Presentation of a group

Question

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 $G= \langle x \rangle \times \langle y \rangle \simeq C_b\times C_a$.

Answer 1

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 $x^b=y^{-a}=1$.