Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Can one explain to me why is this true? $$\mathbb{Z}*\mathbb{Z}/\langle\alpha \beta \alpha \beta^{-1}\rangle \cong \langle\alpha, \beta: \alpha \beta = \beta^{-1} \alpha\rangle$$

share|improve this question
add comment

1 Answer

Note that $$\langle\alpha,\beta\mid\alpha\beta=\beta^{-1}\alpha\rangle\cong\langle\alpha,\beta\mid\beta\alpha=\alpha^{-1}\beta\rangle,$$ and that $$\beta\alpha=\alpha^{-1}\beta$$ if and only if $$\alpha\beta\alpha=\beta$$ if and only if $$\alpha\beta\alpha\beta^{-1}=e.$$ Can you get the rest of the way?

share|improve this answer
    
thanks a lot, i think i can get the rest thank u :) –  Ronald Apr 11 '13 at 22:33
1  
No, that's definitely not useful. The big thing to keep in mind, here, is that $$\Bbb Z_\alpha*\Bbb Z_\beta/\langle\alpha\beta\alpha\beta^{-1}\rangle=\langle\alpha,\beta\mid\alpha \beta\alpha\beta^{-1}=e\rangle.$$ That's just what the $\langle\text{generators}\mid\text{relations}\rangle$ notation means. –  Cameron Buie Apr 12 '13 at 16:31
1  
Instead, take the isomorphism $\Bbb Z_\alpha*\Bbb Z_\beta\to\Bbb Z_\alpha*\Bbb Z_\beta$ given by $\alpha\mapsto\beta,\beta\mapsto\alpha.$ Then compose it with the canonical projection $\Bbb Z_\alpha*\Bbb Z_\beta\to\Bbb Z_\alpha*\Bbb Z_\beta/ \langle\beta\alpha\beta\alpha^{-1}\rangle.$ (cont'd) –  Cameron Buie Apr 12 '13 at 16:38
1  
This is readily seen to be a surjective homomorphism with kernel $\langle\alpha\beta\alpha\beta^{-1}\rangle,$ and since $$\begin{align}\Bbb Z_\alpha*\Bbb Z_\beta/\langle\beta\alpha\beta\alpha^{-1}\rangle &= \langle\alpha,\beta\mid\beta\alpha\beta\alpha^{-1}=e\rangle\\ &= \langle\alpha,\beta\mid\beta\alpha\beta=\alpha\rangle\\ &= \langle\alpha,\beta\mid\alpha\beta=\beta^{-1}\alpha\rangle,\end{align}$$ we're done. This is the approach that I was describing in my answer, incidentally. –  Cameron Buie Apr 12 '13 at 16:40
1  
Note that $$\beta^{-1}(\alpha\beta\alpha\beta^{-1})\beta=\beta^{-1}\alpha\beta\alpha$$ is not an element of the cyclic subgroup generated by $\alpha\beta\alpha\beta^{-1}$, so that one can't be normal. It must be intended to represent the normal subgroup generated by $\alpha\beta\alpha\beta^{-1}$, then. Excellent! That means that I didn't tell you a bunch of wrong stuff! –  Cameron Buie Apr 13 '13 at 15:08
show 6 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.