This is a reference request for the following statement:

Fact: Let $G$ be a reduced abelian $p$-group with an element of order $p^2$. Then $G$ contains an element of order $p$ and of finite height at least $1$.

The proof of this uses only basics properties of the height function $h_p(-)$ (and the result is trivial if $G$ has no elements of infinite height). However, instead of writing down the proof, I would prefer to make a reference to some textbook or paper, as I'm sure that this fact is well-known to specialists of abelian groups.


1 Answer 1


This might be using a sledgehammer to crack a nut, but a theorem of Kaplansky quoted on p. 200 of L. Fuchs, Abelian Groups, Springer, 2015 states:

Let $A$ be an abelian $p$--group of length $\tau$, $\sigma_0 < \sigma_1 < \cdots < \sigma_{n-1} < \tau$ a sequence of ordinals, then $(\sigma_0, \dots, \sigma_{n-1}, \sigma_n = \infty)$ is the indicator of some $a\in A$ of order $p^n$ if and only if it satisfies the gap condition: if $\sigma_{i+1} > \sigma_i + 1$ then the $\sigma_i$ Ulm--Kaplansky invariant of $A$ is non--zewro.

  • $\begingroup$ Sorry, I meant p. 300 $\endgroup$ Commented May 11, 2022 at 1:07
  • $\begingroup$ This is indeed a big sledgehammer, as in my context it would propably be better to avoid to talk about the Ulm-Kaplansky inviriants. Nevertheless, this is still a reference and I will take a look at Fuchs' book. $\endgroup$
    – PHL
    Commented May 11, 2022 at 11:11

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