If \(G\) is a finite group, and \(p\) is a prime number, and \(p^k\) is the largest power of \(p\) dividing \(\order{G}\text{,}\) then a Sylow-\(p\) subgroup of \(G\) is a subgroup of \(G\) with order \(p^k\text{.}\)
This definition allows \(k\) to be \(0\text{,}\) so \(p\) might not divide \(\order{G}\text{.}\) There is a theorem that gives information about how many Sylow-\(p\) subgroups a finite group might have.
If \(G\) is a finite group and \(p\) is a prime dividing \(\order{G}\text{,}\) then the number of Sylow-\(p\) subgroups of \(G\) divides \(\order{G}\) and is congruent to \(1\) modulo \(p\text{.}\)
Write \(n_3\) to represent the number of Sylow-\(3\) subgroups. We factor \(45=9\cdot5\text{,}\) separating the maximal power of \(3\) from its complement. Since \(n_3\) divides \(45\) and \(n_3\equiv1\) modulo \(3\text{,}\) it follows that \(n_3\) divides \(5\text{.}\) So either \(n_3=1\) or \(n_3=5\text{.}\) But of those two options, only \(1\) is congruent to \(1\) mod \(3\text{.}\) So \(G\) must have \(1\) Sylow-\(3\) subgroup.
Write \(n_7\) to represent the number of Sylow-\(7\) subgroups. We factor \(63=9\cdot7\text{,}\) separating the maximal power of \(7\) from its complement. Since \(n_7\) divides \(63\) and \(n_7\equiv1\) modulo \(7\text{,}\) it follows that \(n_7\) divides \(9\text{.}\) So either \(n_7=1\text{,}\)\(n_7=3\text{,}\) or \(n_7=9\text{.}\) But of those, only \(1\) is a real possibility, since only \(1\) is congruent to \(1\) mod \(7\text{.}\) So \(G\) has \(1\) Sylow-\(7\) subgroup.
Write \(n_3\) to represent the number of Sylow-\(3\) subgroups. We factor \(63=9\cdot7\text{,}\) separating the maximal power of \(3\) from its complement. Since \(n_3\) divides \(63\) and \(n_3\equiv1\) modulo \(3\text{,}\) it follows that \(n_3\) divides \(7\text{.}\) So either \(n_3=1\) or \(n_3=7\text{.}\) Both of those are real possibilities, since both are congruent to \(1\) mod \(3\text{.}\) So \(G\) either has \(1\) Sylow-\(3\) subgroup or it has \(7\text{.}\)
