Question Let $G$ be a countably infinite reduced abelian $p$-group. Is it always possible to write it has an infinite direct sums of non-trivial groups? If it is not true, how far is $G$ from being an infinite direct sum?

On one hand, by Prüfer's second theorem, the answer is true if $G$ has no elements of infinite height. On the other hand, if $G$ has an element of infinite heigth and $G\cong\bigoplus_{i\geq 1} L_i$, at least one of the $L_i$ has an element of infinite height (and is decomposable).

Now, any reduced $G$ with an element of infinite height can be factored has $G= G_1\times H_1$ with $H_1$ cyclic. In fact, on can take $H_1=C_{p^n}$ whenever $G$ has an element of order $p$ and height $n-1$. Since $G_1$ contains an element of infinite height, we can repeat this procedure. That is, for all $j\geq 1$ we have a decomposition $$G=G_j\oplus\bigoplus_{i=1}^jH_i.$$ Let $K$ be the intersection of the decreasing sequence $(G_i)_{i\geq 1}$. Then $G$ contains the direct sum $K\oplus\bigoplus_{i\geq1}H_i$. If the $H_i$ have bounded orders, $\bigoplus_{i\geq1}H_i$ is a direct factor of $G$ and we are done. We can assume that this is the case as soon as $G$ has infinitely many elements of order $p$ and of bounded height.

We also have an obvious map $\varphi\colon G\to\prod_{i\geq 1}H_i$ with kernel $K$. So one can hope that $G$ sits somewhere between $K\oplus\bigoplus_{i\geq1}H_i$ and $K\oplus\prod_{i\geq1}H_i$. However, in general it is not true that $\varphi(G)=\bigoplus_{i\geq1}H_i$ and it is unclear if $1\to K\to G\to\varphi(G)\to 1$ splits.

I claim that morally it is enough to understand what happens for the following classical example of a reduced group with an element of infinite height: $$A=\langle x_0,x_1,\dots\ |\ px_0=0,p^ix_i=x_0,[x_i,x_j]=0\rangle$$.

Indeed, if $G$ is reduced and has an element of infinite height, it has an element $a$ of infinite height such that if $p^nb=a$ with $n\geq 1$, then $b$ has finite height (as otherwise one can find a divisible subgroup). Intuitively, this $a$ will be our $x_0$. For any $b$ with $pb=a$ there exists $c$ with $pc=a$ and $h(c)>h(b)=k$. Hence $b-c$ has order $p$ and height $h(b)$. Such an element gives us a direct cyclic summand $C_{p^{k+1}}$. If for a given $k$ we have infinitely many $b_i$ with $pb_i=a$ and $h(b_i)=k$, then $G$ contains an infinite direct sum of copies $C_{p^{k+1}}$, which is a direct summand (as the orders are bounded). Hence, we can assume that for every $k$ there is only finitely many $b$ with $pb=a$ and $h(b)=k$. The group $A$ is the prototipical example of such a behaviour.


1 Answer 1


Let $G$ have an element of infinite height. By Kaplansky's Theorem (Fuchs (2015) p 300) there are infinitely many $k\in\mathbb N$ such that $G$ has an element $b$ with indicator $(0,\dots, k,\infty)$.

  • $\begingroup$ Could you please explain why this is enough to conclude? As far as I understand, such a $b$ gives us a cyclic direct factor $C_{p^{k+1}}$, but I don't see how this is enough to obtain an infinite direct sum. $\endgroup$
    – PHL
    May 12, 2022 at 6:24
  • 1
    $\begingroup$ You are right, this is not enough. In fact a countable reduced abelian $p$--group is not necessarily an infinite direct sum. Counterexample: Pruefer's group $G=\langle a_i\colon i\in \mathbb N\rangle$ with $pa_0=0= p^ia_{i-1}$ for all $i$. Then $G$ is not $\Sigma$--cyclic but $G/\langle a_0\rangle$ is. $\endgroup$ May 12, 2022 at 8:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.