Suppose \(|G| = 240 = 2^4 \cdot 3 \cdot 5\).
How many Sylow-\(p\) subgroups does \(G\) have for \(p\in \{2, 3, 5\}\)?
Show that if \(G\) has a subgroup of order 15, it has an element of order 15.
Show that if \(G\) does not have such a subgroup, the number of Sylow-\(3\) subgroups is either 10 or 40.
Solution
Write \(n_p\) for the number of Sylow \(p\)-subgroups of \(G\).
By Sylow’s theorem, \[ n_2\mid15,\quad n_2\equiv1\pmod2, \] so \[ \boxed{n_2\in\{1,3,5,15\}}. \] Likewise, \[ n_3\mid80,\quad n_3\equiv1\pmod3, \] so \[ \boxed{n_3\in\{1,4,10,16,40\}}, \] and \[ n_5\mid48,\quad n_5\equiv1\pmod5, \] so \[ \boxed{n_5\in\{1,6,16\}}. \]
Now let \(H\le G\) have order \(15\). In \(H\), \[ n_5(H)\mid3,\quad n_5(H)\equiv1\pmod5, \] so \(n_5(H)=1\), and \[ n_3(H)\mid5,\quad n_3(H)\equiv1\pmod3, \] so \(n_3(H)=1\). Thus both Sylow subgroups are normal, and \[ H\cong C_3\times C_5\cong C_{15}. \] Hence \(G\) contains an element of order \(15\).
Finally suppose \(G\) has no subgroup of order \(15\). Fix \(Q\in\operatorname{Syl}_5(G)\) and let \(Q\) act by conjugation on \(\operatorname{Syl}_3(G)\). Every orbit has size \(1\) or \(5\). A fixed Sylow \(3\)-subgroup \(P\) would be normalized by \(Q\), so \(PQ\) would be a subgroup of order \(15\), contrary to hypothesis. Hence every orbit has size \(5\), so \(5\mid n_3\). From the candidate list above, \[ \boxed{n_3\in\{10,40\}}. \]