Show that every group of order $48$ has a nontrivial normal subgroup.
Proof
Let $G$ denote a group of order $48$. Let $P_k$ denote the k-sylow subgroup, and $n_k$ denote the number of conjugates of $P_k$. Let $N(P_k)$ be the normalizer of $P_k$. Then,
$n_3 \equiv 1 \mod 3$ and $n_3 | 16 \implies n_3=1,4,16$
$n_2 \equiv 1 \mod 2$ and $n_2|3 \implies n_2=1,3$
Suppose that $n_3 = 4$ or $16$, and let $n_2=3$ for contradiction.
Then $n_3 = 3 \implies |N(P_3)|=16$. So we have a subgroup of order 16 in $G$.
There is a theorem in Rotman's textbook (Advanced Modern Algebra):
(Representation of Cosets). If H is a subgroup of finite index n in a group G, then there exists a homomorphism $\phi: G \rightarrow S_n$ with $\ker \phi \subseteq H$.
Using this theorem for $N(P_2)$, we know that there is a homomorphism $\phi: G \rightarrow S_3$ with kernel in $N(P_2)$. $P_3$ can only be sent to the 3-sylow subgroup of $S_3$, since the image of an element must divide the order of the pre-image. Since the 3-sylow subgroup of $S_3$ is normal, its image must also be normal. So $P_3$ is normal, a contradiction. So $n_2=1$ and we're done.
Do you think my proof is correct?
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Best Answer
The second to last line is where you reach a problem: We don't have an "image" of $S_3$, we have a pre-image. This pre-image contains $P_3$, but is not in necessity equal to it. In fact, since the Slyow 3-group of $S_3$ contains the identity, the pre-image of the subgroup of $S_3$ contains the kernel, properly. But the pre-image of a normal subgroup is normal. So we must only really show that the pre-image is a proper subgroup.
A simpler proof though, notes that we may let $\phi$ be the representation by conjugation on the Slyow 2-groups (which are of order 16). If this is trivial, then we see that $n_1=1$ by Slyow's Theoroms. If not, then Ker $\phi$ is the desired proper normal subgroup (Since Ker $\phi=0$ makes no sense considering the order).