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).
If $G$ has $|G|$ many elements, then the set
$$\{a, a^2, a^3, \cdots, a^{|G|}\}$$
either has a repetition or exhaust the group. If it exhaust the group, one of these elements is $1$. If there is a repetition, so $a^{r} = a^{s}$ for $r>s$, then $a^{r-s} = 1$. In any event, at least one of these elements must equal $1$, say $a^n = 1$, and choose $n$ to be the smallest such exponent. If we can show that $n$ divides $|G|$, we are done. But $n$ is the order of the subgroup
$$\{a, a^2, \cdots , a^n(=1)\}$$
And by Lagrange's theorem, the order of a subgroup must divide the order of the group.
To rephrase here a bit: If $a$ is a generator of $G$, then raising it to the power of the order of the group guarantees that it will cycle through all the elements and return to the identity. If $a$ is not a generator of $G$ then Lagrange's theorem guarantees that the order of the subgroup generated by $a$ divides $|G|$ and therefore if $a^{|a|}=1$ then $(a^{|a|})^{u}=1$ where $|G|=u|a|$.
In response to your questions about definitions: A group is an abstract object. We do not know anything about the sorts of objects inside the group. All we know about the group is that it satisfies certain axioms and has a (binary) group operation $\ast$. Given a pair of elements $a,b \in G$, we write $a \ast b$ or $ab$ to denote the group operation acting on the pair of elements. The exponent notation means: $a^{2} = aa, a^{7} = aaaaaaa$.
Best Answer
An element $g \in G$ has order $n$ if $g^n = e$ ($n$ is the smallest positive integer for which this is true). Where $e$ is the identity. See this wikipedia articles for more detail.