This is a follow up to this question.
Let $G = \langle x,y,z\mid{x^{{p^2}}} = {y^p} = {z^p} = 1,{x^y} = {x^{1+p}},[x,z] = [y,z] = 1\rangle$.
By euler phi function, the number of elements of order $p$ is $p-1$.
For each generator, the number of elements of order $p$ is in the following.
generator $x$ = $p-1$.
generator $y$ = $p-1$.
generator $z$ = $p-1$.
There are $3(p-1)$ elements of order $p$ from the generators.
But, how to find the rest of the elements of order $p$? Is it there are more simple way to find the number of elements of order $p$ in a finite group?
I hope someone can give me some idea regarding this question? Thank you
$\newcommand{\Span}[1]{\left\langle #1 \right\rangle}$First of all, note that $G$ is an (internal) direct product $$ G = \Span{x, y} \times \Span{z}. $$ Since $\Span{z}$ is cyclic of order $p$, you have only to find the number $N$ of elements of order dividing $p$ in $H = \Span{x, y}$. Then the total number of elements of order $p$ of $G$ will be $N p - 1$.
Now $N$ is different for $p = 2$ and for $p$ odd.
For $p = 2$, the group $H$ is dihedral of order $8$, and so you should know that it has $5$ elements of order $2$, so $N = 6$ here.
For $p > 2$, a direct computation yields $$ \tag{power} (x^{i} y^{j})^{p} = x^{i p} y^{j p} = x^{i p}, $$ so that $x^{i} y^{j}$ has order dividing $p$ iff $p$ divides $i$. Thus $N = p^{2}$ here.
You get (power) from the standard identity, valid in a group where commutators are central $$ (a b)^{n} = a^{n} b^{n} [b, a]^{\binom{n}{2}}. $$