# Is every group the semidirect product of its center and inner automorphism group?

For every group $$G$$, we have $$G/Z(G)\simeq \operatorname{Inn}(G).$$

I wonder whether the quotient projection has a right inverse. I suspect it doesn’t have one in general. But I can’t find a counter example. Thanks in advance.

## 2 Answers

Take $$G=\operatorname{SL}_2(\mathbb{Z})$$. Then $$Z(G)=\left\{\begin{pmatrix} 1&0\\0&1\end{pmatrix}, \begin{pmatrix} -1&0\\0&-1\end{pmatrix}\right\}.$$ If there was a right inverse, this would mean that there is a subgroup $$H$$ of $$G$$ such that for all $$A\in G$$ exactly one of $$A$$ or $$-A$$ is in $$H$$ ($$H$$ would be the image of the alleged right inverse). In particular, since $$\begin{pmatrix} 1&0\\0&1\end{pmatrix}\in H$$ we have $$\begin{pmatrix} -1&0\\0&-1\end{pmatrix}\notin H$$. Since $$(-A)^2=A^2$$, this means that $$A^2\in H$$ for all $$A\in G$$. But $$\begin{pmatrix} 0&-1\\1&0\end{pmatrix}\begin{pmatrix} 0&-1\\1&0\end{pmatrix}=\begin{pmatrix} -1&0\\0&-1\end{pmatrix} \in H$$, a contradiction.

Suppose that a group $$G$$ is an internal semidirect product $$Z(G) \rtimes H$$ for some subgroup $$H$$ of $$G$$. Then, since every element of $$Z(G)$$ commutes with every element of $$H$$, this semidirect product must in fact be a direct product, so $$Z(G)$$ must in fact be a direct factor of $$G$$.

But not every group has its center as a direct factor. For example, the center $$Z(Q_8)=\{\pm{1}\}$$ of the quaternion group $$Q_8$$ is not a direct factor.

• Even more so, every non-Abelian $p$-group is a counterexample for the same reasons. Commented Jun 17 at 0:14