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Let $a, b \in \mathbb{N},\ a, b\neq 0$ such that $(a,b)=1.$ Suppose $G$ is a group with presentation $$ G=\langle x, y \mid x^{-1}y^{-1}xy^{a+1}=1,\ y^{-1}x^{-1}yx^{b+1}=1\rangle. $$ Prove that $G= \langle x \rangle \times \langle y \rangle \simeq C_b\times C_a$.

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Presumably, that's what you have been asked to do. Any thoughts? –  Alex B. Dec 11 '11 at 9:11
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We see that $G=<x><y>, \ <y^{a}>\ \leq \ <y>\cap <x>,\ <x^{b}>\ \leq \ <y>\cap <x>\ $ but then I don't know how to prove that y and x have finite order... –  WLOG Dec 11 '11 at 9:56

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You have $x^{-1}y^{-1}xy = y^{-a} = x^b$, so this element centralizes both $x$ and $y$ and hence is in the centre of the group. Now $y^{-1}xy = x^{1+b}$ implies $y^{-1}x^by = x^{b(1+b)}$ so $x^{b^2}=1$ and similarly $y^{a^2}=1$. Now use $(a,b)=1$ to deduce $x^b=y^{-a}=1$.

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Nice solution ! –  WLOG Dec 11 '11 at 12:12

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