Number of Subgroups of Prime Power Order is Congruent to 1 modulo Prime

Theorem
Let $G$ be a finite group whose order is $n$.

Let $p$ be a prime number such that $p^k$ is a divisor of $n$.

Then the number of subgroups of order $p^k$ is congruent to $1$ modulo $p$.

Proof
Let $r_k$ denote the number of subgroups of $G$ of order $p^k$.

From Number of Order p Elements in Group with m Order p Subgroups:


 * The number of elements of order $p$ is $r_1 (p - 1)$.

Since there exists $mp$ elements whose order divide $p$, for some $m \in \Z$:


 * The number of elements of order $p$ is $mp - 1$.

Thus,


 * $r_1 \equiv 1 \mod p$.

Let $H_1, H_2, \dots, H_{r_{k - 1}}$ be the $r_{k - 1}$ subgroups of order $p^{k - 1}$.

Let $K_1, K_2, \dots, K_{r_k}$ be the $r_k$ subgroups of order $p^k$.

Suppose $H_i$ is contained in $a_i$ of the subgroups $K_1, K_2, \dots, K_{r_k}$.

Suppose $K_j$ contains $b_j$ of the subgroups $H_1, H_2, \dots, H_{r_{k - 1}}$.

Then:


 * $a_1 + a_2 + \dots + a_{k - 1} = b_1 + b_2 + \dots + b_k$

is the number of distinct pairs of subgroups $H_i, K_j$ for which $H_i$ is a subgroup of $K_j$.

Let $H$ be one of the subgroups of order $p^{k - 1}$.

Let $K_1, K_2, \dots, K_a$ be the subgroups of order $p^k$ which contain $H$.

Then from Composition Series of Group of Prime Power Order:


 * $a > 0$

and:


 * $H$ is normal in $K_1, K_2, ..., K_a$.

Hence, $H$ is also normal $G$.

By correspondence theorem:


 * $G / H$ contains the subgroups $K_1 / H, K_2 / H, \dots, K_a / H$.

These are the only subgroups of $G / H$ of order $p$.

Then from above:


 * $a \equiv 1 \mod p$.

Since this is the case for each $a_i$, we have:


 * $a_1 + a_2 + \dots + a_{r_{k - 1}} \equiv r_{k - 1} \mod p$.

From Number of Order p^(n-1) Subgroups in Order p^n Group, we have also that:


 * $b \equiv 1 \mod p$.

and:


 * $b_1 + b_2 + \dots + b_{r_k} \equiv r_k \mod p$.

Therefore:


 * $r_{k - 1} \equiv r_k \mod p$.

But $r_1 \equiv 1 \mod p$.

So we have, by induction:


 * $1 \equiv r_1 \equiv r_2 \equiv \dots \equiv r_k \mod p$.

Hence the result.

Historical Note
This proof is due to Ferdinand Georg Frobenius.

He showed this result and other generalizations of the Sylow Theorems in a $1895$ paper Verallgemeinerung des Sylow’schen Satzes.