The number of $p$-subgroups of a group Let $G$ be a finite group. Sylow's Third Theorem claims that the number of Sylow $p$-subgroups of $G$ is $\,\equiv 1\mod p$ (I am supposing here that the trivial subgroup $1$ is a $p$-subgroup of order $p^0$). I have heard the same thing occurs for all $p$-subgroups: if $p^n$ divides the order of $G$, the number of $p$-subgroups $G$ of order $p^n$ is $\,\equiv 1\mod p$, but I am not sure that it is true.
If it is true, I wish for a reference of the proof; if it is not, I wish for a counterexample.
Thank you for paying me attention.
 A: This is Theorem of Frobenius (1895). The following proof is mixture of arguments of Wielandt, Krull and Gallagher. 
Let $p^r$ be any divisor of $|G|$.  Write $|G|=p^r.m$. Here $p$ may or may not divide $m$. 
Let  $X$ be the collection of all subsets of $G$ of order $p^r$. Then 
$|X|=\binom{p^rm}{p^r} $. 
By orbit-stabilizer theorem 
$$ p^rm=|G|=|\mathcal{O}(A)|.|Stab(A)|.$$
Claim 1. If $1\in A$ then $Stab(A)\subseteq A$:
\begin{align*}
g\in Stab(A)\hskip2mm  \Longrightarrow\hskip2mm gA=A \hskip2mm \Longrightarrow\hskip2mm g.1\in g.A=A \hskip5mm (\mbox{as }1\in A) \hskip2mm\Longrightarrow\hskip2mm Stab(A)\subseteq A.
\end{align*}
Claim 2. $|Stab(A)|$ divides $|A|=p^r$. 
Since for $g\in Stab(A)$ we have $gA=A$ hence 
$Stab(A).A=A$
which implies that $A$ (on RHS) is union of right cosets of $Stab(A)$, say
$$Stab(A).a_1 \cup Stab(A) a_2 \cup \cdots \cup Stab(A).a_r=A.$$
Hence $|Stab(A)|$ divides $|A|=p^r$. This proves Claim 2.
Write $X$ as disjoint union of orbits,
$$X= \mathcal{O}(A_1) \cup  \mathcal{O}(A_2)\cup \cdots \cup \mathcal{O}(A_t),$$
and we may assume that $1\in A_i$ for each $i$ (for, $A_i$, being subset of $G$, contains some $x\in G$ and then 
$\mathcal{O}(x^{-1}A_i)=\mathcal{O}(A_i)$ with $1=x^{-1}x\in x^{-1}A_i$.) 
Claim 3. $\mathcal{O}(A_i)=m\Longleftrightarrow A_i$ is subgroup of order $p^r$.
Since $1\in A_i$, hence $Stab(A_i)\subseteq A_i$. Then, with Claim 1, 
$$ |\mathcal{O}(A_i)|=m \Longleftrightarrow |Stab(A_i)|=p^r=|A_i| \Longleftrightarrow Stab(A_i)=A_i.$$
Claim 4. $\mathcal{O}(A_i)=m \Longleftrightarrow \mathcal{O}(A_i)$ is coset space of subgroup $A_i$ (i.e. $\mathcal{O}(A_i)=G/A_i$).
If $\mathcal{O}(A_i)$ is coset space $G/A_i$, then $|\mathcal{O}(A_i)|=|G/A_i|=m$. Conversely, if  $|\mathcal{O}(A_i)|=m$ then by Claim 3, $A_i$ is subgroup of order $p^r$, 
hence $\mathcal{O}(A_i)=\{gA_i\colon g\in G\}=G/A_i.$
By Claim 3 and 4, we have a bijection 
$$
\begin{Bmatrix}
\mbox{orbits of order $m$}\end{Bmatrix} \longleftrightarrow \begin{Bmatrix}\mbox{ subgroups of order $p^r$}
\end{Bmatrix}.
$$
Note that if $|\mathcal{O}(A_i)|\neq m$ then $Stab(A_i)$ is proper subset of $A_i$, hence $|Stab(A_i)|$ strictly divides $p^{r}$, hence $|\mathcal{O}(A_i)|=|G|/|Stab(A_i)|=(p^rm)/|Stab(A_i)|$, 
which is divisible by $pm$. 
Thus, if there are $k$ subgroups of order $p^r$, then there will be $k$ orbits of size $m$, say  $\mathcal{O}(A_1)$, $\mathcal{O}(A_2)$, $\cdots$, $\mathcal{O}(A_k)$ and the other orbits have size divisible by $pm$. 
Then 
\begin{align*}
 X= \begin{Bmatrix} \mathcal{O}(A_1) \cup \cdots \mathcal{O}(A_k)\end{Bmatrix} \cup 
\begin{Bmatrix} \mathcal{O}(A_{k+1}) \cup \cdots \mathcal{O}(A_t)\end{Bmatrix}
\end{align*}
The first $k$ orbits have size $m$ and the remaining have size divisible by $pm$, hence 
$$\binom{p^rm}{p^r}=|X|\equiv km \pmod {pm}.$$
This congruence relation depends only on order of $G$; not on structure of $G$. Hence, in particular, if $G$ is cyclic (of order $p^rm$), 
then the number of subgroups of order $p^r$ is $1$ i.e. $k=1$,  we get 
\begin{align*}
|X|\equiv m \pmod {pm}.
\end{align*}
Thus 
$$km\equiv m \pmod {pm}, k\equiv 1 \pmod p.$$
(Thanks for the patience :))
A: I assume that you have proved Exercise 1C.8 of Isaacs' book Finite Group Theory that says that you can reduce the problem to $p$-groups.
Lemma 1. Let $H$ be a proper subgroup of a finite $p$-group $G$ and suppose $|H|=p^b$. Then the number of subgroups of order $p^{b+1}$ which contain $H$, is congruent 1 mod $p$.Proof. Fix a subgroup $H$ of order $p^b$ of the $p$-group $G$. Since $H$ is normal and proper in $N_G(H)$, one can find a subgroup $K$ with $|K:H|=p$ (look at the center of $N_G(H)/H$, which is non-trivial, pick an element of order $p$, say $\overline{x}$ and put $K=\langle x \rangle$). On the other hand, if $K$ is a subgroup of $G$ with $H \subset K$ and $|K:H|=p$, then $H \lhd K$ (because $p$ is the smallest prime dividing $|K|$), and hence $K \subseteq N_G(H)$. We conclude that |{$K \leq G: H \subset K$ and  $|K:H|=p$}| = |{$\overline{K} \leq N_G(H)/H: |\overline{K}|=p$}|. The lemma now follows from the fact that in the group $N_G(H)/H$ the number of subgroups of order $p$ is congruent to $1$ mod $p$ (in any group, which order is divisible by the prime $p$, this is true and follows easily from the McKay proof of Cauchy’s Theorem). $\square$
Lemma 2. The number of subgroups of order $p^{b-1}$ of a $p$-group $G$ of order $p^b$ is congruent 1 mod $p$.
Proof. Let $G$ be a non-trivial group of order $p^b$. Let $\mathcal{S}$ be the set of all subgroups of $G$ of order $p^{b-1}$. Observe that this set is non-empty and fix an $H \in \mathcal{S}$. We are going to do some counting on the set $\mathcal{S}$, by defining an equivalence relation $\sim$ as follows: $K\sim L$ iff $H \cap K=H \cap L$ for $K, L \in \mathcal{S}$. Observe that for $K, L \in \mathcal{S}$ with $K \neq L$, $G=KL$, $K \cap L \lhd G$ and $|K \cap L|=p^{b-2}$. It is easy to see that the equivalence class $[H]$ is a singleton. In addition, $H=K$ iff $H \in [K]$ for $K\in \mathcal{S}$. Now fix a $K \in \mathcal{S}$, $K \neq H$. Counting orders one can see that if $L \in \mathcal{S}$, then $H \cap K\subset L$ iff $L \in [K]$. Hence the number of elements in $[K]$ is exactly the number of subgroups of order $p^{b-1}$ containing $H \cap K$, minus 1, namely $H$. Owing to the previous lemma, we conclude that $|[K]| \equiv 0$ mod $p$. The lemma now follows. $\square$.
Using these two lemma's you now can prove the statement by cleverly counting the number of subgroup pairs $H$ and $K$ of $G$ having order $p^b$ and $p^{b+1}$ respectively. Now back to your original question
Theorem. Let $G$ be a group of order $p^a$, $p$ prime. Let $0 \leq b \leq a$ and $n_b=$|{$H \leq G: |H|=p^b$}|. Then $n_b \equiv 1$ mod $p$.
Proof. Let $H$, $K \leq G$, with $|H|=p^b$ and $|K|=p^{b+1}$. Define a function $f$ as follows: $f(H,K)=1$ if $H \subset K$ and $f(H,K)=0$ otherwise. Let us compute $\sum_{H} \sum_{K} f(H,K)$ in two different ways: $\sum_{H}$$\sum_{K} f(H,K) = \sum_{H} \sum_{H \subset K}1$ $\equiv \sum_{H} 1$ mod $p$, according to Lemma 1 above. Similarly, by reversing the order of summation of $H$ and $K$, the sum equals $\sum_{K} 1$ mod $p$, using Lemma 2. In other words, for all $b$, the number of subgroups of $G$ of order $p^b$ is congruent mod $p$ to the number of subgroups of $G$ of order $p^{b+1}$. The theorem now follows from the fact that the number of subgroups of order $p^a$ equals 1, namely $G$ itself. $\square$.
Remark. The theorem counts all subgroups of fixed order $p^b$. If we restrict ourselves to the normal subgroups of order $p^b$ the same holds: |{$H \unlhd G: |H|=p^b$}|$\equiv 1$ mod $p$. Sketch of proof: let $G$ act by conjugation on the set of all subgroups $H$ of order ${p^b}$. The fixed point are exactly the normal subgroups. Now apply the Theorem above.
Finally, also see this StackExchange entry.
A: The number $\left|S \right|$ of subsets of order $p^{r}$ in a set of order $n = p^{r}m$, where $p$ does not divide $m$, and $r > 0$, is the binomial coefficient:
$\left|S\right| = \binom{n}{p^{r}} = \frac{n(n-1)...(n-k)...(n-p^{r} + 1)}{p^{r}(p^{r} -1)...(p^{r} -k)...1}$.
Note that each time $p$ divides a term $(n-k) = (p^{r}m - k)$ in the numerator of $\left|S \right|$, it also divides the term $(p^{r} - k)$ of the denominator the same number of times. To see this, write $k = p^{s}l$, where $p$ does not divide $l$, then $s < r$ so that $(p^{r} - k) = p^{s}(p^{r-s} - l)$ and $(n-k) = p^{s}(p^{r-s}m-l)$ are both divisible by $p^{s}$ but not by $p^{s+1}$. 
It follows that $\left|S\right| \equiv 1 \mod p$
A: first odf all you can't take the trivial subgroup as  a $p$-subgroup of order $p^0$ because in the theorem $n\in \Bbb N^{*}$,
if $p^{n}$ devides the order of $G$ then the number of $p$-Sylow subgroups is  $\,\equiv 1\mod p$ .
