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Are the equivalent class of split extension of $G$ by $K$ really in one to one correspondence with homomorphisms $G \to \mathrm{Aut}(K)$? When I am trying to prove it, I find it may be not the case. I only know that $$1\to K\to K\rtimes_{\rho_1}G\to G\to 1 \quad\text{and}\quad 1\to K\to K\rtimes_{\rho_2}G\to G\to 1$$

are equivalent if and only if there is a nonabelian 1-cocycle $\beta:G\to K$ such that $\rho_1=\mathrm{Ad}_{\beta}\circ \rho_2$. When $\rho_1=\rho_2$, $\beta$ is an abelian 1-cocycle. Thus, the automrophism group of $1\to K\to K\rtimes_{\rho}G\to G\to 1$is isomorphic to $Z^{1}_{\rho}(G,C_K)$.

Mike Pierce
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Xuexing Lu
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1 Answers1

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A split extension may indeed be equivalent to several different extensions with middle term a semi-direct product.

Mariano Suárez-Álvarez
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    For example, there are semi-direct products in which the action is non-trivial, and yet the group is isomorphic to a direct product. This happens if you let $H$ act on $G$ by inner automorphisms of $G$: if $H$ acts through a map $f:G \rightarrow H$, then the isomorphism $G \rtimes H \rightarrow G \times H$ sends $(g,h)$ to $(g f(h), h)$. – Dan Ramras May 04 '10 at 17:41