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