Congruence condition on number of subgroups of given prime power order

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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.
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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:

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.

Related facts

Stronger facts

Corollaries

Other related facts

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