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Let $p$ a prime. Prove that the group $G=\langle x,y: x^{p}=y^{p}=x^{-2}y^{-1}xy=1\rangle$ is cyclic of order $p$.

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    @stef: Welcome to MathSE. I wanted to let you know a few things about MathSE. We like to know the sources of questions. We also like to know what you've tried on a problem or what your thoughts are, so that the answer does not re-invent the wheel. Also, many users find questions posted in the imperative ("Show that", "Prove", "Do", "Find") unpleasant and somewhat rude; please consider rephrasing your three questions. These sort of pleasantries usually result in more and better answers. Thank you!2011-12-13

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A slightly different approach (I think):

  1. Since G' = \langle [x,y]\rangle and $[x,y]=x$, |G'| is $p$ or $1$.
  2. Clearly, G/G'\cong C_p. (Substitute $[x,y]=1$ into the relations.)
  3. Therefore $|G| = p$ or $p^2$. In particular $G$ is abelian.
  4. Thus G'=1 and $G\cong C_p$.
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    I like this answer very much. Just thought I'd say...2011-12-14
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There is a morpihsm $G\to \mathbb Z_p$ sending $x$ to $0$ and $y$ to $1$, as you can check from the relations. On the other hand, since $y$ has order $p$ in $G$, there is a morphism $\mathbb Z_p\to G$ mapping $1$ to $y$.

Can you check they are mutually inverse ?

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    That's easy! It's a little harder to show that $o(x)=1$. I would recommend considering $y^{-p}xy^p$.2011-12-11