Show that a group of order \(30\) must have a subgroup of order \(15\).
Solution
Let \(G\) be a group of order \(30\).
1. At least one of the Sylow \(3\)- and Sylow \(5\)-subgroupsis normal in \(G\).
Sylow’s theorems give \[ n_3\mid10,\quad n_3\equiv1\pmod3, \qquad n_5\mid6,\quad n_5\equiv1\pmod5 \] [@DF04]. Among the divisors \(1,2,5,10\) of \(10\), only \(1,10\) satisfy the first congruence. Among \(1,2,3,6\), only \(1,6\) satisfy the second. Thus \(n_3\in\{1,10\}\) and \(n_5\in\{1,6\}\).
If both counts exceeded \(1\), the ten subgroups of order \(3\) would supply \(10(3-1)=20\) distinct elements of order \(3\), and the six subgroups of order \(5\) would supply \(6(5-1)=24\) distinct elements of order \(5\). Distinct subgroups of the same prime order intersect only in the identity, since any common nonidentity element generates both. An element cannot have both orders \(3\) and \(5\). Hence these are \(44\) distinct nonidentity elements, impossible in a group of order \(30\). At least one count is \(1\); its unique Sylow subgroup is normal because conjugation preserves subgroup orders.
2. The product of a Sylow \(3\)-subgroup and a Sylow\(5\)-subgroup is a subgroup of order \(15\).
Choose Sylow subgroups of orders \(3\) and \(5\). By step 1, at least one is normal; call a normal one \(N\) and the other \(H\). Their intersection is trivial, since its order divides both \(3\) and \(5\).
The set \(NH\) is a subgroup. It contains the identity, and normality of \(N\) gives \[ (n_1h_1)(n_2h_2)=n_1(h_1n_2h_1^{-1})h_1h_2\in NH, \qquad (nh)^{-1}=(h^{-1}n^{-1}h)h^{-1}\in NH. \] Multiplication \(N\times H\to NH\) is surjective by definition and injective: \(n_1h_1=n_2h_2\) implies \(n_2^{-1}n_1=h_2h_1^{-1}\in N\cap H=\{1\}\). Therefore \(|NH|=|N||H|=3\cdot5=15\).