fun with "necessarily" abelian

I have this problem:

Suppose that $\displaystyle X$ is a group with $\displaystyle x^2=1$ for all $\displaystyle x \in X$. Show that $\displaystyle X$ is necessarily abelian.

Prove that if $\displaystyle X$ is finite, then $\displaystyle |X|=2^n$ for some $\displaystyle n \geq 0$ and $\displaystyle X$ needs at least $\displaystyle n$ generators.

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For the first part I did:

$\displaystyle \forall x \in X$ , $\displaystyle x(1)=x=(1)x$, so: $\displaystyle e_X=1$

and: $\displaystyle x^2=e_X \, \forall x \in X \Rightarrow x = x^{-1} \, \forall x \in X$

$\displaystyle \forall x,y \in X:$

$\displaystyle xy=(xy)^{-1}=y^{-1}x^{-1}=yx$

So $\displaystyle X$ is abelian. Is that the same as *necessarily* abelian?

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Not sure how to go about the second part...