Prove that no group of order \(90\) is simple.
Solution
Suppose for contradiction that \(G\) is simple of order \[ 90=2\cdot 3^2\cdot 5. \]
Let \(n_5\) be the number of Sylow \(5\)-subgroups. Sylow’s theorem gives \[ n_5\equiv1\pmod5, \qquad n_5\mid18, \] so \(n_5\in\{1,6\}\). Simplicity rules out \(n_5=1\), hence \(n_5=6\).
Conjugation on the six Sylow \(5\)-subgroups gives a homomorphism \[ \rho:G\to S_6. \] Its kernel is normal. The action is nontrivial, so simplicity implies that \(\rho\) is injective. Thus we may regard \(G\) as a subgroup of \(S_6\) of order \(90\).
The sign homomorphism restricted to \(G\) has normal kernel. Since \(G\) is nonabelian simple, it admits no nontrivial quotient of order \(2\), so the sign is trivial on \(G\). Hence \[ G\le A_6. \] But then \[ [A_6:G]=\frac{360}{90}=4. \] The action of \(A_6\) on the four left cosets of \(G\) gives a homomorphism \[ A_6\to S_4. \] Since \(A_6\) is simple and the coset action is nontrivial, this map would be injective, impossible because \[ |A_6|=360>|S_4|=24. \] Therefore no simple group of order \(90\) exists.