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If G is a finite group such that $o(x)=2$ for all $x \neq e$, $\ $

(a) Prove that $|G|=2^n$ for some $n\in N$.

(b) Since G is abelian, use that to prove that

$G \cong \mathbb{Z_2}$ x $...$ x $\mathbb{Z_2}$ ($n $ factors)

Any help is appreciated. Thanks!

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Please clarify: are you given that $G$ is abelian, or is that something you need to prove? –  vadim123 May 4 '13 at 2:20
    
Presumably we can't use Cauchy's Theorem, but the proof in the abelian case is quite a bit easier than the general proof. –  André Nicolas May 4 '13 at 2:22
    
@vadim I read the linked question, and its answers, and the question simply asserts knowledge of the finite case: the question is asking about infinite groups...and if the conclusions about the finite case can be extended to the infinite direct product. So the question, though the titles look like duplicate, is not at all a duplicate. –  amWhy May 4 '13 at 2:42
    
uh1: do you know, or have you encountered, Cauchy's Theorem for groups? If not, then it's probably not admissible for a proof of $(a)$. If you have encountered it, then $(a)$ follows directly from that theorem. –  amWhy May 4 '13 at 2:55
    
thanks @amWhy! Got it! –  uh1 May 4 '13 at 3:09

2 Answers 2

up vote 2 down vote accepted

Let $a,b\in G$. Then $e=(ab)^2=abab=aba^{-1}b^{-1}\Rightarrow ba=ab$. So $G$ is abelian.

Since $G$ is abelian, $\langle a\rangle$ and $\langle b\rangle$ normalises each other. Note that $\langle a\rangle\cap\langle b\rangle=e$, so $\langle a,b\rangle\cong\langle a\rangle\times\langle b\rangle\cong\mathbb{Z}_2^2$. By induction you can show that $G\cong\mathbb{Z}_2^n$.

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thanks i was thinking along the same lines for (b). do u have any ideas for proving (a) without using the direct product proved in (b) –  uh1 May 4 '13 at 2:33
    
@uh1, do you know Cauchy's Theorem? Assume $|G|$ is not a 2-power, then there exists an odd prime dividing $|G|$, and so $G$ has an element of order an odd prime, a contradiction. –  Easy May 4 '13 at 2:36
    
thanks! appreciate your help –  uh1 May 4 '13 at 3:08
    
I wish I could do every group theory question with using GAP, Easy. @Easy: +1 –  Babak S. May 4 '13 at 4:26
    
@BabakS., if all group theory question can be done by GAP, then all those problems are in some sense solved, however this is not the situation...But thanks for your support! –  Easy May 4 '13 at 4:52

As has been noted, the condition that $o(x)=2$ for all nonidentity $x$ implies that $G$ is abelian. Being finite, we know then that $G$ is the sum of cyclic groups. Any summand that is not $\mathbb{Z}_2$ will contribute a nonidentity element of order larger than $2$.

We don't need to know $G$ is abelian to conclude that $|G|=2^n$. Cauchy's theorem for groups says that if another prime $p\ne 2$ divided $|G|$, there would be an element of order $p$, a contradiction.

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