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May 5, 2013 at 6:18 comment added Mark Grant I guess my stumbling block is I don't yet know exactly how to pass between extensions $N\to E\to G$ with a given induced outer action and central extensions $C\to A\to G$.
May 5, 2013 at 6:17 comment added Mark Grant @HW (in response to your second-to-last comment): Is it so? If $N$ is non-abelian, the equivalence classes of extensions with kernel $N$ and trivial outer action are non-canonically isomorphic to $H^2(G;C)$. We know that one such extension is split (indeed, the trivial extension is amongst them). Could it be that they all split?
May 3, 2013 at 19:50 comment added HJRW (Where the action of $G$ on $C$ is trivial.)
May 3, 2013 at 19:49 comment added HJRW Indeed, I think you answered your own question: there are such examples whenever $H^2(G,C)\neq 0$.
May 3, 2013 at 19:48 comment added HJRW For your second question, consider for instance $1\to 2\mathbb{Z}\to\mathbb{Z}\to\mathbb{Z}/2\to 1$.
May 3, 2013 at 14:20 history edited Mark Grant CC BY-SA 3.0
Added follow-up question
May 2, 2013 at 18:35 answer added Peter Michor timeline score: 1
May 2, 2013 at 15:45 comment added Derek Holt The assumptions are equivalent to $E = NC_E(N)$.
May 2, 2013 at 12:32 comment added HJRW If the outer action is trivial then the centre of $N$ is indeed central in $E$.
May 2, 2013 at 12:29 comment added Colin Reid I don't see any reason why $N$ should be abelian unless you impose some additional conditions.
May 2, 2013 at 12:25 comment added HJRW If $N$ is non-abelian, then it can't be central in $E$. On the other hand, $E=G\times N$ has the property that the outer action of $G$ is trivial.
May 2, 2013 at 12:22 history asked Mark Grant CC BY-SA 3.0