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....
 A: 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 $G$.
A: 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.
A: Suppose G is simple. Since G is simple, we must have that $n_2 = 3$. Let P be a Sylow 2-subgroup of G. Then $|G:N_G(P)|=3$. 
Theorem: If G is a simple group, $|G|\neq 2$, and if $H<G, |G:H|=n > 1$ then $G\cong$ subgroup of $A_n$.
So in our case we have that $G\cong$ subgroup of $A_3$ but $|A_3|=3$ and $|G|=48$, which is a contradiction. Thus G cannot be simple.
Proof of Theorem:
Let G be a simple group, $|G|>2$. Let $H<G$ and $|G:H|=n>1$. There exists a homomorphism $\theta :G \to S_n$ with $ker(\theta)\subseteq H$ (the permutation representation of the G acting on cosets H of G by left multiplication).
Since G is simple, we must have that $ker(\theta) = 1$. So $\theta (G) \cong G$ by the isomorphism theorem.
Suppose that the result is not true, that $\theta(G) \nsubseteq A_n$. Then we must have $A_n\theta(G) = S_n$. 
Since $\frac{S_n}{A_n} = \frac{A_n\theta(G)}{A_n}\cong \frac{\theta(G)}{A_n\cap\theta(G)}$ by the second isomorphism theorem, and G is simple, we must have $A_n\cap \theta(G)=1$ or $\theta(G)$.
If $A_n\cap \theta(G)= \theta(G)$, then $S_n/A_n=1$, which is a contradiction.
In the case $A_n\cap \theta(G)=1$ we have that $S_n/A_n \cong \mathbb{Z}/2\mathbb{Z}\cong \theta(G)\cong G$, i.e. that $|G|=2$, which is a contradiction.
Thus we must have that $\theta(G)\subseteq A_n$
