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Definition: A group is called hopfian if every surjective homomorphism $f: \mathcal{G} \rightarrow \mathcal{G}$ is an isomorphism. Clearly every finite group is hopfian.

Problem:

Prove that

\[\mathcal{G} = \langle x, y: y^{-1} x^2 y = x^3 \rangle\]

is not hopfian.

Well, we define $f:\mathcal{G} \rightarrow \mathcal{G}$ by $f(x)=x^2$ and $f(y)=y$ and extend it to $\mathcal{G}$ homomorphically. Since $\mathcal{G}$ is well defined then $f$ is a surjective because

$$f\left ( y^{-1} xy x^{-1} \right ) = x$$

but not an isomorphism because if we take $z=y^{-1} x y$ then we note that $xz \neq z x  $ but

$$f\left ( xz \right ) = f \left ( zx \right ) = x^5$$