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.
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Statement
Version for a group of prime power order
Let be a group of prime power order and suppose . The following are true:
- The number of subgroups of of order is congruent to modulo .
- If is itself a power of , i.e., if is a group of prime power order, then the number of normal subgroups of of order is congruent to 1 mod .
- If is itself a power of , i.e., if is a group of prime power order, then the number of p-core-automorphism-invariant subgroups of of order is congruent to 1 mod .
In particular, the collection of groups of order is a collection of groups satisfying a universal congruence condition.
Version for a general finite group
Let be a finite group and be a prime power dividing the order of . Then, the number of subgroups of of order is congruent to 1 mod .
Related facts
Stronger facts
Corollaries
- Congruence condition on number of subgroups of given prime power order satisfying any given property weaker than normality
- Congruence condition on number of normal subgroups with quotient in a specific variety in a group of prime power order
- Congruence condition on number of subgroups of given prime power order and bounded exponent in abelian group: In an abelian group of prime power order, the number of subgroups of a given order and bounded exponent is either zero or congruent to one modulo .
- Jonah-Konvisser abelian-to-normal replacement theorem: Jonah and Konvisser establish a congruence condition on the number of abelian subgroups of order for odd primes and small values of .
- Jonah-Konvisser elementary abelian-to-normal replacement theorem: Jonah and Konvisser establish a congruence condition on the number of elementary abelian subgroups of order for odd primes and small values of .
- Equivalence of definitions of universal congruence condition
- Congruence condition fails for subgroups of given prime power order and bounded exponent
Facts used
Proof
Equivalence between congruence condition on subgroups and on normal subgroups
If is not normal, the number of conjugates of in equals , which is a power of , and hence a multiple of . Thus, the non-normal subgroups of order are a union of conjugacy classes whose orders are multiples of . Thus, the number of non-normal subgroups of order is a multiple of .
Thus, the total number of subgroups of order equals the number of subgroups of order modulo .
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 of order .
To prove: For any , the number of subgroups of order is congruent to modulo .
Proof: If , the statement is vacuously true, otherwise, the given fact yields:
.
By induction, we know that for each maximal subgroup of , the number of subgroups of is congruent to modulo . Thus, it suffices to show that the number of maximal subgroups is itself congruent to modulo . This number in turn equals the number of maximal subgroups in the elementary abelian group , which is congruent to modulo . This completes the proof.
Hands-on proof for a group of prime power order
Given: A group of prime power order .
To prove: For any , the number of subgroups of order in is congruent to modulo . Also, the number of normal subgroups of order in is congruent to modulo .
Proof: First, note that if is not normal, the number of conjugates of in equals , which is a power of , and hence a multiple of . Thus, the non-normal subgroups of order are a union of conjugacy classes whose orders are multiples of . Thus, the number of non-normal subgroups of order is a multiple of .
So, proving that the number of normal subgroups of order is modulo is equivalent to proving that the number of subgroups of order is modulo .
Let's prove this, by induction first on and then on . The case is clear. For the case , note that any subgroup of order is normal and contains the Frattini subgroup. More precisely, the subgroups of order correspond to subgroups of index in .
The subgroup is elementary Abelian, and by an easy computation, the number of subgroups of index in this group is modulo . Thus, the number of subgroups of order is modulo .
It is also clear that when , the result is true -- there is only one subgroup of order .
We now prove the result in general. We need to count the number of subgroups of order for . For this, we instead count the number of pairs of subgroups where and .
This number can be counted in two ways. First, we could list all the subgroups of order in , and for each of these, list the subgroups of index . The number of subgroups of order is modulo , and for each of them, the list of subgroups of index is modulo -- hence the total number of pairs is modulo .
Second, we could count the number of pairs starting with . When is not a normal subgroup, then its number of conjugates is a multiple of , 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 is a multiple of .
For normal, the number of containing , and of order , equals the number of subgroups of order in , which has smaller order. Hence, that number is mod . Thus, the number of pairs is a sum of numbers, one for each normal subgroup, each of which is mod . Since we earlier concluded that the number of pairs is mod , this forces that the number of normal subgroups is mod , in turn forcing that the number of subgroups is mod .
Proof for a finite group
Given: A finite group , a prime power dividing the order of .
To prove: The number of subgroups of order in is modulo .
Proof: Let be a -Sylow subgroup of . Let be the set of all subgroups of of order . Consider the action of on by conjugation: sends a subgroup to the subgroup .
Now, the orbits in under this action have sizes equal to powers of . Thus, either a subgroup has an orbit of size one, or its orbit has size a multiple of . It suffices to show that the number of elements whose orbits have size one, is modulo .
For the orbit of to have size one is equivalent to saying that . This implies that is a subgroup, and if is not contained in , the order of this subgroup is a prime power strictly greater than the order of . This contradicts the assumption that is a -Sylow subgroup, so . Thus, is a normal subgroup of . Conversely, if is a normal subgroup of , its orbit has size one.
Thus, the problem reduces to showing that the number of normal subgroups of of size for given , is mod -- 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 -groups; his proof is similar to the proof presented on the wiki page.