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
Canonical name	GroupsWithAbelianInnerAutomorphismGroup
Date of creation	2013-03-22 17:25:30
Last modified on	2013-03-22 17:25:30
Owner	rm50 (10146)
Last modified by	rm50 (10146)
Numerical id	7
Author	rm50 (10146)
Entry type	Topic
Classification	msc 20A05