The order of the kernel of the $p$th power map of an abelian and noncyclic group of order $p^3$

Question

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.

Answer 1

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, $|\ker\phi|\neq p$.

Answer 2

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 $p^3$.

Answer 3

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.