16
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Steve D Aug 15 '17 at 21:01
  • $\begingroup$ What happens if you take the intersection of the subgroups containing all the elements of order $4$?. $\endgroup$ – Mark Bennet Aug 15 '17 at 21:07
12
$\begingroup$

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.

$\endgroup$
11
$\begingroup$

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.

$\endgroup$
6
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 $e$? $\endgroup$ – user46652 Aug 16 '17 at 15:38
  • $\begingroup$ @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. $\endgroup$ – Henning Makholm Aug 16 '17 at 21:15
  • $\begingroup$ thanks! I'll see if there's a systematic way to update such choices of $x$ to make it work. $\endgroup$ – user46652 Aug 16 '17 at 22:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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