# Presentation of cyclic group

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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Wouldn't it be useful to specify where $x,y$ live? – Marc van Leeuwen Dec 11 '11 at 13:45
@MarcvanLeeuwen: This is a group presentation. $x$ and $y$ don't "live" anywhere per se - if anything, they represent the free group on two generators. Look up group presentation for more details, or check out the first chapter of the book "Combinatorial Group Theory" by Magnus, Karrass and Solitar. – user1729 Dec 12 '11 at 11:41
@user1729: Thanks. Yes I know about presentations. Just the question does not mention them, and I did not recognise the notation used. Actually the title mentions presentation, but not in a way that made the coin drop for me. – Marc van Leeuwen Dec 12 '11 at 12:26
@stef: Welcome to MathSE. I wanted to let you know a few things about MathSE. We like to know the sources of questions - if it's homework, please add the [homework] tag. People will still help, so don't worry. 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! – Arturo Magidin Dec 13 '11 at 4:17

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... – user1729 Dec 14 '11 at 17:22

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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But we know that $o(y)\ |\ p\$ not that $o(y)=p$ – WLOG Dec 11 '11 at 12:41
Well, you have to check that too :) – Mariano Suárez-Alvarez Dec 11 '11 at 12:43
That's easy! It's a little harder to show that $o(x)=1$. I would recommend considering $y^{-p}xy^p$. – Derek Holt Dec 11 '11 at 15:18