[Math] Let a finite group $G$ have $n(>0)$ elements of order $p$(a prime) . If the Sylow p-subgroup of $G$ is normal, then does $p$ divide $n+1$

Question

Suppose $G$ is a finite group and $p$ is a prime that divides $|G|$. Let $n$ denote the number of elements of $G$ that have order $p$ . If the Sylow p-subgroup of $G$ is normal, then is it true that $p$ divides $n+1$ ? I know that $p-1=\phi(p)|n$ but I cannot approach further . Please help . Thanks in advance

Answer 1

Every element of order $p$ is contained in the Sylow $p$ subgroup (this is a slight extension of Sylow II), so you want to know that the number of elements of order $p$ in a $p$-group, plus one, is divisible by $p.$ This is true for a cyclic group of order $p.$ Now, note that a $p$ group has non-trivial center, so has a central element of order $p,$ which should be enough for the induction step.

EDIT Here is a complete argument, along slightly different lines. As above, we only need to consider $p$-groups. Now, consider the set $X$ of elements of order $p$ in a $p$-group $G,$ and the conjugation action of $G$ on that set. We have a variant of the class equation: $$ |X| = n_Z + \sum |G|/C(x),$$ where the sum is over conjugacy classes of non-central elements, and $n_Z$ is the number of central elements of order $p.$ Every summand is divisible by $p,$ so we need to only consider $n_Z.$ Now, since the center is an abelian group, we need to show the result for abelian groups. By the fundamental theorem of abelian groups, we know that an abelian group is a product of cyclic groups. For cyclic group, the result is clear. Now, suppose $G = H_1 \times H_2,$ then, it is easy to see that $n(G) = n(H_1) + n(H_2) + n(H_1) n(H_2).$ Indeed, the elements of order $p$ in $G$ are those of the form $(1, x),$ where $x$ is of order $p$ in $H_2,$ and those of the form $(y, 1),$ with $y$ of order $p$ in $H_1,$ and those of the form $(x, y)$ (as above. Now use induction on the number of direct summands in $G.$