Proof that group with a minimal generating set consisting of $n$ elements of order $2$ has order $2^n$ Let $G$ be a group that is generated by $n$ elements with $n\in \mathbb{N}$, where all $n$ elements have order $2$ and generate minimally. To prove that the order of $G$ is $2^n$ we must use induction. For $n=0$ the only element in the group is the identity element. So the order of $G$ is $1=2^0$. The statement holds for $n=0$. Now we presume that the statement holds for a $n\in \mathbb{N}_{\geq 0}$. We must prove that the statement holds for $n+1$ to finsh the proof. How do I continue?
 A: Let me first give the argument assuming $G$ is commutative.
You have shown the statement for $n=0$ and assume it for $n$. 
Now let $g_1, \dots , g_{n+1} \in G$ some $n+1$ elements that generated the group (that is not generated by $n$). 
Consider the subgroup $H$ generated by $g_1, \dots ,g_n$. This is generated by $n$ elements (and cannot be generated by less, since otherwise you could als reduce the number of generators for $G$).
Thus you can apply the induction hypothesis to conclude $|H| = 2^{n}$. 
Now, $H$ is a proper sugroup of $G$. Consider the quotient group $G/H$. 
Since $g_{n+1}$ has order two, it consists only of the classes $H$ and $g_{n+1}H$. [Note you use commutativity.]
So $|G/H|= 2$ and since $|G|= |H| \ |G/H|$, the claim follows.

As said I assumed $G$ commutative. But a group where each element has order (at most) two is always commutative as $(ab)^{-1}$ is on the one hand $ab$ and on the other hand $b^{-1}a^{-1}=ba$.
So you can either assume $G$ commutative or all elements have order (at most) two. (But you need one of the two assumptions.)
