# groups with abelian inner automorphism group

The inner automorphism group $\mathrm{Inn}(G)$ is isomorphic^{} to the central quotient of $G$, $G/Z(G)$. if $\mathrm{Inn}(G)$ 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.

$$G=\u27e8r,s\mid {r}^{4}={s}^{2}=1,rs=s{r}^{3}\u27e9$$ |

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 $\mathrm{Inn}(G)$ is non-trivial, then $G$ is nonabelian^{}, for $\mathrm{Inn}(G)$ 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, $\mathrm{Inn}(G)$, 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 |