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 one Sylow $2-$subgroup of order $16$, so it is normal.
In the second case, there are $3$ Sylow $2$-subgroups of order $16$. let $G$ act by conjugation on the Sylow $2$-subgroups. This produces a homomorphism from $G$ into $S_3$. The image cannot consist of just $2$ elements. Also, 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 $48 / 6 = 8$.
My first question is the first sentence of the bolded part: "The image cannot consist of just $2$ elements." Why can't it have $2$ elements? The homomorphism will map $24$ elements of $G$ to the identity permutation of $S_3$, and the homomorphism will map $24$ elements of $G$ to the a $2-$cycle permutation of $S_3$. I don't see any contradiction here. If mapping $24$ elements of $G$ to a $2-$cycle permutation of $S_3$ implies that there is a normal subgroup of order $8$ or $16$, then I don't see why this implication is true.
My second question is about: "since no Sylow $2$-subgroup is normal, the kernel cannot have $16$ elements." I don't understand this statement either. I don't see any relationship at all between a Sylow $2$-subgroup not being normal and the kernel of the homomorphism. If the kernel of a homomorphism has $16$ elements, then I agree that these $16$ elements constitute a normal subgroup of $G$, but I don't see what that has anything to do with a Sylow $2$-subgroup being normal or not.
Best Answer
You can try this way:
$48=2^4.3$
Now Number of Sylow $2$ subgroups of $G$ =$1+2k$ .Either $1+2k=1 or 3$ .If it is $1$ we are done.If it is $3$ then we have $3$ Sylow $2$ subgroups of $G$ of order $16$ say $H_1,H_2,H_3$ .then
$o(H_1H_2)=\dfrac{o(H_1)o(H_2)}{o(H_1\cap H_2)}$.Possible orders of $o(H_1\cap H_2)=2,4,8$ by lagrange's theorem/.
The only one which works here is $8$ (verify).
Now use the result that