Question about p-Sylow subgroups being maximal

Question

I am having some trouble wrapping my head around p-Sylow subgroups at the moment. I am given that for p, prime, a p-Sylow subgroup of G is a maximal p-subgroup that is not a proper subgroup of any other p-subgroup of G

Give this defintion, which states that these subgroups are maximal, how can there be more than one p-Sylow subgroup for a given prime p unless they are all the same?

Answer 1

An example may clarify. Clearly, a Sylow $2$-subgroup of $S_3$ must be cyclic of order $2$. But there are several such subgroups, each generated by a single transposition. There are three such subgroups, namely $\{e, (12)\}$, $\{e, (13)\}$ and $\{e, (23)\}$. All of them are isomorphic, of course.

In general, the conjugate of any subgroup $H$ is a subgroup isomorphic to $H$ (so if $H$ were Sylow $p$, so is its conjugate). There is no reason for this conjugate to equal $H$ necessarily.

Answer 2

I would take almost the same exemple than Randall,in S4 (symmetric group) You have that 2 is a divisor of 24 (order of S4) and you have multiple 2 Sylows. Another way to convince you would be: try to show that if G admits a p-sylow that is not normal, then G must have at least 2 of them.