Prove that a group where $a^2=e$ for all $a$ is commutative Defining a group $(G,*)$ where $a^2=e$ with $e$ denoting the identity class....
I am to prove that this group is commutative. To begin doing that, I want to understand what exactly the power of 2 means in this context. Is the function in the group a power or something?

The trick with these types of problems is to evaluate the 'product' of group elements in two different ways. 
So for this problem, we interpret $(ab)^2$ two different ways, where $a,b \in G$.
First, we have this rule in $G$ that an element 'squared' is the identity. So we know that
$$
(ab)^2=e
$$
But 
$$
(ab)^2=abab
$$
Also note that
$$
e=e\cdot e=a^2b^2
$$
So we must have
$$
a^2b^2=abab
$$
But then that gives us
$$
\begin{align}
a^{-1}a^2b^{2}b^{-1}&=a^{-1}ababb^{-1}\\
ab&=ba
\end{align}
$$
since $a,b \in G$ were arbitrary, $G$ is commutative. Later we use the same trick for rings by evaluating $(a+b)^2$ two different ways.
