Let $G$ be a finite abelian $p$-group, $p$ a prime. I say that a pair $(G',\varphi)$ is a maximal cyclic quotient (please excuse me if this definition already exists and refers to a different concept) of $G$ if $G'$ is a cyclic group and $\varphi\colon G\to G'$ is a surjective map with the following property: if $H\leq \ker\varphi$ is a subgroup such that $G/H$ is cyclic, then $H=\ker\varphi$.

My first question is the following: is it true that if $(G',\varphi)$ is a maximal cyclic quotient of $G$ then $G\simeq G'\times\ker\varphi$?

Let now $G=\prod_iG_i$ be the decomposition of $G$ into cyclic $p$-groups and let $H$ be a subgroup of $G$. Does there exist an automorphism of $G$ which maps $H$ to a product $\prod_iH_i$ with $H_i\leq G_i$ for all $i$? An affirmative answer to this question would imply an affirmative answer to the first one, but I have no clue about its truth; it sounds as a pretty strong statement indeed...


Let $G = C_p \times C_{p^3} = \langle x \rangle \times \langle y \rangle$, and let $H = \langle z\rangle$, with $z = xy^p$. Then no automorphism of $G$ can map $H$ into one of the direct factors of $G$. Since $|H|=p^2$, it would have to map it into the $C_{p^3}$ factor, and hence to $\langle y^p \rangle$. But $z$ has order $p^2$ and is not the $p$-th power of an element of $G$ of order $p^3$, so $z$ cannot be mapped to $y^p$, or to any other element of $\langle y \rangle$ of order $p^2$, because these elements are all $p$-th powers. That answers the second question in the negative.

Also $G/H$ is a maximal cyclic quotient of $G$: it is cyclic of order $p^2$, but $H/\langle z^p \rangle = G/\langle y^{p^2} \rangle$ is not cyclic. But $H$ is clearly not a direct factor of $G$. So the answer to the first question is also no.

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  • $\begingroup$ Thank you for the answer! Do you also think that it is possible to find a counterexample where $H$ and $G$ have the same number of factors in their direct product decomposition? $\endgroup$ – Ferra Oct 7 '15 at 13:05
  • $\begingroup$ I don't see immediately how to do that. $\endgroup$ – Derek Holt Oct 7 '15 at 16:47

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