How many $p$-elements does $G$ have?

Question

Let $G$ be a group with a non normal $p$-sylow subgroup $P$. Is there any information about the number of $p$-elements(an element with $p$-power order) of $G$?

Answer 1

You can find out some general properties, but nothing precise really unless you have more information.

Let $G$ be a finite group. Denote by $a_p(G)$ the number of $p$-elements of $G$.

Frobenius' theorem states that when $n$ divides the order $G$, the number of solutions to $x^n = 1$ in $G$ is a multiple of $n$. Note that $a_p(G)$ is the number of solutions to $x^{p^\alpha} = 1$, where $p^\alpha$ is the largest power of $p$ dividing $|G|$. Hence Frobenius' theorem implies that $a_p(G)$ is a multiple of $p^\alpha$, say $a_p(G) = p^\alpha r$.

Now the number of elements of order $k$ in $G$ is a multiple of $\varphi(k)$, where $\varphi$ is the Euler totient function. Thus for any $1 \leq l \leq \alpha$, it follows that $p-1$ divides the number of elements of order $p^l$ in $G$. Hence $p-1$ divides $a_p(G) - 1$. In particular $p-1$ divides $r-1$, and thus

$$a_p(G) = p^{\alpha}(t(p-1) + 1)$$

for some integer $t \geq 0$.

Let $n_p(G)$ be the number of Sylow $p$-subgroups of $G$, so $n_p(G) \equiv 1 \mod{p}$ by Sylow's theorem. It is possible to deduce some small things about $a_p(G)$ if we know $n_p(G)$, see this question (SE) and this one (MO). For example, G. A. Miller proved that

• If $n_p(G) = 1$, then $a_p(G) = p^\alpha$.
• If $n_p(G) = p+1$, then $a_p(G) = p^{\alpha+1}$.
• If $n_p(G) > p+1$, then $a_p(G) \geq p^\alpha(2p - 1)$.