Prove that if $g^2=e$ for all $g$ in $G$ then $G$ is Abelian. 
Prove that if $g^2=e$ for all $g$ in $G$ then $G$ is Abelian.

This question is from group theory in Abstract Algebra and no matter how many times my lecturer teaches it for some reason I can't seem to crack it. 
(Please note that $e$ in the question is the group's identity.)
Here's my attempt though...
First I understand Abelian means that if $g_1$ and $g_2$ are elements of a group $G$ then they are Abelian if $g_1g_2=g_2g_1$...
So, I begin by trying to play around with the elements of the group based on their definition...
$$(g_2g_1)^r=e$$ 
$$(g_2g_1g_2g_2^{-1})^r=e$$
$$(g_2g_1g_2g_2^{-1}g_2g_1g_2g_2^{-1}...g_2g_1g_2g_2^{-1})=e$$
I assume that the $g_2^{-1}$'s and the $g_2$'s cancel out so that we end up with something like,
$$g_2(g_1g_2)^rg_2^{-1}=e$$
$$g_2^{-1}g_2(g_1g_2)^r=g_2^{-1}g_2$$
Then ultimately...
$$g_1g_2=e$$
I figure this is the answer. But I'm not totally sure. I always feel like I do too much in the pursuit of an answer when there's a simpler way.
Reference: Fraleigh p. 49 Question 4.38 in A First Course in Abstract Algebra. 
 A: Proof: let for all $a,b$ in group $G$. 
claim that To show $ab=ba$ a commutative. 
By using a fact that $a\cdot a=b\cdot b=(ab)\cdot(ab)=e$.
since $(ab)^2=a^2\cdot b^2=e\cdot e=e$. We have $ab\cdot ab=e$. Multiplying on the right by $ba$, we obtain
\begin{align}
ab\cdot ab\cdot ba &= e\cdot ba\\
ab\cdot a(b\cdot b)\cdot a &= ba\\
ab\cdot a\cdot b^2\cdot a &=\\
ab\cdot a\cdot e\cdot a &=\\
ab\cdot a\cdot a &=\\
ab\cdot e &=\\
ab &= ba,
\end{align}
for all $a,b$ in $G$. since $G$ is abelian group. This is proved last.
A: Another proof is by contradiction. 
Let G be a group with operation *. You want to show that:
$(\forall g \in G:g^2=e)\implies(G\text{ Abelian}\Leftrightarrow \forall x,y \in G: x*y = y*x)$. 
(where $g^2$ is shorthand for $g*g$)
Suppose by contradiction that the group is not Abelian, i.e. that ($\exists x,y\in G: x*y\neq y*x)$. Now multiply on the left by $x$ and on the right by $y$.
You get $x^2*y^2 \ne (xy)^2$. But then it means that $e*e \neq e$ which is a contradiction.
A: By construction:
$\quad\begin{align*}
(ab)(ab) &= e = a(bb)a \\
\require{cancel}\cancel{(ab)}(ab) &= \cancel{(ab)}(ba) \qquad\text{by associativity followed by cancellation}\\
ab &= ba
\end{align*}$
Hence, the group is Abelian.
A: Hint: Take $(ab)^2=1$ and multiply both sides on the right with $b$, then again on the right with $a$.
A: given $g^2=e$ for all $g\in G$
So $g=g^{-1}$ for all $g\in G$
Let,$a,b\in G$
Now $ab=a^{-1}b^{-1}
               =(ba)^{-1}
               =ba$
So $ab=ba$ for all $a,b\in G$ .Hence $G$ is Abelian Group.
A: For any $g, h \in G$, consider the element $g\cdot h\cdot h\cdot g.~$
Since $g^2 = g\cdot g= e$ for all $g \in G$, we find that
$$g\cdot h\cdot h\cdot g = g\cdot(h\cdot h)\cdot g = g\cdot e\cdot g = g\cdot g = e.$$
But, $g\cdot h$ has unique inverse element $g\cdot h$, while we have just proved that $(g\cdot h)\cdot (h\cdot g) = e$, and so it must be that $g\cdot h = h\cdot g$ for all $g, h \in G$, that is, $G$ is an abelian group.
A: Whenever you have a condition $g^2=e$ in a group, it's  equivalent to $g=g^{-1}$ (multiply both sides by $g^{-1}$). 
In this case, it applies to every element of the group, so you can add or remove inverses from any expression freely. So the proof is simply $$ab=(ab)^{-1}=b^{-1}a^{-1}=ba.$$
A: Let $a,b\in G$:
\begin{align}
ab\cdot (ab)^{-1} &= ab\cdot b^{-1}a^{-1}\\
&= b\cdot  ba\quad(\because \text{each element is self inverse})\\
&= a(b^{2})a\\
&= a\cdot e\cdot a\\
&= a^{2}\\
&=e
\end{align}
This shows $(ab)^{-1}=ba$. Now $(ab)^2=e$, so $ab\cdot ab=ab\cdot ba=e$, which by the cancellation law gives $ab=ba$ for all $a,b\in G$, since $a$ and $b$ were arbitrary. Hence $G$ is abelian as required.
A: $$xyx^{-1}y^{-1}=(xyx^{-1})^2x^2(x^{-1}y^{-1})^2$$
A: Hint: Note that $g_1g_2=g_2g_1$ if and only if $g_1g_2g_1^{-1}g_2^{-1}=e$ (Why?), and that $g^{-1}=g$ for all $g\in G$ (Why?).
A: Then for all $a,b \in G$: $$ab = (bb)ab(aa) = b(baba)a = ba.$$
A: Alternatively, the map $$\begin{align*}f:G&\rightarrow G\\x&\mapsto x^{-1}(=x)\end{align*}$$ is an automorphism of $G$ and so $G$ is Abelian!
A: $(ab)^{2}=e
$($\because$ $a,b \in G, ab\in G$, due to the closure property of group axiom )$\implies 
(ab)(ab)=e\tag 1$
Pre multiply $a$ on both sides also post multiply $b$ on both sides. The equation (1) becomes,
 $aababb=ab\tag2$($\because$ by associativity and self invertible property.
$ba=ab\tag3$
A: We have for all $a,b\in G$,
$$\begin{align}
\color{red}{abab}&=(ab)^2\\
&=e\\
&=ee\\
&=a^2b^2\\
&=\color{red}{aabb},
\end{align}$$
so that
$$\begin{align}
ba&=e(ba)e\\
&=(a^{-1}a)ba(bb^{-1})\\
&=a^{-1}(\color{red}{abab})b^{-1}\\
&=a^{-1}(\color{red}{aabb})b^{-1}\\
&=(a^{-1}a)ab(bb^{-1})\\
&=e(ab)e\\
&=ab.
\end{align}$$
