# Congruence condition on number of subgroups of given prime power order

This article describes a congruence condition on an enumeration, or a count. It says that in a finite group and modulo prime number, the number of subgroups of given prime power order satisfies a congruence condition.
View other congruence conditions | View divisor relations

## Statement

### Version for a group of prime power order

Let $G$ be a group of prime power order $p^n$ and suppose $0 \le r \le n$. The following are true:

• The number of subgroups of $G$ of order $p^r$ is congruent to $1$ modulo $p$.
• If $n$ is itself a power of $p$, i.e., if $G$ is a group of prime power order, then the number of normal subgroups of $G$ of order $p^r$ is congruent to 1 mod $p$.
• If $n$ is itself a power of $p$, i.e., if $G$ is a group of prime power order, then the number of p-core-automorphism-invariant subgroups of $G$ of order $p^r$ is congruent to 1 mod $p$.

In particular, the collection of groups of order $p^r$ is a collection of groups satisfying a universal congruence condition.

### Version for a general finite group

Let $G$ be a finite group and $p^r$ be a prime power dividing the order of $G$. Then, the number of subgroups of $G$ of order $p^r$ is congruent to 1 mod $p$.

## Facts used

1. Congruence condition relating number of subgroups in maximal subgroups and number of subgroups in the whole group

## Proof

### Equivalence between congruence condition on subgroups and on normal subgroups

If $H \le P$ is not normal, the number of conjugates of $H$ in $P$ equals $[P:N_P(H)]$, which is a power of $p$, and hence a multiple of $p$. Thus, the non-normal subgroups of order $p^r$ are a union of conjugacy classes whose orders are multiples of $p$. Thus, the number of non-normal subgroups of order $p^r$ is a multiple of $p$.

Thus, the total number of subgroups of order $p^r$ equals the number of subgroups of order $p^r$ modulo $p$.

### Proof for a group of prime power order using the stronger result

This proof uses fact (1): congruence condition relating number of subgroups in maximal subgroups and number of subgroups in the whole group, combined with an induction on order.

Given: A group $P$ of order $p^k$.

To prove: For any $r \le k$, the number of subgroups of order $p^r$ is congruent to $1$ modulo $p$.

Proof: If $r = k$, the statement is vacuously true, otherwise, the given fact yields:

$n(G) \equiv \sum_{M \operatorname{max} P} n(M) \pmod p$.

By induction, we know that for each maximal subgroup $M$ of $P$, the number of subgroups of $M$ is congruent to $1$ modulo $p$. Thus, it suffices to show that the number of maximal subgroups is itself congruent to $1$ modulo $p$. This number in turn equals the number of maximal subgroups in the elementary abelian group $P/\Phi(P)$, which is congruent to $1$ modulo $p$. This completes the proof.

### Hands-on proof for a group of prime power order

Given: A group $P$ of prime power order $p^k$.

To prove: For any $r \le k$, the number of subgroups of order $p^r$ in $P$ is congruent to $1$ modulo $p$. Also, the number of normal subgroups of order $p^r$ in $P$ is congruent to $1$ modulo $p$.

Proof: First, note that if $H \le P$ is not normal, the number of conjugates of $H$ in $P$ equals $[P:N_P(H)]$, which is a power of $p$, and hence a multiple of $p$. Thus, the non-normal subgroups of order $p^r$ are a union of conjugacy classes whose orders are multiples of $p$. Thus, the number of non-normal subgroups of order $p^r$ is a multiple of $p$.

So, proving that the number of normal subgroups of order $p^r$ is $1$ modulo $p$ is equivalent to proving that the number of subgroups of order $p^r$ is $1$ modulo $p$.

Let's prove this, by induction first on $k$ and then on $k - r$. The case $k = r$ is clear. For the case $k - r = 1$, note that any subgroup of order $p^{k-1}$ is normal and contains the Frattini subgroup. More precisely, the subgroups of order $p^{k-1}$ correspond to subgroups of index $p$ in $P/\Phi(P)$.

The subgroup $P/\Phi(P)$ is elementary Abelian, and by an easy computation, the number of subgroups of index $p$ in this group is $1$ modulo $p$. Thus, the number of subgroups of order $p^{k-1}$ is $1$ modulo $p$.

It is also clear that when $k = 0$, the result is true -- there is only one subgroup of order $1$.

We now prove the result in general. We need to count the number of subgroups of order $p^r$ for $0 < r < k - 1$. For this, we instead count the number of pairs of subgroups $H \le K \le P$ where $|H| = p^r$ and $|K| = p^{r + 1}$.

This number can be counted in two ways. First, we could list all the subgroups of order $p^{r+1}$ in $P$, and for each of these, list the subgroups of index $p$. The number of subgroups of order $p^{r+1}$ is $1$ modulo $p$, and for each of them, the list of subgroups of index $p$ is $1$ modulo $p$ -- hence the total number of pairs is $1$ modulo $p$.

Second, we could count the number of pairs starting with $H$. When $H$ is not a normal subgroup, then its number of conjugates is a multiple of $p$, and for each of these conjugates, the number of pairs involving that conjugate is equal. Hence, the total number of pairs contributed by non-normal subgroups $H$ is a multiple of $p$.

For $H$ normal, the number of $K$ containing $H$, and of order $p^{r+1}$, equals the number of subgroups of order $p$ in $P/H$, which has smaller order. Hence, that number is $1$ mod $p$. Thus, the number of pairs is a sum of numbers, one for each normal subgroup, each of which is $1$ mod $p$. Since we earlier concluded that the number of pairs is $1$ mod $p$, this forces that the number of normal subgroups is $1$ mod $p$, in turn forcing that the number of subgroups is $1$ mod $p$.

### Proof for a finite group

Given: A finite group $G$, a prime power $p^r$ dividing the order of $G$.

To prove: The number of subgroups of order $p^r$ in $G$ is $1$ modulo $p$.

Proof: Let $P$ be a $p$-Sylow subgroup of $G$. Let $A$ be the set of all subgroups of $G$ of order $p^r$. Consider the action of $P$ on $A$ by conjugation: $g \in P$ sends a subgroup $H \in A$ to the subgroup $gHg^{-1}$.

Now, the orbits in $A$ under this action have sizes equal to powers of $p$. Thus, either a subgroup $H$ has an orbit of size one, or its orbit has size a multiple of $p$. It suffices to show that the number of elements whose orbits have size one, is $1$ modulo $p$.

For the orbit of $H$ to have size one is equivalent to saying that $P \le N_G(H)$. This implies that $PH$ is a subgroup, and if $H$ is not contained in $P$, the order of this subgroup is a prime power strictly greater than the order of $p$. This contradicts the assumption that $P$ is a $p$-Sylow subgroup, so $H \le P$. Thus, $H$ is a normal subgroup of $P$. Conversely, if $H$ is a normal subgroup of $P$, its orbit has size one.

Thus, the problem reduces to showing that the number of normal subgroups of $P$ of size $p^r$ for given $r$, is $1$ mod $p$ -- this is what we did in the proof for a group of prime power order.

## References

### Journal references

• A contribution to the theory of groups of prime power order by Philip Hall, Proceedings of the London Mathematical Society, ISSN 1460244X (online), ISSN 00246115 (print), Page 29 - 95(Year 1934): In this paper, Philip Hall summarizes and extends many of the known results about groups of prime power order.More info, Theorem 1.51: Hall proves the statement only for $p$-groups; his proof is similar to the proof presented on the wiki page.