How do they know that the Sylow 3 subgroups are cyclic?

Question

I have this exercise:

Find all Sylow 3 subgroups of $S_4$ and demonstrate that they are all conjugate.

This is the answer given in the manual:

Now, Sylows first theorem says that every Sylow 3 subgroup must have order 3. And by the definition of a p-subgroup every element must have an order that is a power of 3. So I solved with writing up all 24 possibilities we see that the only possibilities we are left with are the 3 cycles.

But can you see this before you start without writing every possibility, like they did in the solutions manual? Is it possible to see that the elements in the Sylow 3 subgroup in this case must be cyclic, and consist of the 3 cycles, without checking each case?

Answer 1

Any group with prime order is cyclic. Suppose $|G|=p$ with $p$ prime. Let $a\in G$ and consider the subgroup $H$ generated by $a$. Since the order of $H$ must divide the order of $G$, $|H|=1$ or $p$. If $|H|=1$, the $a$ is the identity element. If $|H|=p$, then $H=G$ and $G$ is cyclic.

Answer 2

Consider a generic Sylow-$3$ subgroup in $S_4$ which has order $3$ say $\{e,a,b\}$ where $e$ denotes the identity. Notice that $a^2 =b$ then $b^2=e \Rightarrow a^3 = a^2a = ba = e \Rightarrow A = \langle a \rangle$; hence $A$ must be generated by a $3$-cycle.