A group has either 0, ∞, or an odd number of involutions, and every such number occurs. A group either has 0, ∞, or (p−1) mod 2p elements of order p, p odd. Does every such number occur?
In AoPS–group elements of order three, gammaduc asked to prove that every number congruent to 2 mod 6 is the number of elements of order 3 in some group.
I’m pretty sure there is no group with exactly 68 elements of order 3. However, 2 ≡ 68 mod 6.
A minimal such group is finite: By Dicman’s lemma, the (finitely many) elements of order 3 generate a finite subgroup, and that subgroup has the same number of elements of order 3.
Such a group has cyclic Sylow 3-subgroups: If the Sylow 3-subgroup is not cyclic, then Hall’s 1936 (“Hall enumeration”) shows that the number of subgroups of order 3 is congruent to 4 mod 9, and so the number of elements of order 3 is congruent to 8 mod 18. Since 14 ≡ 68 mod 18, non-cyclic Sylows give a contradiction.
Such a group has order divisible by 9: Otherwise the group has 34 Sylow 3-subgroups, but no finite group has 34 Sylow p-subgroups.
Let G act on the 34 subgroups of order 3 by conjugation. A (cyclic) Sylow 3-subgroup P acts on them as well, and cannot normalize any but one, lest it contain it. Let Q be a subgroup of order 3, not contained in P. Let R be the stabilizer in P of Q. Then R normalizes Q, and RQ is a 3-group, so contained in some Sylow 3-subgroup, so cyclic, so is either Q or R (depending on whether R=1 or not). Since Q is not contained in P, but R is, we must have R=1, and P acts semi-regularly on the other subgroups of order 3. Since P has order at least 9, this means 1 ≡ 34 mod 9, but in fact 7 ≡ 34 mod 9, a contradiction.
The number of elements of order 2 in a group is fairly restricted: 0, odd, or infinity. All such possibilities occur already in the trivial group and in dihedral groups.
The number of elements of order 3 in a group can be shown to be similarly restricted: 0, 2 mod 6, or infinity. However, something strange happens: not all possibilities can be realized.
Even worse, there is a fairly small number that I cannot decide whether it is or is not the number of elements of order 3 in a finite group:
Is there a group with exactly 92 elements of order 3?
More boldly, I would like to know (but feel free to answer only the first question):
Exactly which numbers occur as the number of elements of order 3 in a group?
Background: Such questions were studied a bit by Sylow and more heavily by Frobenius. The theorem that the number of elements of order p is equal to −1 mod p is contained in one of Frobenius’s 1903 papers. Since elements of order 3 come in pairs, this doubles to give 2 mod 6 for p=3.
However, Frobenius’s results were improved some 30 years later by P. Hall who showed that if the Sylow p-subgroups are not cyclic, then the number of elements of order p is −1 mod p2.
If the Sylows are cyclic of order pn, then the number of subgroups of order p is congruent to 1 mod pn by the standard counting method. If the Sylow itself is order p, then the subgroup generated by the elements of order p acts faithfully and transitively on the Sylow subgroups, so for small enough numbers, the subgroup can just be looked up.
In all cases, we can assume the group is finite since the subgroup generated by the elements of a fixed order is finite (assuming there are only finitely many elements of that fixed order).
Easier example: For instance there is no group with exactly 68 elements of order 3, since such a group would have cyclic Sylow 3-subgroups by Hall, order 3 Sylows by the counting, but then would have 34 Sylow 3-subgroups, and so (the subgroup generated by the elements of order 3 would) be a primitive group of degree 34. One checks the list of primitive groups of degree 34 (that is, A34 and S34, both with ginormous Sylow 3-subgroups) to see no such group exists.
One could also try 140, but the action need not be primitive so the table lookup is harder. Such a group has Sylows of order 3, but is not solvable, so is somewhat restricted.
Some references:
- not all numbers are Sylow numbers, OEIS has a list.
- Dicman’s lemma is in Robinson’s CitToG, 14.5.7, page 425 in the 1st edition. A subgroup generated by a finite normal subset is itself finite.
- Hall’s 1936 paper is doi:10.1112/plms/s2-40.1.468, and the most quotable version is theorem 4.6 page 500.
- Berkovich (book and 1994 Arch. Math.) and Mann (unpublished?) have a similar result to Hall for p=2.
- The argument for the “counting” part is how Isaacs’s FGT 1.16 simplifies when applied subgroups of order p, instead of Sylows.
Here is a proof that no group has exactly 68 elements of order 3:
Let H be a group with exactly 68 elements of order 3.
Let G be the subgroup of H generated by the elements of order 3.
G is finite by Dicman’s lemma (Robinson, CTG, 14.5.7, page 425ish).
G has exactly 34 subgroups of order 3 (each subgroup contains 2 elements of order 3, each element of order 3 is a generator).
The Sylow 3-subgroups of G are cyclic (Hall 1936 Th 4.6: otherwise the number of subgroups is congruent to 4 mod 9, but 34 is congruent to 7 mod 9).
The Sylow 3-subgroups have order 3 (Let P be a Sylow 3-subgroup. Consider the action of P by conjugation on the 34 subgroups Q of order 3. Let R ≤ P be the stabilizer of Q in P, that is, R = N_P(Q). RQ is a 3-group, so RQ is contained in some Sylow 3-subgroup, so RQ is cyclic. Hence either R ≤ Q or Q ≤ R. Q ≤ P iff Q ≤ R, since P is cyclic and Q ≠ 1. If Q is not a subgroup of P, then R ≤ Q ∩ P = 1, since P contains exactly one subgroup of order 3. In summary: P acts semi-regularly on the 33 subgroups of order 3 not contained within it. Hence |P| divides 33, so |P| = 3.) (This is very similar to Isaacs FGT 1.16)
Contradiction, no group has 34 Sylow 3-subgroups (Let N = N_G(P) be the Sylow normalizer. If N ≤ M ≤ G, then N is also a Sylow normalizer of M, so [M:N] ≡ 1 mod 3. However, [G:N] = 34 has very few divisors: 1, 2, 17, 34, and the only ones that are ≡ 1 mod 3 are 1 and 34, so M=N or M=G. Hence N is a maximal subgroup, and the action of G on its Sylow 3-subgroups is not only transitive but primitive. There are exactly 2 primitive groups of degree 34, the alternating and the symmetric group. Both have more than 34 Sylow 3-subgroups, and of course both have Sylow 3-subgroups much larger than order 3.)