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....

You can do it without Sylow, if you want to. Consider conjugacy classes. The allowed sizes must divide $48$: $1$, $2$, $4$, $8$, $16$, $3$, $6$, $12$, $24$, and $48$. The identity must be in a class by itself. So now we have $47$ elements to pack into classes; and because that's odd, there must be a class of size $1$ or $3$.
Say there's an element $x$ with conjugacy class of size $1$, then it's in the center. Then $\langle x \rangle$ is normal. We either have a normal subgroup, in which case $G$ is not simple, or the entire group, meaning $G$ is cyclic. But a cyclic group of size $48$ can't be simple.
Say there's an element $x$ with conjugacy class of size $3$, then we can create a homomorphism to $S_3$. The size of the image is at most $6$, so the kernel is at least size $8$. Verify that the kernel isn't the whole group. Kernels of homomorphisms are normal subgroups, so we have a non-trivial normal subgroup, making $G$ not simple.
Source: Forced to find all simple groups up to order $60$, no Sylow allowed. Don't try $30$ or $56$ like that.
