$|G|=24$ prove that $G$ is not simple

Question

Let $G$ be a group of order 24, and we shall assume there there exist a non-normal 2-sylow group in $G$. i want to show that it's not simple.

first i have showed that there are exactly three 2-sylow group's.

and now i am trying to show that there is a non trivial homomorphism from $G$ to $S_3$. i know that $S_3\cong D_3$ and indeed there are three 2-sylow groups in $D_3$ , the reflection's. and there is one $3-$sylow group in $D_3$ so i wanted to creat a homomorphism that take the different p-sylow groups in $G$ and set them in right $p-$sylow group's in $D_3$ but for that i need to show that the intersection of every two $2-$sylow gouts in $G$ is trivial. and i can't manage to do that....

Answer 1

A simple group $G$ of order $24$ would have three Sylow $2$-subgroups, and so there would be a non-trivial homomorphism from $G$ to $S_3$ (see the comment) which will be injective by simplicity and therefore $24$ would divide $6$ which is absurd.

Answer 2

If $n_2=1$ we are done. If $n_2=3$, then the intersection of the Sylow $2$-subgroups is normal (it's the normal core in $G$ of any of them); so, if such intersection is non-trivial, then we are done again; on the other hand, if it is trivial, then the two elements left out of the $2$-Sylow's must be the nontrivial elements of the unique $3$-Sylow, which is then normal. So, in any case there is a nontrivial, proper, normal subgroup of $G$, which is then non-simple.