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).
Best Answer
I urge you to read this entry, it is all explained there. From the proof (1959) of James McKay of Cauchy's Theorem (the existence of an element of prime power order $p$, if $p$ divides the order of $G$) (a) follows.
If there are $l$ subgroups of order $p$, then $l=1$ or any pair of different subgroups of order $p$ have trivial intersection (apply Lagrange's Theorem on the intersection, being a subgroup of each of the subgroups of order $p$).
Hence, leaving out the identity element, there are $l(p-1)$ elements of order $p$, yielding (b). Combining (a) and (b) gives $l \equiv 1$ mod $p$.