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A colleague recently mentioned a "Prüfer Decomposition Theorem", claiming that every abelian group $A$ could be expressed as the direct sum $A = T(A) \oplus F(A)$, where $T(A)$ is the torsion subgroup and $F(A)$ is the torsion-free subgroup (=A/T). Is this true?

I realize that unlike finite abelian groups, the classification of abelian groups is unfinished (as others note).

And while this decomposition easily holds for finitely generated abelian groups (which decompose like $A=\mathbb {Z} ^{n}\oplus \mathbb {Z} /q_{1}\mathbb {Z} \oplus \cdots \oplus \mathbb {Z} /q_{t}\mathbb {Z}$), and for divisible groups (which decompose to $A=\left(\bigoplus_{p\in {\mathbf {P} }}{\mathbb {Z} }[p^{\infty }]^{(I_{p})}\right)\oplus {\mathbb {Q} }^{(I)}$ ), Wikipedia at least claims it does not hold generally.

However, a recent question asked about verifying a proof of this, so I am suspicious.

Also, absent a complete classification, what can we say about the sorts of things A/T may be? For example, a torsion-free abelian group of finite rank $r$ is a subgroup of $\mathbb{Q}_r$, and there are also the p-adic integers $\mathbb{Z}_p$ of infinite rank.

Shaun
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alexchandel
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1 Answers1

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The linked question has an important additional assumption: that $A/T$ is free. In fact given a short exact sequence

$$0\to T\to A\to F\to 0$$

with $F$ free, we have that $A\simeq T\oplus F$. Regardless of what those groups are.

Is this true?

Your colleague is unfortunately wrong.

The counterexample is as follows: $A=\prod \mathbb{Z}_p$ over the set of all primes. Then $T(A)=\bigoplus\mathbb{Z}_p$. But there is no subgroup of $A$ isomorphic to $A/T(A)$. Because $A/T(A)$ is divisible (and nontrivial). However the equation $px=C$ has no solution in $A$, if $p$ is prime and $C$ has non-zero $p$-th coordinate.

freakish
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