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

Solution 1:

Hint: Take $(ab)^2=1$ and multiply both sides on the right with $b$, then again on the right with $a$.

Solution 2:

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.

Solution 3:

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 $$ab=(ab)^{-1}=b^{-1}a^{-1}=ba.$$

Solution 4:

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?).