## A non-hopfian group

Posted: December 13, 2010 in Elementary Algebra; Problems & Solutions, Groups and Fields
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Definition. A group is called hopfian if every surjective homomorphism $G \longrightarrow G$ is an isomorphism. Clearly every finite group is hopfian.

Problem. Prove that $G = \langle x,y: \ y^{-1}x^2y=x^3 \rangle$ is not hopfian.

Solution. Define $f: G \longrightarrow G$ by $f(x)=x^2$ and $f(y)=y$ and extend it to $G$ homomorphically. Note that $G$ is well-defined because

$f(y^{-1}x^2y)=y^{-1}x^4y=(y^{-1}x^2y)^2=x^6=f(x^3).$

We also have $f(y^{-1}xyx^{-1})=x$ and so $f$ is surjective. To prove that $f$ is not an isomorphism, let $z=y^{-1}xy.$ Then $xz \neq zx$ but $f(xz)=f(zx)=x^5. \ \Box$