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?
On
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 $$(1,0,0,1,\ldots) + (1,1,0,0,\ldots) = (0,1,0,1,\ldots) $$.
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 $2$.