# Does there exist an abelian $2$-group of finite exponent that is not a direct sum of cyclic groups?

Does there exist an abelian $2$-group (an abelian group, all of whose elements have order a power $2$) of finite exponent that is not isomorphic to a direct sum of $2$-cyclic groups?

The exponent of G , denoted expG , is the smallest positive integer $m$ such that, for every $g\in G$ , $g^m=e_{G}$.

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Please try to make your question self-contained, so that we know what you're asking without referring back to the title. And try to make it clearer too. – Alex Becker Jan 19 '12 at 4:47
What exactly do you mean by "2-group"? – Alon Amit Jan 19 '12 at 4:51
I am guessing that $2$-group means a group such that every element has order a power of $2$ (in recent questions Ali has asked about $p$-groups with this meaning). I am confused on "$2$-cyclic". Do you mean "cyclic $2$-groups"? Then you would be looking for counterexamples that are not finitely generated. (But they don't exist! See Arturo's answer.) – Jonas Meyer Jan 19 '12 at 4:56
Please try to make the posts self-contained. Your post right now does not contain any questions, just a definition. The subject should not be an integral part of the post, without which it makes no sense, just like you shouldn't have to read the title of a book on the spine in order to understand the book. – Arturo Magidin Jan 19 '12 at 5:06

If by "$2$-cyclic groups" you mean "cyclic group whose order is a power of $2$", then the answer is "no." It is a theorem of Prufer and of Baer that an abelian group of bounded exponent is necessarily isomorphic to a direct sum of cyclic groups (at least, in standard set theory with the Axiom of Choice; since there are models of set theory without AC in which there are vector spaces over $\mathbf{F}_2$ without a basis, as shown by Asaf Karagila in this question, and such a vector space would be an abelian $2$-group of bounded exponent that cannot be realized as isomorphic to a direct sum of cyclic $2$-groups, since such a realization would immediately provide us with a basis).

See this previous question for some discussion on the structure of abelian groups that are not necessarily finitely generated.

Here's a proof of the result of Prufer and Baer; it uses the Axiom of Choice rather fundamentally. It is very similar to the argument used in the finite abelian $p$-group case.

Since a torsion abelian group is isomorphic to the direct sum of its $p$-parts, it suffices to handle the case of a $p$-group. Let $G$ be an abelian $p$-group of bounded exponent, say $p^n$. We proceed by induction on $n$.

If $G$ is of exponent $p$, then $G$ is a vector space over $\mathbb{F}_p$, hence it has a basis (AC); the basis affords a representation of $G$ as a direct sum of cyclic groups of order $p$, and we are done.

Assume the result holds for abelian groups of exponent $p^n$, and say $G$ has exponent $p^{n+1}$. Let $H=G^p$. Then $H$ is an abelian group of exponent $p^n$, so there exist $\{h_i\}_{i\in I}$ elements of $H$, independent, such that $H$ is the (internal) direct sum of the cyclic subgroups $\langle h_i\rangle$. For each $i$, let $g_i\in G$ be such that $g_i^p = h_i$ (AC again). I claim that the $g_i$ are also independent. Indeed, if some product $g_{i_1}^{a_1}\cdots g_{i_m}^{a_m}$ is trivial, then raising it to the $p$th power we obtain $h_{i_1}^{a_1}\cdots h_{i_m}^{a_m}=1$; since the $h_i$ are independent, this means $p|a_i$ for each $i$, so writing $a_i=pb_i$, we get $1 = g_{i_1}^{a_1}\cdots g_{i_m}^{a_m} = h_{i_1}^{b_1}\cdots h_{i_m}^{b_m}$, and by the independent of the $b_i$ we get $g_{i_j}^{a_j}=h_{i_j}^{b_j}=1$ for each $j$. Thus, $\langle g_i\rangle\cong \oplus \langle g_i\rangle$.

Now let $K=\mathrm{ker}(f)$, where $f\colon G\to G$ is $f(g)=g^p$. Then $K$ is a vector space over $\mathbf{F}_p$; if we let $M=K\cap \langle g_i\rangle$, then there is a complement $N$ to $M$ in $K$ (AC again). Let $\{k_j\}_{j\in J}$ be a basis for $N$ (AC again). Then $N$ is a direct sum of the cyclic groups $\langle k_j\rangle$ of order $p$.

I claim that $G=N\oplus \langle g_i\rangle$. Indeed, $\langle g_i\rangle\cap N =\langle g_i\rangle \cap (K\cap N) = (\langle g_i\rangle \cap K)\cap N = M\cap N=\{1\}$; and if $g\in G$, then $g^p\in H$, so we can write $g^p = h_{i_1}^{a_1}\cdots h_{i_m}^{a_m}$ for some $h_i$; let $x = g_{i_1}^{a_1}\cdots g_{i_m}^{a_m}\in\langle g_i\rangle$. Then $gx^{-1}$ lies in $K$, since $(gx^{-1})^p = g^px^{-p} = 1$; then $gx^{-1}$ can be written as $mn$, with $m\in M=K\cap\langle g_i\rangle$, $n\in N$. Hence $$g = (gx^{-1})x = (mn)x = n(mx)\in N\langle g_i\rangle.$$ Thus, $N\cap\langle g_i\rangle = \{1\}$ and $N\langle g_i\rangle = G$, so $G=N\oplus \langle g_i\rangle$. Since each of $N$ and $\langle g_i\rangle$ are direct sums of cyclic groups, so is $G$. $\Box$

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What book or paper do i can see "theorem of Prufer and of Baer that an abelian group of bounded exponent is necessarily isomorphic to a direct sum of cyclic groups" thank you very much – Ali Gholamian Jan 19 '12 at 5:07
@AliGholamian: As noted in the answer I linked to, you can find the material in Rotman's "Introduction to the Theory of Groups", 4th Edition, Chapter 10. – Arturo Magidin Jan 19 '12 at 5:16
Thank you very much – Ali Gholamian Jan 19 '12 at 5:23
+1: An exciting conclusion. Admittedly I find it disturbing that a result in pure algebra depends on the choice of set theory. I just want to "believe in naïve set theory, Russell's paradox and Zorn's lemma" (too much algebra depends on Zorn's lemma for me to be comfortable without:-) – Jyrki Lahtonen Jan 19 '12 at 16:20