Is there a finite abelian group $G$ such that $\textrm{Aut}(G)$ is abelian but $G$ is not cyclic?

Question

Is there an example in which $G$ is a finite abelian group and $\textrm{Aut}(G)$ is abelian but $G$ not cyclic?

Answer 1

If $G$ is finite abelian and not cyclic, it can be written as direct sum $G=G_1\oplus G_2\oplus\ldots \oplus G_n$ of $n\ge2$ nontrivial cyclic groups of orders dividing each other. Say $G_1\cong \mathbb Z/m\mathbb Z$ and $G_2\cong \mathbb Z/md\mathbb Z$. As $G_1\oplus G_2$ is a direct summand, $\operatorname{Aut}(G_1\oplus G_2)$ is a subgroup of $\operatorname{Aut}(G)$. The group $\mathbb Z/m\mathbb Z\oplus\mathbb Z/md\mathbb Z$ has especially the automorphisms $$\phi\colon (x+m\mathbb Z,y+md\mathbb Z)\mapsto (x+y+m\mathbb Z,y+md\mathbb Z)$$ and $$\psi\colon(x+m\mathbb Z,y+md\mathbb Z)\mapsto(x+\mathbb Z,y+dx+md\mathbb Z).$$ We have $\psi(\phi(m\mathbb Z,1+md\mathbb Z))=(1+m\mathbb Z,1+d+md\mathbb Z)$ and $\phi(\psi(m\mathbb Z,1+md\mathbb Z))=(1+m\mathbb Z,1+md\mathbb Z)$. As $m>1$, this shows that $\operatorname{Aut}(G_1\oplus G_2)$ is not abelian.