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Please accept my apologies for another elementary group theory question.

Let $$N\hookrightarrow E \twoheadrightarrow G$$ be a group extension such that the induced outer action $\psi\colon\thinspace G\to \mathrm{Out}(N)$ is trivial. Is it necessarily true that $N$ is central in $E$?

I have read comments on this site to this effect, but have been unable to come up with a proof. All I am seeing is that the conjugation action $\psi\colon\thinspace E\to \mathrm{Aut}(N)$ has image in the group $\mathrm{Inn}(N)$ of inner automorphisms.

On the other hand, work of Eilenberg and Mac Lane (as summarized in Brown's book "Cohomology of Groups", Section IV.6) shows that extensions as above are classified by $H^2(G;C)$, where $C$ denotes the centre of $N$ regarded as a trivial $G$-module, and therefore by central extensions of the form $$C\hookrightarrow A \twoheadrightarrow G.$$ This seems to suggest that the above statement is indeed true.

Edit: Thanks very much for the answer and comments so far, which have been very enlightening. In particular, comments of HW and others show clearly that the answer to my question as stated is "No". I have a follow-up question, which I'd like to ask here rather than start a new thread.

It is stated as Theorem 15.21 (3) in Peter Michor's book (linked in his answer below) that there is a split extension inducing a given outer action $\psi\colon\thinspace G\to\mathrm{Out}(N)$ if and only if $\psi$ lifts through the epimorphism $\mathrm{Aut}(N)\to\mathrm{Out}(N)$. This I agree with (and can even prove!). However, it seems to me that non-equivalent extensions can induce the same outer action (if $H^2(G;C)\neq 0$). My question is then:

Do there exist non-split extensions $$ N\hookrightarrow E \twoheadrightarrow G $$ such that the induced outer action $\psi\colon\thinspace G\to \mathrm{Out}(N)$ is trivial? Does anybody know any "natural" examples?

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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. –  HJRW May 2 '13 at 12:25
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I don't see any reason why $N$ should be abelian unless you impose some additional conditions. –  Colin Reid May 2 '13 at 12:29
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If the outer action is trivial then the centre of $N$ is indeed central in $E$. –  HJRW May 2 '13 at 12:32
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The assumptions are equivalent to $E = NC_E(N)$. –  Derek Holt May 2 '13 at 15:45
    
For your second question, consider for instance $1\to 2\mathbb{Z}\to\mathbb{Z}\to\mathbb{Z}/2\to 1$. –  HJRW May 3 '13 at 19:48
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1 Answer

Edit: Diagram added, more details added.

See 15.21 in pages 177-190 of here, where I collected the results on extensions of groups and Lie groups that I could find. 15.24 summarizes your situation quite clearly: We have $\text{Inn}(N)= N/Z(N)$ where $Z(N)$ is the center of $N$. Then you have a mapping of extensions $$ \begin{array}{ccccc} Z(N) & \xrightarrow{i|_{Z(N)}} & E & \xrightarrow{\theta} & G\times\text{Inn}(N) \newline \downarrow & & \downarrow & & \downarrow \newline N & \xrightarrow{i} & E & \xrightarrow{p} & G \end{array} $$ where the down arrows are inclusion, identity, and first projection, and where $\theta(x)=(p(x),\text{Conj}_x|_N)$. The first line is a central extension since $G\times \text{Inn}(N)$ acts trivially on $Z(N)$.

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This answer and the reference to your book are hugely helpful, thanks. Could you clarify how the central extension is "described by the cohomology class in $H^2(G,Z(N))$"? Are you pulling back said cohomology class using the projection map $G\times\mathrm{Inn}(N)\to G$? –  Mark Grant May 3 '13 at 6:18
    
It is not so simple, since it also involves the central extension $Z(N)\to N\to \text{Inn}(N)$. I did not work it out. –  Peter Michor May 3 '13 at 7:05
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