Prove a group of order $48$ must have a normal subgroup of order $8$ or $16$.
Solution: The number of Sylow $2$-subgroups is $1$ or $3$. In the first case, there is a normal subgroup of order $16$ so we are done.
In the second case, let $G$ act by conjugation on the Sylow $2$-subgroups. This produces a homomorphism from $G$ into $S_3$. Because of the action, the image cannot consist of just $2$ elements. On the other hand, since no Sylow $2$-subgroup is normal, the kernel cannot have $16$ elements. The only possibility is that the homomorphism maps $G$ onto $S_3$, and so the kernel is a normal subgroup of order $48 / 6 = 8$.
I don't understand parts of the second paragraph. I understand we can choose to let $G$ act on the $3$ Sylow $2$-subgroups by conjugation. There are $3 \cdot (2-1) = 3$ non-identity elements in total, or $4$ elements in total including the identity that belong to a Sylow $2$-subgroup. But why is the homomorphism from $G$ into $S_3$ when there are only $4$ unique elements to get mapped to? Also, why cannot the image not consist of just $2$ elements because of the action? If $G$ does map onto $S_3$, then why is the order of the kernel $48/6$?