Group of order $48$ must have a normal subgroup of order $8$ or $16$ 
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 $48/6$?
 A: Look in a slightly different way. Let $H$ be a Sylow-2 subgroup. It has 3 cosets in $G$, and by picking an element $x$ of order $3$, the three cosets of $H$ will be $\{H,xH,x^2H\}$. 
$G$ acts on this set of cosets naturally. Now how $x$ permutes these cosets? It will move them like $3$-cycle:
$$H\mapsto x.H \mapsto x.(xH)\mapsto x.(x^2H)=H.$$
On the other hand, how element of $h\in H$ will acts? It will take first coset $H$ to itself. Suppose $h$ takes $xH$ also to itself (and hence $x^2H$ to itself). Then this means 
$$(*) \,\,\,\,\,\,\,\,\,\,\,\,h(xH)=xH \mbox{ i.e. } xhx^{-1}\in H.$$


*

*If this (*) is true for every $h\in H$, then we get $xHx^{-1}\subseteq H$, i.e. $H$ will be normal. We have a normal subgroup of order $16$.

*If (*) is not true for some $h'\in H$, then this $h'$ will permute remaining cosets $xH$ and $x^2H$. This forces that the image of $G$ under homomorphism into $S_3$ will contain an element of order $2$, and it already contains an element of order $3$. Thus, image is $S_3$, and so kernel must have order $8$, a normal subgroup you expected.
A: Now $G$ acts on the 3 Sylow 2-subgroups by conjugation. Note that we assumed here that all three Sylow 2-subgroups are not normal (or we are done). This implies in particular that the image $G \to S_3$ has more than 2 elements since:


*

*If the image has only one elements, all Sylow 2-subgroups are normal, and 

*If the image has two elements, the image must be generated by a transpose (either $(12), (23)$ or $(31)$), in any case there will be one normal Sylow 2-subgroup.
Now the solution keeps on showing that the image cannot have exactly three elements. In this case the kernel of the homorphism $G \to S_3$ will be Sylow 2-subsubgroup as it has $16$ elements. But a kernel is always normal. This contradicts the assumption that all 3 Sylow 2-subgroups are not normal. 
As a result, the image of $G\to S_3$ must have $6$ elements. That is, the homomorphism $G\to S_3$ is surjective. Let $N$ be the kernel of this homomorphism. Then $N$ has $|G|/|S_3| = 48/6 = 8$ elements and is a normal subgroup of $G$. 
