Calculate the number of order $p$ elements in a group (using Sylow)

Question

I'm having trouble understanding how to calculate the number of elements of order $p$ (being $p$ prime number) in a group $G$ so that $$|G|=p^a\cdot m,$$ being $a\in\mathbb{N}$, $a\geq 1$, and $p\nmid m$. I need this for problems of the form: "Prove that there aren't simple groups of order $x$" and so. For these, I first calculate the minimum number of Sylow p-subgroups of $G$ to verify that $G$ is simple (that's the smallest number possible greater than $1$). I do that for every prime factor in $G$, and calculate the number of elements of each prime order (This is the operation I don't know), and if I've done everything right, the sum of all them plus $1$ (the neutral element) must be greater that $|G|$, thus a contradiction.

What is the "formula" for calculating these number of elements of each prime order $p$ in $G$ (knowing |G| and the number of Sylow $p$-subgroups)?

Any help will be appreciated, thanks in advance.

Example:

Prove that groups of order $132$ aren't simple.

For this problem, I know that $132=2^2\cdot 3\cdot 11$, and, if it's not simple, then the minimum nomber of Sylow subgroups are $n_11=12$, $n_3=4$ and $n_2=3$. So I know there are $12\cdot (11-1)=120$ elements of order $11$ and $4\cdot (3-1)=8$ elements of order 3. My book then states that there are at least $4$ elements of order $4$ (but I don't know why is this, the exponent of $2$ is $2$ and I don't know how to calculate the minimum number of elements of order $2$). Finally he sees that $$120+8+4+1=133>132$$ so $G$ is not simple.

My question is: why a minimum of $4$ order $2$ elements?

Answer 1

This technique is possible only when the exponent $a=1$. In which case the $p$-Sylow subgroups will be cyclic of order $p$ and there can't be any overlap between the $p$-Sylow subgroups other than the identity element i.e. $P_i \cap P_j =\{e\}$ because the cardinality of the intersection of those two subgroups $|P_i \cap P_j|$ should divide $p=|P_i|=|P_j|$. In general if $a>1$ there can be overlap between the subgroups.

A good exercise for the above technique the following problem

Let $G$ be a finite group of order $pqr$ where $p,q,r$ are distint primes. Show that $G$ cannot be simple

$\mathbf{Hint}$: Show that there is always a normal $p$ or $q$ or $r$ Sylow subgroup