Number of elements of order $p$ in the sylow $p$-subgroup of $S_{p^2}$

Question

I have the group $G = \mathbb{Z}_p \wr \mathbb{Z}_p$, it's well known that $G$ is isomorphic to a sylow $p$-subgroup of $S_{p^2}$, it has order $p^{p+1}$ and there are 2 possible orders for a non-identity element: $p$ or $p^2$. I want to know if it's possible to find out the exactly number of elements of order $p$ or $p^2$ in $G$, so far I just know that the number of elements of order $p$ must be divisible by $p-1$, but since I'm working with large groups it doesn't seem to help much. For the cases $p=2,3,5,7$ a brute force algorithm solved my problem but I also need for $p=11$ and $p=13$. Can someone help?

Answer 1

Here's a different approach, which doesn't involve any computation inside the $p$-group (but does require knowing its normalizer). First, compute the number of Sylow $p$-subgroups $P$ of $G=S_{p^2}$. This is the index $|G:N_G(P)|$, but $N_G(P)=P.(C_{p-1})^2$, so we can obtain this. Then we need to know how many elements of order $p^2$ there are in $G$. This is easy, since they are all conjugate and have centralizer of order $p^2$, so we have $|G|/p^2$ many. Finally, we need to know how many different Sylow $p$-subgroups a given $p^2$-cycle belongs to. But a $p^2$-cycle stabilizes a unique system of imprimitivity with $p$ different blocks of size $p$, so a $p^2$-cycle cannot lie in two different Sylow $p$-subgroups.

So we may simply divide the number of elements of order $p^2$ by the number of Sylow $p$-subgroups to obtain $$ (|G|/p^2) / (|G|/p^{p+1}(p-1)^2)=p^{p-1}(p-1)^2.$$

Answer 2

View the group as $(C_p)^p \rtimes (C_p)$, writing the elements of the subgroup $(C_p)^p$ as vectors $(x_1, \dots, x_p)$, and letting $\sigma$ denote a generator of the last $C_p$. Then each element can be expressed uniquely in the form $g = (x_1, \dots, x_p) \cdot \sigma^a$, where $x_i, a \in \mathbb Z/p\mathbb Z$. If $a = 0$, then $g$ of course has order $\leq p$. If instead $a \neq 0$, then $g^p$ is the cyclic sum $(\sum_i x_i, \dots, \sum_i x_i)$. So $g$ has order $p^2$ (i.e. $g^p$ is not the identity) if and only if $a$ and $\sum_i x_i$ are both nonzero modulo $p$.

There are $p$ ways to choose each of $x_1, \dots, x_{p-1}$, and then $p-1$ ways to choose each of $x_p$ and $a$. Therefore, this group has exactly $(p-1)^2 p^{p-1}$ elements of order $p^2$, and the remaining $(2p-1) \cdot p^{p-1} - 1$ nonidentity elements have order $p$.