# groups with abelian inner automorphism group

The inner automorphism group $\operatorname{Inn}(G)$ is isomorphic to the central quotient of $G$, $G/Z(G)$. if $\operatorname{Inn}(G)$ is abelian, one cannot conclude that $G$ itself is abelian. For example, let $G=\mathcal{D}_{8}$, the dihedral group of symmetries of the square.

 $G=\langle r,s\mid r^{4}=s^{2}=1,rs=sr^{3}\rangle$

and $Z(G)=\{1,r^{2}\}$. Representatives of the cosets of $Z(G)$ are $\{1,r,s,rs\}$; these define a group of order $4$ that is isomorphic to the Klein $4$-group (http://planetmath.org/Klein4Group) $V_{4}$. Thus the central quotient is abelian, but the group itself is not.

However, if the central quotient is cyclic, it does follow that $G$ is abelian. For, choose a representative $x$ in $G$ of a generator for $G/Z(G)$. Each element of $G$ is thus of the form $x^{a}z$ for $z\in Z(G)$. So given $g,h\in G$,

 $gh=x^{a_{1}}z_{1}x^{a_{2}}z_{2}=x^{a_{1}}x^{a_{2}}z_{1}z_{2}=x^{a_{1}+a_{2}}z_% {1}z_{2}=x^{a_{2}}x^{a_{1}}z_{2}z_{1}=x^{a_{2}}z_{2}x^{a_{1}}z_{1}=hg$

where the various manipulations are justified by the fact that the $z_{i}\in Z(G)$ and that powers of $x$ commute with themselves.

Finally, note that if $\operatorname{Inn}(G)$ is non-trivial, then $G$ is nonabelian, for $\operatorname{Inn}(G)$ nontrivial implies that for some $g\in G$, conjugation by $g$ is not the identity, so there is some element of $G$ with which $g$ does not commute. So by the above argument, $\operatorname{Inn}(G)$, if non-trivial, cannot be cyclic (else $G$ would be abelian).

Title groups with abelian inner automorphism group GroupsWithAbelianInnerAutomorphismGroup 2013-03-22 17:25:30 2013-03-22 17:25:30 rm50 (10146) rm50 (10146) 7 rm50 (10146) Topic msc 20A05