Suppose that $G$ is a group with $g^2 = I$ for all $g \in G$. Show that $G$ is necessarily abelian. Prove that if $G$ is finite, then $|G| = 2^k$ for some $k>=0$ and $G$ needs at least $k$ generators.
Well for the first part - G has order two so it must be the cyclic group with two elements. So we have $gI = Ig$ as the only ways of multiplying the element so G has to be abelian?
I don't know how you would go about proving the second part. I can't see how $G$ being finite means $G$ will necessarily have order of $2^k$ and will need at least $k$ generators?