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: 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.
A: $$a^2 = a*a$$ 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 $$\begin{align} abab & = e \\ \\  \iff a(abab)b &= aeb \\ \\ \iff a^2(ba) b^2 &= ab \\ \\ \iff ebae &= ab \\ \\ \iff ba &= ab.\end{align}$$
A: If $a^2 =e$ for all $a\in G$, then $a^{-1}(a^2) = a^{-1}e = a^{-1}$ for all $a\in G$, but since $a^{-1}(a^2) = (a^{-1}a)a =ea = a$, we have $a^{-1} = a$ for all $a \in G$.  If we now take any $a, b \in G$, we see that $(ab)^{-1} = ab$.  But $(ab)^{-1} = b^{-1}a^{-1} = ba$; thus $ab = ba$ for all $a,b \in G$ and so $G$ is commutative. QED.
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
A: $ab=aeb=a\left(ab\right)\left(ab\right)b=\left(aa\right)ba\left(bb\right)=ebae=ba$
A: Let $x,y \in G$.  Then $(xy)^2=e$, and $x^2 y^2=e e =e$, whence $xyxy=xxyy$.  By left-cancellation of $x$ and right-cancellation of $y$, we get that $yx=xy$. Hence $G$ is commutative.
