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

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