groups (but there may be one- just I don't see it yet). – Geoff Robinson Aug 28 '11 at 8:56

Here is a proof for a nice case of Ted's answer:

Suppose G is a group with normal subgroup N such that both G/N and N are simple groups of order 60. Let C be the centralizer of N in G. Then G/C embeds in Aut(N), a group of order 120. CN = C×N since CN=1 and C and N centralize each other. In particular, CCN/NG/N embeds as a normal subgroup into the simple group of order 60, so has order 60 or 1. Hence G/C has order 60 or order at least 60⋅60 > 120 = |Aut(N)|, a contradiction. Hence C is a simple group of order 60, and G = CN = C×N is a direct product of isomorphic simple groups and so N is not characteristic in G.

I believe the same holds for any non-abelian simple group NG/N, but I've forgotten how to prove G = CN in general without using a little machinery. If G is not a direct product of non-abelian simple groups, then Fit*(G) = N, and so Bender's theorem shows C = 1, a contradiction.

• Is there some way to see the contradiction from $C=1$ besides knowing the order of $\mbox{Aut}(N)$ and then comparing it with the order of $G$? – Ted Aug 28 '11 at 5:15
• @Ted: If C = 1, then G ≤ Aut(N), so we need to know something about the automorphism groups of simple groups. I like to phrase it as $G/CN$ ≤ Out(N), where the left hand side is a quotient of a simple group and the right hand side is solvable, so the left hand side is trivial. How do we know Out(N) does not contain a subgroup isomorphic to N? Well, in all known simple groups this is easy, but in general it is called Schreier's conjecture and so far requires the CFSG to prove. Notice that Out(2×2×2×2) contains 2×2×2×2, so it is not completely clear why N is not in Out(N). – Jack Schmidt Aug 28 '11 at 14:35
• (Oh, this is the same as Geoff's earlier comment on Ted's answer.) – Jack Schmidt Aug 28 '11 at 15:55