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Let $f:G\to G$ when $f=x^2$ if $f$ is homomorphism then $G$ is abelian group.

suppose $f$ is homomorphism so $f(xy)=f(x)f(y)$ $\forall x,y \in G$

then $xy=yx$

how to prove this sentence.

José Carlos Santos
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jinasakwa wuiska
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  • If $f(x) = xx$ and is a homomorphism then by definition of the function $f(xy) = (xy)(xy)$ but by definition of the homomorphism $f(xy)=f(x)f(y)=(xx)(yy)$. This holds for all $x,y\in G$. Since clearly $f(xy)=f(xy)$ then $(xy)(xy)=(xx)(yy)$. Im assuming it is accepted that $G$ is a group, so we have associativity and cancellation, thus $xy=yx$, proving Abelianiciousness. – CogitoErgoCogitoSum Apr 01 '18 at 16:49

1 Answers1

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$(xy)^2=xyxy$ in general. But your function being a homomorphism gives us the equation $(xy)^2=f(xy)=f(x)f(y)=x^2y^2=xxyy$ for any $x,y\in G$. So that way we get xyxy=xxyy. Can you finish from here?

Mark
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