How to prove the number of the elements of order $p$ in a group of order $p^4$ is ${p^3}-1$

Question

This is a follow up to this question.

Let $G = \langle x,y,z\mid{x^p} = {y^p} = {z^{p^2}} = 1,[x,y] = z,[x,z] = [y,z] = 1\rangle$.

To find the number of elements of order $p$, can i use the same approach as question before? By using the commutator identities? $$(ab)^n = a^nb^n[b,a]^{\binom{n}{2}}$$

I'm a little bit confuse since in this case ${z^{p^2}}=1$ not ${x^{p^2}}=1$.

Answer 1

This group does not have order $p^4$; it has order $p^3$ and is the Heisenberg group. I’m assuming $p$ is odd.

To verify that $z$ in fact has order (dividing) $p$, note because $z$ is central, the group is nilpotent of class (at most) $2$, and so the commutator bracket is bilinear. From this, you have that $$z^p = [x,y]^p = [x^p,y] = [1,y] = 1.$$ So even though the presentation tells you that the order of $z$ divides $p^2$, in fact it divides $p$.

It is clear that this group has the Heisenberg group as a quotient (mod out by $z^p$). But since $z^p$ is trivial, the group has order at most $p^3$ (every element can be written, by collection, as $x^ay^bz^c$, with $0\leq a,b,c\lt p$), and since it has the Heisenberg group as a quotient, it has order exactly $p^3$ and is isomorphic to the Heisenberg group.

The number of elements of order $p$ in this group is $p^3-1$.

Now, you could have a group of order $p^4$ with generators of the given order if you had the group $$G = \langle x,y,z\mid x^p=y^p=z^{p^2}=[x,z]=[y,z]=1, [x,y]=z^p\rangle$$ (that is $z$, is a $p$th root of the commutator $[x,y]$, instead of being equal to the commutator). That is a group obtained by adding a central $p$th root to the commutator in the Heisenberg group. Could that be the group you were supposed to look at?

In that case, we have that elements are of the form $x^ay^bz^c$ with $0\leq a,b\lt p$, $0\leq c\lt p^2$. Then you would have $$\begin{align*} (x^ay^bz^c)^n &= (x^ay^b)^nz^{nc}\\ &= x^{an}y^{bn}[y^b,x^a]^{\binom{n}{2}}z^{nc}\\ &= x^{an}y^{bn}[y,x]^{ab\binom{n}{2}}z^{nc}. \end{align*}$$ If $n=p$ and $p$ is odd, then $x^{ap}=y^{bp}=[y,x]^{ab\binom{p}{2}}=1$ (since $\binom{p}{2}$ is a multiple of $p$). Thus, $x^ay^bz^c$ has order dividing $p$ if and only if $z^{pc}=1$, if and only if $p|c$. Thus, the elements of order dividing $p$ are of the form $x^ay^bz^{kp}$, with $0,\leq a,b,k\lt p$. This gives you $p^3$ elements. But if $a=b=k=0$ you get the identity, so this leads to exactly $p^3-1$ elements of order $p$.

Answer 2

The group is nilpotent of class 2 so the commutator is linear in both arguments. In particular $[x,y]^p=[x^p, y]=1=z^p$. Since every element is uniqely represented as $x^ky^mz^n$, $0\le k,m,n\le p-1$, the number of elements of your group is $p^3$ and all nontrivial elements have order $p$. This is the Heisenberg group of order $p^3$.