Let $G$ be a finite abelian group and let $M(G)$ be the set of all elements of $G$ that fix with any automorphism of $G$. Then prove

$$M(G)=\langle1\rangle \text{ or } Z_{2}$$

Attempt: We know that $M(G)$ is elementary abelian $2$-group.

Also if $G$ is an abelian group of finite odd order, then $M(G)=1$.

  • 2
    $\begingroup$ So you can assume $G$ is a 2-group and a direct product of cyclic groups. Note that $|M(G)|=2$ if and only if there is unique cyclic direct factor of maximal order. $\endgroup$ – Derek Holt Apr 12 '13 at 8:18

It is a good idea to prove this for the cyclic groups $\mathbb Z / p^n \mathbb Z$, then to use the decomposition of finite abelian groups.


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