The number of elements of order $p$ in a $p$-group is $-1 \bmod p$ I am reading a proof and it takes the following statement - but it is not immediate to me why it is true at all:

The number of elements of order $p$ ($p$ prime) in a $p$-group is $-1 \bmod p$.

 A: if the group is abelian, then $a \to a^p$ is a homomorphism and the order of the kernel is a power of $p$ - as a subgroup of a $p$-group. however this kernel consists of the identity and all the elements of order $p$, hence the number of the latter is congruent to $-1 (\mod p)$.
since any $p$-group has a non-trivial center, the required result follows if it can be shown that the number of non-central elements of order $p$ is a multiple of $p$. but this follows by considering conjugacy classes, since every centralizer has order a positive power of $p$.
A: Here is another method.
Let $G$ be a $p$-group and $X=\{g\in G\mid |g|=p\}$ be the set of all the elements of order $p$.
Let $G$ act on $X$ by conjugation.
Note that
\begin{eqnarray*}
&& |orbit(x)|=1\\
&\Leftrightarrow& orbit(x)=\{x\}\\
&\Leftrightarrow& \{g\cdot x=gxg^{-1}\mid g\in G\}=\{x\}\\
&\Leftrightarrow& \forall g\in G, ~gxg^{-1}=x\\
&\Leftrightarrow& x\in Z(G).
\end{eqnarray*}
Then by the class equation,
\begin{eqnarray*}
|X|
&=&\sum_{\substack{x\in X\\|orbit(x)|=1}}|orbit(x)|+\sum_{\substack{i=1\\ |orbit(x_i)|\neq 1\\ orbit(x_i)\neq orbit(x_j)\text{ if }i\neq j}}^{n}|orbit(x_i)|\\
&=&\sum_{\substack{x\in X\\ x\in Z(G)}}1+\sum_{\substack{i=1\\ [G:stab(x_i)]\neq 1}}^{n}[G:stab(x_i)]\\
&=& |Z(G)\cap X|+\sum_{\substack{i=1\\ [G:stab(x_i)]\neq 1}}^{n}[G:stab(x_i)].
\end{eqnarray*}
Then $$p\text{ divides }\sum_{\substack{i=1\\ [G:stab(x_i)]\neq 1}}^{n}[G:stab(x_i)]=|X|-|Z(G)\cap X|$$
(because $H$ is a $p$-group)
and
$$|X|\equiv |Z(G)\cap X|\pmod{p}.$$
(Tricky.) Verify that $(Z(G)\cap X)\cup \{e\}$ is a subgroup of $G$.
Then by Lagrange's Theorem and $e\notin Z(G)\cap X$, 
$|Z(G)\cap X|+1=|(Z(G)\cap X)\cup \{e\}|$ divides $|G|$.
It follows that $$|Z(G)\cap X|\equiv -1\pmod{p}.$$
A: This question is from Rotman's An Introduction to the Theory of Groups 4/e, p.75, Lemma 4.7
or Isaacs's Finite Group Theory, p.7, 1A.8.(b).
There is another exercise in Gallian's Contemporary Abstract Algebra 8/e.
Which is essentially the same as this problem.
See exercise 24.59.
Here is the author's proof.
Corollary of Theorem 4.4. Let $G$ be a group and $|G|=n$.
If $d\mid n$, 
then the number of elements of order $d$ is $\phi(d)k$ for some $k\in \Bbb{Z}$, 
where $\phi$ is the Euler's totient function.
We prove your assertion.
Let $G$ be a $p$-group and $|G|=p^n$.
By Lagrange's Theorem, 
the order of every element in $G$ is a power of $p$.
Thus, 
\begin{eqnarray*}
\#\{g\in G:|g|=p\}
&=&|G|-\#\{g\in G:|g|=p^n\}-\#\{g\in G:|g|=p^{n-1}\}\\
&-&\cdots -\#\{g\in G:|g|=p^2\}-\#\{g\in G:|g|=1\}\\
&=&p^n-\phi(p^n)k_n-\phi(p^{n-1})k_{n-1}-\cdots -\phi(p^2)k_2-1\\
&=&ps-1.
\end{eqnarray*}
Exercise 24.59. 
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, 
prove that $p$ divides $n+1$.
A: Here is my proof.
The idea was obtained by plenty observations.
Let $G$ be a $p$-group and $|G|=p^n$.
By the exercise II.5.3 in Hungerford's Algebra, 
there exists a normal subgroup $N$ of order $p$ in $G$.
If $N\neq H\leq G$ and $|H|=p$,
that is, $H$ is another subgroup of order $p$,
verify that the set $S=\{hN\mid h\in H\}$ is a subgroup of $G/N$.
By Correspondence Theorem,
there exists $K$ such that $N\leq K$ and $K/N=S$.
Note that $|K/N|=|S|=p$ and $|K|=p^2$.
By the exercise II.5.13 in Hungerford's Algebra,
$K$ 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 $G$ 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 $G$ correspond to a same subgroup $K$.
Similarly,
except $N$,
every $p$ subgroups of order $p$ in $G$ correspond to a same subgroup of order $p^2$.
Let $s$ be the number of subgroups of order $p$ in $G$.
Then $p\mid (s-1)$.
Let $r$ be the number of elements of order $p$ in $G$.
Note that
$s=\frac{r}{p-1}$,
as the following figure indicates.

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