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.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

If $G$ is a finite abelian group such that $o(x)=2$ for all $x \neq e$ and $|G|=2^n$ for some $n\in\mathbb N$, prove that $G \cong \mathbb{Z}_2\times\cdots\times\mathbb{Z}_2$ ($n$ factors).

Any help is appreciated. I asked this question before and was suggested to try induction but I haven't been able to write down the inductive proof properly. If anybody can help that will be really nice. Thanks!

share|cite|improve this question
Do you know the fundamental theorem of finite abelian groups? It states that all finite abelian groups are products of cyclic groups. Your problem would follow from this fact fairly immediately. – Jared May 16 '13 at 21:21
@Jared: Overkill, we only need linear algebra. – Martin Brandenburg May 17 '13 at 0:34
@MartinBrandenburg: Could it be assumed that $G$ is a permutation group and then we proved the claim? I think we could. Thanks. – S. Snape May 17 '13 at 5:23
This is too easy. You can leave out "abelian" from the hypothesis, as any group all whose elements are involutions is abelian. – Marc van Leeuwen May 17 '13 at 10:23
@MarcvanLeeuwen You could also leave out $|G|=2^n$...methinks this is an introductory question. – user1729 May 17 '13 at 11:24
up vote 1 down vote accepted

The proposition is true for $n=1$. For $n\geq 2$, take $g_1 \in G$ and let $G_1=\langle g_1\rangle$. For $i=2,\dots, n$ take $g_i \in G \text{ \ } G_{i-1}$ where $G_{i-1}=\langle g_1,\dots,g_{i-1}\rangle$. By the induction hypothesis (or directly), for $i < n, G_i \cong \mathbb{Z}_2^i$. In particular $G_{n-1} \cong \mathbb{Z}_2^{n-1}$.

By construction $g_n \in G\text{ \ }G_{n-1}$, so $\langle g_n \rangle \cap G_{n-1}=\{e\}$, they are both normal since $G$ is abelian and every element in $G$ can be written as a product of elements from $\langle g_n\rangle$ and $G_{n-1}$. (You can see this by noting that the left cosets of $G_{n-1}$ are $G_{n-1}$ and $g_nG_{n-1}$ and partition the set.) Thus $G \cong \langle g_n \rangle \times G_{n-1} \cong \mathbb{Z}_2\times\mathbb{Z}_2^{n-1}\cong\mathbb{Z}_2^n$ as required.

This is a rather brute force approach, but there's only one idea behind it i.e. "keep pulling out $\mathbb{Z}_2$ until there's nothing left". The solution is much simpler if you know about modules (a generalisation of vector spaces) or by treating it as a linear algebra problem (as egreg does below), but there's nothing wrong with this elementary approach.

share|cite|improve this answer
thanks btw, u mean $<g_{n}>$ x $G_{n-1}$ in the last line right? – uh1 May 17 '13 at 2:58
@uh1 No problem. Also, yes I did, thank you, I just changed it. – Tom Oldfield May 17 '13 at 9:45

If you write $G$ in additive notation, it's fairly easy to check that you can make it into a vector space over the field $\mathbb{F}_2$ with two elements, where

$$ 0g=0,\quad 1g=g $$

Then, take a basis of $G$ as vector space: then $G$ becomes the direct sum of $n$ copies of $\mathbb{F}_2$, which, with respect to addition is none else than $\mathbb{Z}/2\mathbb{Z}$.

This can be of course generalizable to abelian groups $G$ such that $x^p=e$ for any $x\in G$, where $p$ is prime.

share|cite|improve this answer

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.