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.
Best Answer
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 :))