Mathematical Reflections 2011, Issue 5 - Problem U206

Problem:
Prove that there is precisely one group with $30$ elements and $8$ automorphisms.

Proposed by Gabriel Dospinescu.

Solution:
Let $G$ be a group such that $|G|=30$ and $|\textrm{Aut}(G)|=8$. Since $\textrm{Inn}(G)$ is a subgroup of $\textrm{Aut}(G)$ and $G/Z(G) \simeq \textrm{Inn}(G)$, then by Lagrange's Theorem $|\textrm{Inn}(G)|$ divides $|G|$ ($Z(G)$ is a normal subgroup of $G$) and $|\textrm{Aut}(G)|$, so $|\textrm{Inn}(G)|=1,2$. In any case $\textrm{Inn}(G)$ is isomorphic to a cyclic group, so $G/Z(G)$ is isomorphic to a cyclic group and so $G$ is abelian. By the Fundamental Theorem of Finitely Generated Abelian Groups, there is only one abelian group of order $30$ and this is $\mathbb{Z}_{30}$.