We can determine exactly which positive numbers occur as the number of Sylow p-subgroups of finite groups, at least modulo some detailed information about simple groups. What can we say about the number of subgroups of order p?

Cauchy’s theorem says that a finite group of order divisible by p has at least one subgroup of order p.

**Is there any restriction on the number τ _{p}(G) of subgroups of order p contained in G?**

If τ

_{p}(G) = n, then the conjugation action of G on its subgroups gives a homomorphism from G to Sym(n) with kernel K consisting of the elements of G that normalize every subgroup of order p. If n ≤ p, then in fact every subgroup of order p is contained in K. If n is strictly less than p this is obvious as Sym(n) has order coprime to p. If n=p, then ever element of order p is “regular”, so that it fixes no element. If P is one of the subgroups of order P, then it normalizes itself, and so maps to the identity element in Sym(p), and so is contained in K. Since τ

_{p}(G) = τ

_{p}(K), we might as well assume G = K, and that every subgroup of order p is normal. The subgroup generated by these normal p-subgroups is a normal p-subgroup with the same value of τ

_{p}, so again we can assume G is a p-group. However, in a p-group every normal subgroup of order p is central, and so Z(G) is another group with the same τ

_{p}and we can assume G is abelian. In fact, the elements of order p form a subgroup of Z(G) with the same τ

_{p}, so we might as well assume G is a vector space over Z/pZ and we can easily count its subgroups of order p to get τ

_{p}(G) = (|G|−1)/(p−1) which is either 1 or at least as big as p+1, and so must be 1 if n ≤ p.

Now of course there is a finite group G with τ_{p}(G) = p+1, namely PSL(2,p). So we have a very clear classification of the small numbers that occur as τ_{p}(G): only 1 and p+1 amongst numbers up to p+1. If q is a prime power congruent to 1 mod p, then τ_{p}(AGL(1,q)) = q ≡ 1 mod p.

Direct products alter τ_{p} in a strange way. The number of solutions of x^{p} = 1 in G is denoted θ_{p}(G). Since (x,y)^{p} = (x^{p},y^{p}), it is clear that θ_{p}(G×H) =θ_{p}(G)⋅θ_{p}(H). Since p is prime, τ_{p}(G) = (θ_{p}(G)−1)/(p−1). If τ_{p}(G) = g, τ_{p}(H) = h, and τ_{p}(G×H) = k, then k⋅(p−1)+1 = (g⋅(p−1)+1)⋅(h⋅(p−1)+1), so k = g⋅h⋅(p−1) + g + h. Note that if both g and h are congruent to 1 mod p, then so is k.

It seems reasonable to ask if one must have 1 ≡ τ_{p}(G) mod p.

**Is there a congruence condition on τ _{p}?**

Frobenius (1895) showed that if the order of the group is divisible by p, then the number of subgroups of order p is in fact equal to 1 mod p, just like for Sylow p-subgroups. An elementary proof similar to “class equation” type proofs is given by Isaacs and Robinson (1992).

One could ask if this is sharp:

**Is every number congruent to 1 mod p a τ _{p}?**

For p=2, things are again quite nice. The number of subgroups of order 2 in a finite group of even order must be odd, and in the dihedral group of order 4n+2, there are exactly 2n+1 subgroups of order 2. Hence the numbers that can be the number of subgroups of order 2 in a finite group are exactly 0 and the odd numbers.

For p=3, things again go badly fairly quickly. While 1, 4, 7, 10, 13, 16, 19, 22, 25, 28, and 31 are all possible numbers of subgroups of order 3 in a finite group, 34 is not. Note how the problem is already a little surprising. No finite group can have 22 Sylow p-subgroups, but it can have 22 subgroups of order p (for p=3 only).

**Do congruences on τ _{p} tell us anything about the group?**

In fact the number τ has some non-obvious interactions with the structure of the group. If τ

_{2}(G) = 1, then a Sylow 2-subgroup P is either cyclic or generalized quaternion and Ω(Z(P)) ≤ Z(G). In fact 1 ≡ τ

_{p}(P) mod 4 even iff P is either cyclic, generalized quaternion, dihedral, or semi-dihedral, a result of Alperin–Feit–Thompson and Taussky, according to page 52 of Isaacs’s CToFG. Gorenstein’s Finite Groups §5.4 contains information on p-groups with few subgroups of order p, including the fact that if τ

_{p}(P) = 1 and P is a p-group of odd order, then P is cyclic. The congruence result for odd p is actually an older result of P. Hall (1936): If the Sylow p-subgroups of G are cyclic of order p

^{n}, then 1 ≡ τ

_{p}(G) mod p

^{n}, and otherwise 1+p ≡ τ

_{p}(G) mod p

^{2}. Slightly better results due to MacWilliams are available if the Sylow p-subgroup is known to have large Frattini rank, but I haven’t found much use for them as the types of groups with such Sylows are unwieldy.

We can use these sorts of results to rule out possible values of τ_{p}. For instance if τ_{5}(G) in { 21, 26, 46, 51, 76, 86, 91 }, then a Sylow 5-subgroup P of G must be cyclic. Since G acts transitively on its Sylow 5-subgroups, and each one has exactly one subgroup of order 5, G acts transitively on its subgroups of order 5. Looking at the indices of N_{G}(Ω(P)) ≤ N_{G}(P) ≤ M ≤ G, we see that N_{G}(Ω(P)) = N_{G}(P) is maximal (for instance with τ=21, [M:N_{G}(P)] is a divisor of [G:N_{G}(Ω(P))] = 21 that is congruent to 1 mod 3). As in M. Hall’s result on Sylow p-subgroups of quotients, one finds that the quotient group H of G mod the normal core of N_{G}(P) also has τ_{5}(H) = τ_{5}(G). Hence we are looking for primitive groups of given degree and given τ_{5}, but a quick glance at the tables of such groups shows none exist.

However, when 1+p ≡ τ_{p} mod p^{2}, things are much trickier. There is a group with 22 subgroups of order 3. I am not sure about 106 subgroups of order 5.

**What about infinite groups?**

We could get greedy and ask about infinite groups too. Obviously infinity is another possibility, but do we suddenly get any more even possibilities? The answer is no, but relies on a result of Dietzmann (1937) that the subgroup generated by a finite normal subset is finite. In other words, if a group has a finite number of subgroups of order p, then the subgroup generated by all of those subgroups is also finite (and obviously has exactly the same number of subgroups of order p). This means we can state our results for possibly infinite groups, but our proofs only need to handle finite groups.

**θ versus τ**

Since θ behaves so nicely with direct products and τ behaves so nicely with respect to example groups, it is hard to tell which is the better invariant to study. The possible values T of θ will form a multiplicative submonoid of pZ. However, this misses the fact that { t – 1 : t in T } is contained in (p−1)Z, that is, that τ is an integer.

**Generating more τ numbers**

Is there a “generating set” for the τ numbers? Do direct products of AGL suffice for solvable groups? Do simple groups suffice for the rest? How does one compute τ of a p-nilpotent group?