Show that there is no simple group of order 148SolutionLet G

Show that there is no simple group of order 148

Let G be a simple group of order 148

148 has prime factors as 2,2, 37

Hence from Sylow theorem, there is atleast one subgroup of order 37 (By Sylow theorem)

Also we must have any subgroup of order 37 must have conjugatge and the number of such subgroups

is x = 1 Mod 37 which divides 148

But 37+8 , 74+1, 111+1 do not divide 148.

Hence it follows that there is only one subgroup of order 37 which is its own conjugate.

Or if H is a subgroup of order 37, then H is its own self conjugate and hence normal.

This contradicts the fact that G is simple and hence no subgroup of G is normal.

So there is no simple group of order 148.