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Every homomorphism $\varphi: K \to \mathrm{Out}(H)$ determines an unique extension of $H$ by $K$. Why is this true for groups $H$ with a trivial center?

Even if we only consider split extensions, as far as I understand, the inner automorphisms $\mathrm{Inn}(H)$, in any case, should set up an equivalence relation on elements in $\mathrm{Aut}(G)$, isn't it? That is, any $\varphi(k_1)$ should be equivalent to a $\varphi(k_2)$ if both $\varphi(k_1)$ and $\varphi(k_2)$ are inner automorphisms.

Moreover, $$H \rtimes_{\varphi(k_1)} K \cong H \rtimes_{\varphi(k_2)} K$$ iff $\varphi(k_1)$ and $\varphi(k_2)$ lie in the same coset of $\mathrm{Inn}(H)$ in $\mathrm{Aut}(H)$. I don't see why this should hold for groups $H$ with trivial center. Is there also a more general condition for when this is true?

  • Why do you think that it holds only for simple groups $H$? In fact it holds for all groups with $Z(H)=1$ (and note that I did not say that it holds only for those groups). – Derek Holt May 04 '20 at 08:40
  • @DerekHolt I just read it somewhere on this website (can't find the link now). Thanks, it's good to know it holds for all groups with $Z(H) = 1$. Could you explain why though? –  May 04 '20 at 08:43
  • By the way, the center of a simple group is either trivial or the whole group. So that doesn't seem to far from this statement either –  May 04 '20 at 08:45
  • I can't make much sense of what you have written about split extensions. An example when this does not hold is when $H$ and $K$ both have order 2, and $\varphi$ is the trivial homomorphism. In this case there are two extensions, the split extension isomorphic to $C_2 \times C_2$ and the nonsplit extension isomorphic to $C_4$. There are other (more complicated) examples where there is no such extension. – Derek Holt May 04 '20 at 08:50
  • Thanks for the example. Could just write a proof for your statement "in fact it holds for all groups with $Z(H) = 1$"? I think that'll suffice as an answer for now –  May 04 '20 at 08:56
  • This should be in Robinson's book. – the_fox May 04 '20 at 09:47
  • @the_fox It would be helpful if you tell me the chapter –  May 04 '20 at 09:49
  • Chapter $11$ "The Theory of Group Extensions". – the_fox May 04 '20 at 10:00

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Let $A = {\rm Aut}(H)$ and $I = {\rm Inn}(H)$, so ${\rm Out}(H) = A/I$. Assume that $Z(H)=1$. Then $I \cong H$.

Given $\varphi:K \to {\rm Out}(H)$, define $$E = \{ (k,a) \in K \times A : \varphi(k) = aI \}.$$ Then you can check that $E$ is a subgroup of $K \times A$.

Define $i:I \to E$ by $i(a) = (1,a)$ and $\rho:E \to K$ by $\rho((k,a)) = k$. Then you can check that

$$ 1 \to I \stackrel{i}\to E \stackrel{\rho}\to K \to 1$$

is a short exact sequence, so $E$ is an extension of $I \cong H$ by $K$. This is the unique extension referred to.

Derek Holt
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