Let $G$ be a finitely generated group.
Assume for all $g\in G, g^2=e$.
Then, how do I show that $G$ is actually finite?
I don't know where to start..
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Hint: Show first that it is abelian and then use the classification theorem of finitely generated abelian groups. You will get that $G=(\mathbb{Z}/2\mathbb{Z})^n$ for some natural number $n$.
Edit: Use of the classification theorem is not necessary as pointed out in the comment below, since $G$ abelian implies that for $n$ generators we can have at most $2^n$ elements (every possibility of multiplying the generators amongst eachother, as soon as we use one twice, it becomes the identity).
J.R.
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6The classification theorem is maybe too big a gun for this. Once $G$ is shown to be abelian, one readily sees that $n$ generators can produce at most $2^n$ elements. – Hagen von Eitzen Oct 16 '14 at 15:43
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@HagenvonEitzen: Agreed. Thank you. – J.R. Oct 16 '14 at 15:44
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The key here is that $G$ is abelian. Once you show that, then if $g_1,...,g_n$ are the generators, every element of $G$ is of the form $g_1^{a_1}...g_n^{a_n}$.
Since each $g_i$ is of finite order there are finitely many distinct words of this form.
Zarrax
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Let $\Bbb K=\Bbb Z/\Bbb Z$. And define $\Bbb K\times G\to G$ by $(k,g)\mapsto g^k$. Hence $G$ is a $\Bbb K$ vector space with finite dimension . so $G$ is finite.
Hamou
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