Not a Hopfian group

Groups, Rings, Domains, Modules, etc, Galois theory
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Riemann
Posts: 176
Joined: Sat Nov 14, 2015 6:32 am
Location: Melbourne, Australia

Not a Hopfian group

#1

Post by Riemann »

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.
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$
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Riemann
Posts: 176
Joined: Sat Nov 14, 2015 6:32 am
Location: Melbourne, Australia

Re: Not a Hopfian group

#2

Post by Riemann »

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

$$f\left ( xz \right ) = f \left ( zx \right ) = x^5$$
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$
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