infinite abelian group where all elements have order 1, 2, or 4 Let $A$ be a (not necessarily finitely generated) abelian group where all elements have order 1, 2, or 4. Does it follow that $A$ can be written as a direct sum $(\bigoplus _\alpha \mathbb Z/4) \oplus (\bigoplus_\beta \mathbb Z/2)$?
 A: Call a set $S\subseteq A$ "good" if


*

*$S$ does not contain $a^2$ for any $a\in A$.

*Whenever $s_1^{m_1}s_2^{m_2}\cdots s_n^{m_n}=e$ where $s_1,s_2,\ldots,s_n$ are different elements of $S$, we have $s_1^{m_1}=e$.


Apply Zorn's lemma to the family of good subsets of $A$ (ordered by inclusion).
Show that a maximal good set corresponds to a direct sum as in the question.
A: What I said in the comments is true, and goes by the name
Prüfer's First Theorem: An abelian $p$-group $G$ with bounded exponent (an integer $k$ such that $g^k=1$ for all $g\in G$) is a direct sum of cyclic subgroups.
The proof is by induction on $k=p^e$, the base case $e=1$ being the vector-space case.
For the inductive step, write $pG=\oplus_\alpha\langle g_\alpha\rangle$, and choose $h_\alpha$ in $G$ with $ph_\alpha=g_\alpha$. Then the $h_\alpha$ generate a subgroup of $G$ (call it $H$) that is a direct sum of $\langle h_\alpha\rangle$. Let $L$ be a subgroup of $G$, maximal with respect to having trivial intersection with $H$. Then $L$ is also a direct sum of cyclic subgroups (by the vector-space case), and you can show $G=H\oplus L$.
Reference: Fundamentals of the Theory of Groups, by Kargapolov and Merzljakov, $\S$10.
A: Yes.  Note that such an abelian group is the same thing as a module over the ring $\mathbb{Z}/4$.  Note also that any direct sum of copies of $\mathbb{Z}/4$ is injective as a module over $\mathbb{Z}/4$ (this is easy to check by Baer's criterion, since the only nontrivial proper ideal in $\mathbb{Z}/4$ is $(2)$).  Now let $A$ be a $\mathbb{Z}/4$-module.  If $A$ has an element of order $4$, that element generates a submodule isomorphic to $\mathbb{Z}/4$.  Since $\mathbb{Z}/4$ is an injective module, $A$ splits as a direct sum $\mathbb{Z}/4\oplus A'$ for some submodule $A'\subset A$.  If $A'$ has an element of order $4$, we can split off a direct summand of $\mathbb{Z}/4$ from it, and so on.
Repeating this process by transfinite induction until there are no elements of order $4$ left, we can write $A$ as a direct sum $B\oplus C$ where $B$ is a direct sum of copies of $\mathbb{Z}/4$ and $C$ has no elements of order $4$.  But then every element of $C$ has order $1$ or $2$, so $C$ is a $\mathbb{Z}/2$-vector space.  Thus $C$ is a direct sum of copies of $\mathbb{Z}/2$, and $A=B\oplus C$ is the direct sum decomposition you ask for.
