Is there a group with exactly 92 elements of order 3?

Question

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.

Answer 1

Let $p,p' \equiv 2\pmod{3}$ be prime.

Suppose that $G$ is a group with the following properties:

(i) The $3$-Sylow subgroup of $G$ is cyclic;

(ii) The number of elements of order $3$ in $G$ is $2 p p'$.

Claim: There exists a group which in addition satisfies either:

(iiia) $G$ is simple; or

(iiib) $3\,||\, \#G$, and $G$ surjects onto $\mathbb{Z}/3\mathbb{Z}$.

Remark: Suppose that $G$ is a group with cyclic $3$-Sylow subgroups. Since all $3$-Sylow subgroups are conjugate, this implies that every subgroup of order $3$ in $G$ is also conjugate.

Proof: We may assume that $G$ is not simple, and hence admits a proper normal subgroup $H$.

Suppose that $H$ is a normal subgroup of $G$. If $3$ divides $\# H$, then $H$ contains a subgroup of $G$ of order $3$. All such subgroups are conjugate in $G$, and since $H$ is normal, they all lie in $H$. Thus $G$ and $H$ have the same number of elements of order $3$. Moreover, the $3$-Sylow subgroup of $H$ is clearly cyclic, and so we may replace $G$ by $H$.

Suppose that $(\# H,3) = 1$. Then $G/H$ still has a cyclic $3$-Sylow subgroup, and hence every element of order $3$ in $G/H$ is conjugate. This implies that every element of order $3$ in $G/H$ lifts to an element of order $3$ in $G$. Thus $G/H$ has at most $2 p p'$ elements of order three. Yet by a theorem of Frobenius, the number $N$ of $g\in G$ of order exactly $3$ is divisible by $\phi(3) = 2$, and the number $N + 1$ of elements of order dividing $3$ is divisible by $3$. Hence $N\equiv 2\pmod{6}$. Thus the number of elements of $G/H$ of order $3$ is either $2$ or $2pp'$. In the latter case, we replace $G$ by $G/H$.

We may now assume that $G/H$ has exactly $2$ elements of order $3$. Clearly $G/H$ has a unique subgroup of order $3$, which must be normal. Thus $G/H$ and $G$ have a quotient of order $\frac{1}{3}\#(G/H)$, and thus $G$ contains a normal subgroup $F$ of order $3\#H$. As above, we may replace $G$ by $F$. Note, however, that $3$ exactly divides $\#F$, and $F$ surjects onto $\mathbb{Z}/3\mathbb{Z}$, and thus, by Schur-Zassenhaus (overkill in this case, of course), $F$ is a semi-direct product.

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My understanding of Jack's argument:

Suppose that $p' = 2$, so $G$ has $4p$ elements of order $3$. We still assume that the $3$-Sylow of $G$ is cyclic. $G$ acts by conjugation on the $2p$ subgroups of order $p$, giving a map $G\to S_{2p}$. Let $Q$ be one of these subgroups, and let $M$ be the normalizer of $Q$. Let $P$ be a $3$-Sylow containing $Q$. Certainly $[G:M] = 2p$, by the orbit-stabilizer formula.

If $X\subset G$ contains $M$, then $[G:X] = 1,2,p, \text{ or } 2p$. Let $N$ be the normalizer of $P$. Clearly $M\supset N$. Thus $P\subset M$, and thus $P$ is a $3$-Sylow of $M$. Similarly, $P$ is also a $3$-Sylow subgroup of $X$. The Sylow theorems applied to $M$ and $X$ thus imply that $[X:N], [M:N] \equiv 1\pmod{3}$, and thus $[X:M]\equiv 1\pmod{3}$. Hence, since $[X:M] = 1,2,p\text{ or }2p$, and, since $p \equiv 2\pmod{3}$, either $X = M$ or $X = G$. Thus $M$ is a maximal subgroup of $G$, and hence the action of $G$ is primitive.

Now we suppose:

Assumption (*): The only primitive subgroups of $S_{2p}$ are $A_{2p}$ and $S_{2p}$.

Then, we deduce that some quotient of $G$ is either $A_{2p}$ or $S_{2p}$. If $G$ is simple, we deduce that $G = A_{2p}$, which doesn't have cyclic $3$-Sylows if $2p > 5$. If $G$ is a semi-direct product of $\mathbb{Z}/3\mathbb{Z}$ with a group of order coprime to $3$, then $G$ cannot surject onto $A_{2p}$ or $S_{2p}$ if $p > 2$. It seems to follow that:

If $p \equiv 5,8 \pmod{9}$ and Assumption (*) holds, then $G$ cannot have $4p$ elements of order $3$. (The congruence conditions on $p$ ensure that the $3$-Sylow of $G$ is cyclic).

Answer 2

Here is my attempt, fixed by Septimus Harding:

If H is a group with 92 elements of order 3, then let G be the subgroup generated by those elements. G has finite order by Dicman's lemma, and so G has a Sylow 3-subgroup P.

P must be cyclic, lest Hall's result imply 46≡4 mod 9. Since P is cyclic, 46≡1 mod |P|, so |P| divides 45, so |P| is 1, 3, or 9. 1 is impossible since there would be 0 elements of order 3. 3 is impossible, since then G has 46 Sylow 3-subgroups but no finite group has 46 Sylow p-subgroups. So P is cyclic of order 9.

Now G acts by conjugation on the 46 subgroups of order 3, and since the Sylow 3-subgroups are cyclic and (all of) their order 3-subgroups are conjugate, G acts transitively on the 46 subgroups. Let Q = Ω(P) ≤ P be one of the subgroups of order 3, and let M = N_G(Q) be its normalizer. Then N = N_G(P) ≤ M and [G:M] = 46. If M ≤ X ≤ G, then both [M:N] and [X:N] are ≡1 mod 3 by Sylow, but [X:N] = [X:M][M:N], so [X:M] ≡ 1 mod 3. However [X:M] divides 46, so [X:M] = 1 or 46, so X = M or G, and so M is a maximal subgroup of G. So G acts primitively on 46 points (the cosets of N, aka, the subgroups of order 3), so G/Core(G,M) is a primitive group of degree 46, that is, G/Core(G,M) is the alternating or the symmetric group on 46 points. Since the Sylow 3-subgroups of the alternating and symmetric groups (on 6 or more points) are not cyclic, we have a contradiction, and no such G exists.

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This works to prove that: if n=pq for primes p,q ≡ {5,8} mod 9, and if the only primitive groups of degree pq have non-cyclic Sylow 3-subgroups (like Alt(n) and Sym(n)), then there is no group with exactly 2n elements of order 3, even though 2n ≡ 2 mod 6. I believe primitive groups of degree pq are classified, so this could probably be made completely explicit. 140 elements is not immediately handled, since 70 = 2*5*7, but I suspect the same idea will work.