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I want to show that

Let $G$ be an Abelian $p$-group of order $p^m$. If ${\rm Aut}(G)$ is Abelian, then $G$ is a cyclic $p$-group.

Any help would be appreciated!

PS: Actually, I was engaged in showing “if ${\rm Aut}(G)$ is Abelian, then the Abelian group $G$ is a cyclic group” without $G$ having prime power order.

I started it with induction, and I’ve finished half of the proof, I do it in this way:

Write $G$ as the product $G_{p_1}\times\cdots\times G_{p_n}$ (see here), where $p_i$ are the prime divisors of $|G|$. It’s rather easy to show the case $n$ under the hypothesis of $n-1$, considering ${\rm Aut}(G)= {\rm Aut}(G_{p_1})\times\cdots \times {\rm Aut}(G_{p_n})$.

All that remains to prove is the trivial case when $n=1$, which is my question.

  • If $G$ is a $p$-group, then its order $|G|$ has exactly one prime divisor, $p$. That is sort of the point! If $G$ is not a $p$-group, then in most cases it is not the direct product of $p$-subgroups for various $p$'s, the example of the symmetric group $S_3$ shows. (You are referring to this question https://math.stackexchange.com/questions/2709850/a-subgroup-of-abelian-group for that, but there you have the extra hypothesis that the group is abelian) – Mariano Suárez-Álvarez Mar 29 '18 at 07:56
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    If $G$ is an abelian $p$-group, then it is a direct product of $k$ cyclic groups, and it is clearly sufficient to prove that the automorphism group is nonabelian in the case $k=2$ - that is when $G = C_{p^a} \times C_{p^b}$ for some $a,b \ge 1$. You can do that by writing down specific automorphisms that do not commute. – Derek Holt Mar 29 '18 at 08:39

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As I said in my comment, it suffices to find two non-commuting automorphisms of $G=\langle x \rangle \times \langle y \rangle$, where $x$ and $y$ have orders $p^a$ and $p^b$ for some $a,b \ge 1$, and we may assume that $a \ge b$.

Define the automorphisms $\alpha$ and $\beta$ of $G$ by $$\alpha:x \mapsto xy, \, y \mapsto y,\ \ \ \ \beta:x \mapsto x, \, y \mapsto x^{p^{a-b}}y.$$

Then $\alpha\beta(x) = xy$, whereas $\beta\alpha(x)= x^{1+ p^{a-b}}y,$ so $\alpha\beta \ne \beta\alpha$.

Derek Holt
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