# How far is a countably infinite reduced abelian $p$-group from being an infinite direct sum?

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.

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 $$(0,\dots, k,\infty)$$.
• 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.
• 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. May 12, 2022 at 8:40