Let \(|G|=240=2^4\cdot3\cdot5\).
- Determine the Sylow-number possibilities for \(p=2,3,5\).
- Show that a subgroup of order \(15\) is cyclic.
- If \(G\) has no subgroup of order \(15\), show that \(n_3\in\{10,40\}\).
Solution
Sylow’s theorem gives \[ n_2\mid15,\quad n_2\equiv1\pmod2, \] so \[ n_2\in\{1,3,5,15\}. \] Also \[ n_3\mid80,\quad n_3\equiv1\pmod3, \] so \[ n_3\in\{1,4,10,16,40\}, \] and \[ n_5\mid48,\quad n_5\equiv1\pmod5, \] so \[ n_5\in\{1,6,16\}. \]
Now let \(H\le G\) have order \(15\). In \(H\), the Sylow \(5\)-subgroup is unique because its number divides \(3\) and is \(1\) modulo \(5\). The Sylow \(3\)-subgroup is also unique because its number divides \(5\) and is \(1\) modulo \(3\). Hence both are normal, commute, and have trivial intersection. Therefore \[ H\cong C_5\times C_3\cong C_{15}. \]
Finally suppose \(G\) has no subgroup of order \(15\). Let \(P\) be a Sylow \(3\)-subgroup. If \(5\mid|N_G(P)|\), Cauchy’s theorem gives an element \(y\in N_G(P)\) of order \(5\). Then \(P\trianglelefteq\langle P,y\rangle\) and \[ |\langle P,y\rangle|=15, \] a contradiction. Thus \[ 5\nmid |N_G(P)|=\frac{240}{n_3}. \] Therefore \(5\mid n_3\). Among \[ \{1,4,10,16,40\}, \] this leaves exactly \[ n_3\in\{10,40\}. \]