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 $n+1$? 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
 A: 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 $G.$
A: This is the Exercise 24.59 in Gallian's Contemporary Abstract Algebra 8/e.
By Sylow 2nd Theorem,
there exists only one Sylow $p$-subgroup $L$.
For any subgroup $H$ of order $p$,
by Sylow 1st Theorem,
$H\leq L$.
By Exercise 24.41,
there exists a subgroup $N$ of order $p$,
which is normal in $G$.
By these two things,
we can conclude that $N\lhd L$.
If $N\not\subseteq H\leq L$ and $|H|=p$,
verify that $H/N=\{hN\mid h\in H\}$ is a subgroup of $L/N$.
By Correspondence Theorem,
there exists $K$ such that $N\leq K$ and $K/N=H/N$.
Note that $|K/N|=|H/N|=p$ and $|K|=p^2$.
By the Corollary of Theorem 24.2,
$K$ is abelian.
Write $K/N=\langle kN\rangle$,
where $k\notin N$.
Since
$$\langle kn_i\rangle/N
\ni(kn_i)^mN
\stackrel{K\text{ is abelian}}{=}k^m n_i^mN
=k^m N
=(kN)^m
\in \langle kN\rangle
=K/N,
$$
we have $\langle kn_i\rangle/N=K/N$ for each $n_i\in N=\{n_1, n_2, ...,n_p\}$.
Note that $|\langle kn_i\rangle/N|=|K/N|=p$.
So $\langle kn_i\rangle$ is a subgroup of order $p$.
Hence,
$\langle kn_1\rangle, \langle kn_2\rangle, ..., \langle kn_p\rangle$
are all distinct subgroups of $L$ whose order is $p$
such that $\langle kn_i\rangle/N=K/N$.
The $p$ subgroups $\langle kn_1\rangle, \langle kn_2\rangle, ..., \langle kn_p\rangle$ of $L$ correspond to a same subgroup $K$.
Similarly,
except $N$,
every $p$ subgroups of order $p$ in $L$ correspond to a same subgroup of order $p^2$.
Let $s$ be the number of subgroups of order $p$ in $L$.
Then $p\mid (s-1)$.
Note that
$s=\frac{n}{p-1}$,
as the following figure indicates.

Therefore,
\begin{eqnarray*}
p\mid (s-1)=\left(\frac{n}{p-1}-1\right)
&\Rightarrow& p(p-1)\mid n-(p-1)\\
&\Rightarrow& p\mid n-p+1\\
&\Rightarrow& p\mid n+1
\end{eqnarray*}
