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Are the equivalent class of split extension of G by K really in one to one correspondence with homomorphisms from G to Aut(K)? When I am trying to proof it, I find it may be not the case. I only get that $1\to K\to K\rtimes_{\rho_1}G\to G\to 1$ and $1\to K\to K\rtimes_{\rho_2}G\to G\to 1$ are equivalent if and only if there is a nonabelian 1-cocycle $\beta:G\to K$ such that $\rho_1=Ad_{\beta}\circ \rho_2$. 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 $Z^{1}_{\rho}(G,C_K)$.

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

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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 $f:G\to H$, then the isomorphism $G\rtimes H \to G\times H$ sends (g,h) to (g f(h), h). – Dan Ramras May 4 '10 at 17:41

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