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?

$$a^2 = a*a$$ where '$*$' is the group operation.
In general,  $$a^n = \underbrace{a*a*\cdots * a}_{\large n\;\text{times}}$$
For the proof that $G$ must be commutative, if you're stuck, look at the product of two elements. To simplify, I'll omit the '$*$' symbol for the group operation and simply use juxtaposition of two elements to denote the group operation. 
Take $a, b \in G.$ Then we know that $a^2 = b^2 = e.$ Since $G$ is a group, $ab \in G$, since $G$ must be closed under $*$. Furthermore $(ab)^2 = e$, since $a*b \in G$. So we can see that $$(ab)^2 = abab = e$$
Now, left multiply each side of the equation by $a$, and right multiply each side of the equation by $b$: that gives us $$\begin{align} abab & = e \\ \\  \iff a(abab)b &= aeb \\ \\ \iff a^2(ba) b^2 &= ab \\ \\ \iff ebae &= ab \\ \\ \iff ba &= ab.\end{align}$$
