Let $G$ be a group such $H$ is a normal subgroup of $G$ and $Z(H)=1$ and $Inn(H)=Aut(H)$. Then prove there exists a normal subgroup $K$ of $G$ such that $G=H\times K$.
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Assuming you have the book An Introduction to the Theory of Groups by J.J.Rotman. Please have a look at the following Theorem. He greatly constructed the normal subgroup you are looking for. I suggest you to read the $N/C$ lemma there as well. This lemma is a key-like tool for going forward during the theorem. Theorem 7.15: If $K\vartriangleleft G$ and $K$ is complete, then $K$ is a direct factor of $G$; that is, there is a normal subgroup $Q$ of $G$ with $G=K\times Q$. |
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First, define the map $\psi: G \to Inn(G)$ by $\psi_g(x) = gxg^{-1}$. Since $H$ is a normal subgroup, $\psi_g(H) = H$, so $Inn(G) = Aut(H) = Inn(H)$. From $Z(H) = 1$, we know that $Inn(H) \cong H$. Compose this isomorphism with $\psi$ to get a surjective map $G \to H$. The kernel of this map is $K$. |
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