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Let $G$ a finite group and suppose that all non-identity elements has order 2. Show that $G$ is an Abelian group and it can be writed as a direct product of cyclic groups with order 2.

I've already red some answers to related questions, but I can't see how to prove the part of the direct product of cyclic groups, I understand that $o(G)=2^k$ but I don't know if all the finite groups has a generating set... Some hint or solution?

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    $\begingroup$ Let $H \le G$ be maximal such that $H$ is a direct product of groups of order $2$. Show that if $H \ne G$ then you can find a larger subgroup $H \times \langle z \rangle$ for $z \in G \setminus H$, contradiction. $\endgroup$
    – Derek Holt
    Commented Oct 11, 2016 at 9:18
  • $\begingroup$ Maybe you gonna laugh of me, but i quite don't understand what you mean with 'maximal', but another question is correct say that every group of order 2 are cyclic right? $\endgroup$
    – Ragnar1204
    Commented Oct 11, 2016 at 9:22
  • $\begingroup$ I am not going to laugh at you because there are two possible meanings of "maximal". It could mean eaither largest order, or maximal with respect to inclusion. But it doesn't matter which meaning you use - they will both work! Yes every group of order $2$ is cyclic - in fact every group of order $p$ is cyclic for any prime $p$. $\endgroup$
    – Derek Holt
    Commented Oct 11, 2016 at 9:52
  • $\begingroup$ There's always exist a subgroup like that? (thanks by the help, you make work my brain faster...) I'm thinking in construct the subgroup $H=\{e,g_1\}\{e,g_2\}\cdots\{e,g_n\}$ with the $n$ elements of $G$ you think that this going to work? $\endgroup$
    – Ragnar1204
    Commented Oct 11, 2016 at 9:56
  • $\begingroup$ The point is that you don't have to construct $H$ yourself. You just say let $H$ be a subgroup of largest possible order that is a direct product of groups of order $2$. So then you have $H$.. You then say, if $H \ne G$, then there exists an element $z \in G \setminus H$. Now you have to prove that $H \times \langle z \rangle$ is a subgroup of $G$. That is the only work you have to do. $\endgroup$
    – Derek Holt
    Commented Oct 11, 2016 at 11:10

2 Answers 2

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Consider the set of all $n$-tuples of the form $(g_1, g_2, \ldots, g_n)$, where each $g_i$ is one of the two elements of cyclic-2 group $\{0,1\}_+$ with group law $0+0=1+1=0$ and $0+1=1+0=1$. By definition of direcct product (though direct sum would be a better name here) the addition is taken component-wise, that is, the first component is the sum of the first component of each tuple, the second component is the sum of the second component of each tuple. Example $$(1,0,0,1,\ldots) + (1,1,0,0,\ldots) = (0,1,0,1,\ldots) $$.

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You already know that $G$ is abelian. By the fundamental theorem on finite abelian groups, $G$ is a direct product of cyclic groups. For the order $n$ of any cyclic factor, $G$ contains an element of order $n$. So all the cyclic factors are of order $2$.

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  • $\begingroup$ Or, using a similar result that is usually encountered earlier: It is a vector space over the field with $2$ elements, so it is a direct sum (i.e. direct product) of copies of that field. $\endgroup$
    – Tobias Kildetoft
    Commented Oct 12, 2016 at 7:55

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