Thread: Algebra, Problems For Fun (7)

1. Algebra, Problems For Fun (7)

Definition: A group $G$ is called hopfian if every surjective homomorphism $G \longrightarrow G$ is injective. Clearly every finite group is hopfian.

Problem: Prove that $G=$ is not hopfian.

Suggestion:
Spoiler:
You may define the homomorphism $f: G \longrightarrow G$ by $f(x)=x^2, \ f(y)=y.$

2. Originally Posted by NonCommAlg
Definition: A group $G$ is called hopfian if every surjective homomorphism $G \longrightarrow G$ is injective. Clearly every finite group is hopfian.

Problem: Prove that $G=$ is not hopfian.

Suggestion:
Spoiler:
You may define the homomorphism $f: G \longrightarrow G$ by $f(x)=x^2, \ f(y)=y.$

In the homomorphism you give what is mapped to x, or to odd powers of x? If you hadn't given it a name I would be sceptical such groups existed: in a homomorphic image you add a relation into the presentation, xy=yx or whatever. However, that makes xy and yx equivalent and so you don't map to one of them.

Just a thought...

3. Originally Posted by Swlabr
In the homomorphism you give what is mapped to x, or to odd powers of x? If you hadn't given it a name I would be sceptical such groups existed: in a homomorphic image you add a relation into the presentation, xy=yx or whatever. However, that makes xy and yx equivalent and so you don't map to one of them.

Just a thought...
$f$ 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 $f(y^{-1}xyx^{-1})=x,$ which proves that $f$ is surjective.

so the only thing left is to prove that $f$ is not injective!

4. Interesting. I stand corrected. By rights I should probably stare at this group for the next hour or two, but I actually have to study group rings (this site is wonderful - I can procrastinate by studying maths...). Anyway - to complete the solution:

$(x[y,x^{-1}](x^{-1})^y)\phi = x\phi [y,x^{-1}]\phi (x^{-1})^y\phi = x^2 x (x^{-2})^y = x^2 x x^{-3} = 1$ and as $x[y,x^{-1}](x^{-1})^y \neq 1$ is not trivial we have a non-trivial kernel, and so no injection.

5. Originally Posted by Swlabr

$x^2 x (x^{-1})^y = x^2 x x^{-3} = 1$
$(x^{-2})^y$ not $(x^{-1})^y$. apart from that your answer is correct.

6. Edited.