Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $g\in G$ , $g^m=e_{G}$.

share|improve this question
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

1 Answer 1

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 $G$. $\Box$

share|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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