We have that $x^p=y^2=e$, $xy=yx$ and $(xy)^{2p}=e$, where $p$ is a prime. I want to show that the order of $xy$ is $2p$. To do that we have to show that $2p$ is the smallest power of $xy$ so that it is equal to $1$, right?

I have done the following:

Suppose that $2p$ is not the smallest power, then say $n$ is the smallest power such that $(xy)^n=e$. That means that $n\mid 2p \Rightarrow n\mid 2 \text{ or } n\mid p \Rightarrow n\in \{1,2,p\}$.

If $n=1$, then $xy=e\Rightarrow x=y^{−1}=y$. Since the order of $y$ is $2$, we would have that the order $x$ is also $2$. That is true only when $…

$\sigma_1^0 : \Delta^0 \rightarrow RP^2$ \ \ $\sigma_2^0 : \Delta^0 \rightarrow RP^2$
$1 \mapsto [(0,0)]$ $1 \mapsto [(0,0)]$

$\sigma_1^1 : \Delta^1 \rightarrow RP^2$ $\sigma_2^1 : \Delta^1 \rightarrow RP^2$ $\sigma_3^1 : \Delta^1 \rightarrow RP^2$

where the maps are defined as follows:


$\sigma_1^1((t_0,t_1)) = [(0,0)t_0 + (1,0)t_1]$
$ \sigma_2^1((t_0,t_1)) = [(1,1)t_0 + (1,0)t_1]$
$\sigma_3^1((t_0,t_1)) = [(0,0)t_0 + (1,1)t_1]$

$ \sigma_1^2 : \Delta^2 \rightarrow RP^2$ $ \sigma_2^2 : \Delta^2 \rightarrow RP^2$