A Group is Abelian if and only if Squaring is a Group Homomorphism

Abelian Group problems and solutions

Problem 325

Let $G$ be a group and define a map $f:G\to G$ by $f(a)=a^2$ for each $a\in G$.
Then prove that $G$ is an abelian group if and only if the map $f$ is a group homomorphism.

 
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Proof.

$(\implies)$ If $G$ is an abelian group, then $f$ is a homomorphism.

Suppose that $G$ is an abelian group. We prove that $f:G \to G, a \mapsto a^2$ is a group homomorphism.
Let $a, b$ be arbitrary elements in $G$. Then we have
\begin{align*}
f(ab)&=(ab)^2 && (\text{ by definition of } f)\\
&=(ab)(ab)\\
&=a^2 b^2 && (\text{ since $G$ is abelian})\\
&=f(a)f(b) && (\text{ by definition of } f).
\end{align*}
Therefore, we obtain $f(ab)=f(a)f(b)$ for any $a, b\in G$.
Hence $f$ is a group homomorphism from $G$ to $G$.

$(\impliedby)$ If $f$ is a homomorphism, then $G$ is an abelian group.

Suppose that $f:G\to G$ is a group homomorphism. We prove that $G$ is an abelian group.
Let $a, b\in G$. We want to prove that $ab=ba$.

Since $f$ is a group homomorphism, we have
\[f(ab)=f(a)f(b).\] As a result we have
\[(ab)^2=a^2b^2,\] or equivalently
\[abab=aabb.\] Multiplying this by $a^{-1}$ on the left and by $b^{-1}$ on the right, we obtain
\[ab=ba.\] Since $a$ and $b$ are arbitrary, this implies that $G$ is an abelian group.

Related Question.

Another problem about the relation between an abelian group and a group homomorphism is:
A group homomorphism and an abelian group.


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  1. 07/15/2017

    […] Another problem about the relation between an abelian group and a group homomorphism is: A group is abelian if and only if squaring is a group homomorphism […]

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