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 $G/N\cong H$?
 A: 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 $N$.
A: 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 C∩N=1 and C and N centralize each other.  In particular, C ≅ CN/N ≤ G/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 N ≅ G/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.
