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Let $p$ be an odd prime, and $G$ an abelian group of order $p^3$ which is not cyclic. The $p$th power map $$\phi:G\to G,\qquad g\mapsto g^p$$ is a group homomorphism. Is it true that $\ker\phi$ has order $p^2$ or $p^3$ ?

This is equivalent to asking whether $\operatorname{im}\phi$ has order $1$ or $p$. Since $G$ is not cyclic, every nonidentity element of $G$ has order $p$ or $p^2$, so every nonidentity element of $\operatorname{im}\phi$ has order $p$. But this does not imply that $\operatorname{im}\phi$ has order $p$, so I'm stuck.

Note: I can show this when $G$ is nonabelian.

Note: I would prefer a proof which does not use the fundamental theorem of finitely generated abelian groups.

ashpool
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  • It’s certainly possible that the image is trivial. Thus two cases for the image, ${e}$ or cyclic of order $p$. And that’s it. – Lubin Jul 18 '20 at 02:05
  • @Lubin If the image is not trivial, why does it have order $p$? – ashpool Jul 18 '20 at 02:09
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    Well, it’s so much easier to talk about the kernel of the $p$-th power map, as the problem was set up. You’re counting the elements killed by $p$, so to speak. Could be all, could be $p^2$, but can’t be $p$, ’cause that would mean that the original group was cyclic. Can’t be trivial ’cause you’re in a $p$-group, and every element is of order $p^m$ for some $m$. – Lubin Jul 18 '20 at 02:14
  • @Lubin Why does $|\ker|=p$ imply that the original group is cyclic? – ashpool Jul 18 '20 at 02:25
  • https://math.stackexchange.com/q/1132336/688539 – Clemens Bartholdy Mar 10 '22 at 00:47

3 Answers3

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Your group is $C_p\times C_{p^2}$ or $C_p\times C_p\times C_p$ by the Fundamental theorem about finitely generated abelian groups. In the first case the kernel is of size $p^2$, in the second case it is $p^3$.

markvs
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  • Agree totally, but in my comment I was trying to explain without reliance on the Theorem. – Lubin Jul 18 '20 at 02:15
  • Without that you need to explain that if the image is of size $p$, then the group is cyclic. – markvs Jul 18 '20 at 02:16
  • "kernel", not "image" in the comment. – markvs Jul 18 '20 at 02:27
  • @JCAA You meant that if the image is of size $p^2$, then the group is cyclic? – ashpool Jul 18 '20 at 02:27
  • Yes, the kernel is of size $p$. – markvs Jul 18 '20 at 02:28
  • @JCAA And can you explain that without the theorem? – ashpool Jul 18 '20 at 02:32
  • Some non-trivial properties of abelian groups should be used anyway (like the fact that if every divisor of the order of the group is the size of a subgroup then the group is cyclic). But what is the point of doing this? – markvs Jul 18 '20 at 02:37
  • I'm just a little reluctant to use a fundamental theorem to answer such a (seemingly) simple question... – ashpool Jul 18 '20 at 02:39
  • The question is not simpler than a description of all abelian groups of order $p^3$. You need to consider cases if there are cyclic subgroups of order $p^2$ or not. In the second case the kernel is of size $p^3$. In the first case you need to show, for example, that this cyclic subgroup is a direct factor. This is true, but not obvious. – markvs Jul 18 '20 at 02:43
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By Cauchy's theorem, there is an element of order $p$ .Call it $x$ (this means the kernel can't be trivial)

if $x,...,x^{p-1}$ are the only elements of order $p$ in $G$, then, any other non-identity element in $G$ has to be of order $p^{2}$. But, if $y\in G$ and has order $p^{2}$, then, $|y^{p}|=p$ So, you can show that there can't be exactly one subgroup of order $p$ thus $|ker\phi|\neq p$

For the remaining cases, I guess you have to use the Fundamental theorem of finitely generated abelian groups show that the cases for $p^2$ and $p^3$ are valid.

Citizen313
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This answer is inspired and guided by (and also copied from) everyone who helped me: Lubin, JCAA, and M.darwich.

By Cauchy's theorem (or, as Lubin mentioned, by the fact that $G$ is a $p$-group), $G$ has an element of order $p$, so $\ker\phi\neq1$.

Suppose that $|\ker\phi|=p$. Then, as M.darwish noted, every element of $G\backslash\ker\phi$ has order $p^2$, so if $x\in G\backslash\ker\phi$, then $|x^p|=p$. It follows that $x^p\in\ker\phi$ for all $x\in G$, i.e., $\operatorname{Im}\phi\subseteq\ker\phi$. Hence, $|\operatorname{Im}\phi|\leq|\ker\phi|$. Now, $$|\operatorname{Im}\phi|=\frac{|G|}{|\ker\phi|}=\frac{p^3}{p}=p^2>p=|\ker\phi|,$$ a contradiction. Hence, $|\ker\phi|\neq p$.

ashpool
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