If g^2 = e for all g in G, then G is abelian

If $g^2 = e$ for all $g$ in $G$ , then $G$ is abelian

Proof. Assume that for all

g

G

that

g^2 = e

holds. Let

a,b \in G

. Then since

G

is a group, we know that

ab \in G

. Further, we know that

a^2 = e

and that

b^2 = e

. Now take

(ab)^2 = abab

. Because

(ab)^2 = e

, we have that

abab = e

. Now, multiply

b^{-1}a^{-1}

on both sides on the right side of

abab = e

, then we get

ab = b^{-1}a^{-1}

. As

a^2 = e

, we have that

a = a^{-1}

. The same holds argument for

b

. So we get

ab = b^{-1}a^{-1} = ba

. Wrapping everything together, we can write the proof as

\begin{align*} (ab)^2 = e &\iff abab = e \\ &\iff ababb^{-1} = b^{-1} \\ &\iff aba = b^{-1} \\ &\iff abaa^{-1} = b^{-1}a^{-1} \\ &\iff ab = b^{-1}a^{-1} \\ &\iff ab = ba \quad \text{because } a = a^{-1} \text{ and } b = b^{-1}. \end{align*}