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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 $G/N\cong H$?

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    I deleted my answer because I misread the question.2011-08-27
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    @Geoff: It seemed to work to me, what was wrong?2011-08-27
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    The PO seems to want to fix both $H$ and $N.$ I thought $N$ was allowed to be chosen.2011-08-27
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    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.2011-08-28

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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 $N$.

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    there doesn't seem to be a lot of 'proving' going on here.2011-08-27
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    As he said, "try taking...". So, take your simplest, easiest group to work with. Say, $C_2$...Done.2011-08-27
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    @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.2011-08-27
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    @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.2011-08-27
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    @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).2011-08-28
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    Ah, sorry, I had forgotten what the question asked!2011-08-28
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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.

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    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 $G$?2011-08-28
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    @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*).2011-08-28
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    (Oh, this is the same as Geoff's earlier comment on Ted's answer.)2011-08-28