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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.

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2 Answers 2

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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.

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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.

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    $\begingroup$ Even more so, every non-Abelian $p$-group is a counterexample for the same reasons. $\endgroup$
    – Steve D
    Commented Jun 17 at 0:14

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