# 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 $(\bigoplus _\alpha \mathbb Z/4) \oplus (\bigoplus_\beta \mathbb Z/2)$?

• I believe the answer is 'yes', and follows because the maximal order of an element in $A$ is bounded. But there might be some countability restrictions on $A$. When I'm back home tonight, I'll dig through some references. – Steve D Aug 15 '17 at 21:01
• What happens if you take the intersection of the subgroups containing all the elements of order $4$?. – Mark Bennet Aug 15 '17 at 21:07

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.

Yes. Note that such an abelian group is the same thing as a module over the ring $\mathbb{Z}/4$. Note also that the ring $\mathbb{Z}/4$ is injective as a module over itself (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.

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.

• Thanks, @Henning! I'm stuck at the final step. Let $S'$ be a maximal good subset, and let $A'$ be the subgroup it generates. Suppose to the contrary that there existed $x\not\in A'$. Without loss of generality we can assume $x \neq a^2$ for any $a$. How can I show that $S' \cup \{ x\}$ is a good subset? Specifically, given, say $s_1^{m_1} \cdots s_n^{m_n} x^2 = e$, how do I know each factor equals $e$? – user46652 Aug 16 '17 at 15:38
• @user46652: That appears to be a much better question than I had expected when I wrote down the plan. In fact I can't seem to find an argument that works. It is not always the case that $S\cup\{x\}$ is a good subset; for example if $A=\mathbb Z_4\times\mathbb Z_2$ and $S=\{(1,0)\}$ and $x=(1,1)$. But in that case we can find a different $x$ that does work: $S\cup\{(0,1)\}$ is a larger good subset. so $S=\{(1,0)\}$ was not maximal. – hmakholm left over Monica Aug 16 '17 at 21:15
• thanks! I'll see if there's a systematic way to update such choices of $x$ to make it work. – user46652 Aug 16 '17 at 22:51