I know that this question has been asked a lot, but I´m trying with a different approach(I think).
The proof it´s divided in two parts: Let G be a simple group of order 48.
1)If $S_1,S_2,S_3$ are the 2-sylows of $G$, let $H$ $\in \{S_i \cap S_j : i \neq j \}$ so that H has maximum order. Prove that $|H|>1$.
2)Prove that $N_G(H)$ (the normalizer) has only one 2-sylow. It´s suggested to reduce it to the absurd and conclude that H is normal in G.
For 1) I have concluded that $|H|=8$, and I arrived to that conclusion because $|S_iS_j|=\frac{|S_i||S_j|}{|S_i\cap S_j|}$, so if the order of the intersection is $1,2$ or $4$, $|S_iS_j|\ge 48 $.
For 2) I have a lot of arguments but I didn´t arrive to any conclusions. If I assume that there is more that 1 2-sylow ($n_2>1$) , and because $n_2 | m$, where m is coprime with 2 , I conclude that m=3. So, $|N_G(H)|=2^i3$, and $i\in \{1,2,3,4\}$. If $i=4$ then the order of the normalizer is 48 so it´s normal, and that is absurd. So $i\neq 4$. I have also noticed that $n_2 =1(mod 2)$ and $n_2|3$ so $n_2=3$ (because $n_2>1$).
Am I on the right track? Any help would be appreciated.
I think it can be done more easily. Although this does not fully relate to the problems in your proof, I still wanted to share this proof. I hope this answer also helps you
First, we have that $48 = 2^4 \cdot 3$. For the prime $p|48$, let $N_p$ be the number of Sylow-2 groups in $G$.
Now, if we have that $N_p=1$ from some $p$, then we must have that the unique Sylow $p$-group in $G$is normal. Also note that is has order $p$, so it is not $G$ nor $\{e\}$. Therefore, for this case, we conculde that $G$ is not simple.
Then, we have $N \equiv 1\mod{2}$ and $N_2 |3$, so $N_2$ must be equal to either $1$ or $3$. Hopefully, it is clear that if we have that $N_p=1$, we are done, by our previous argument.
Hence, we need to consider the other case, namely $N_p = 3$. Let $Y$ denote the set of Sylow-2 groups. We can write a homomorphism $\varphi : G \to S_Y$ given by sending $a \in G$ to conjugation $\gamma$ by $a$. Then, the kernel of $\varphi$ is a normal subgroup of $G$ and hence if $G$ is simple, it has to either consist of $\{e\}$ or all of $G$. Note, the first case is impossible, since $\#G = 48 > 6 = \#S_Y$. This means that all Sylow-2 groups are fixed by conjugation.
Now, the latter is inconsistent with Sylow's theorem (by simply recalling the theorem). Hence, we can conclude that $G$ is not simple