There is no simple group of order 148

Let \(|G|=148=2^2\cdot37\). If \(n_{37}\) denotes the number of Sylow \(37\)-subgroups, Sylow’s theorem gives \[ n_{37}\equiv1\pmod{37}, \qquad n_{37}\mid4. \] Among the divisors \(1,2,4\) of \(4\), only \(1\) is congruent to \(1\) modulo \(37\). Hence \[ n_{37}=1. \] Therefore the Sylow \(37\)-subgroup is unique and hence normal. It is nontrivial and proper, so \(G\) is not simple.