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

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.

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).

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.

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.