Groups of size 15 are solvable; no simple groups of size 36

Let \(|G|=15=3\cdot5\). The number \(n_5\) of Sylow \(5\)-subgroups divides \(3\) and is \(1\) modulo \(5\), by Sylow’s theorems [@DF04]. Of the divisors \(1,3\) of \(3\), only \(1\) satisfies that congruence. Thus there is a unique subgroup \(P\) of order \(5\); conjugation preserves its order, so \(P\lhd G\).

The group \(P\) is cyclic of order \(5\), and \(G/P\) is cyclic of order \(3\), since a group of prime order is generated by any of its nonidentity elements. Therefore \[ G\triangleright P\triangleright\{1\} \] is a normal series with abelian factors, proving solvability. Equivalently, the commutator subgroup \(G'\) lies in \(P\) because \(G/P\) is abelian, and then \(G''\subseteq P'=\{1\}\) because \(P\) is abelian. This also proves solvability directly from the derived-series definition.

Let \(|G|=36=2^2\cdot3^2\). Sylow’s theorems give \(n_3\mid4\) and \(n_3\equiv1\pmod3\), so \(n_3=1\) or \(4\). If \(n_3=1\), the unique subgroup of order \(9\) is normal and is nontrivial and proper.

Suppose \(n_3=4\). Conjugation on the four Sylow \(3\)-subgroups gives a homomorphism \(\rho:G\to S_4\). Its action is transitive by Sylow conjugacy, so it is not the trivial action and \(\ker\rho\ne G\). It cannot be injective because \(|G|=36>24=|S_4|\), so \(\ker\rho\ne\{1\}\). The kernel is normal, and it is therefore a nontrivial proper normal subgroup in this case as well. Neither possibility allows \(G\) to be simple.