If Every Nonidentity Element of a Group has Order 2, then it’s an Abelian Group

Abelian Group problems and solutions

Problem 212

Let $G$ be a group. Suppose that the order of nonidentity element of $G$ is $2$.
Then show that $G$ is an abelian group.

 
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Proof.

Let $x$ and $y$ be elements of $G$. Then we have
\[1=(xy)^2=(xy)(xy).\]

Multiplying the equality by $yx$ from the right, we obtain
\begin{align*}
yx&=(xy)(xy)(yx)\\
&=xyxy^2x\\
&=xyx^2 \quad (\text{ since } y^2=1)\\
&=xy \quad (\text{ since } x^2=1).
\end{align*}
Thus we obtain $xy=yx$ for any elements $x, y \in G$. Thus the group $G$ is an abelian group.


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