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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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Do you have an access to the book written by J.J.Rotman, maryam? – Babak S. Jan 18 '13 at 7:24
@ Babak Sorouh: Yes I have the book"An Introduction to the Theory of Groups" by Joseph J. Rotman. – maryam Jan 18 '13 at 7:28
What is the source for this problem? Sorry for asking but ,is it given to you or you found it in a book? – Babak S. Jan 18 '13 at 7:30
I found it in a book and think it is nice problem. For this wrote here. – maryam Jan 18 '13 at 7:37
up vote 3 down vote accepted

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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Helpful reference(s): the right tools for the task!+1 – amWhy Feb 14 '13 at 0:11

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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