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

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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?

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thanks a lot, i think i can get the rest thank u :) – Ronald Apr 11 '13 at 22:33
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
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
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
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

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