No simple group of order $40$

Let \(G\) have order \[ |G|=40=2^3\cdot5. \] Let \(n_5\) be the number of Sylow \(5\)-subgroups. Sylow’s theorem gives \[ n_5\mid8, \qquad n_5\equiv1\pmod5. \] The divisors of \(8\) are \(1,2,4,8\), and only \(1\) is congruent to \(1\) modulo \(5\). Hence \[ n_5=1. \] Therefore the Sylow \(5\)-subgroup is unique and hence normal. It is nontrivial and proper, so \(G\) is not simple.