showing that there is no simple group of order 48 I tried to solve this prob. by using Sylow theorems.
But i can't solve it because it is not seen in the text as a example or exercise.
I even use a diagram of sets and consider all possible cases,  but it is not effective....

If $|G|=48=2^4 3$, then $n_2 \in \{1,3\}.$ If $n_2 = 1,$ the group is not simple. If $n_2 = 3$ and $G$ is simple, then the action of $G$ on the Sylow 2 subgroups induces a homomorphism into $S_3$ which must have a trivial kernel. This implies that $G$ is isomorphic to a subgroup of $S_3,$ which is of course too small of a group to contain $G$.
