Sylow $p$-subgroups and their normalizers I am trying to solve this problem:
Let $H$ and $K$ be Sylow $3$- and $5$-subgroups, respectively, of a finite group $G$.  Suppose $H$ and $K$ have orders $3$ and $5$, respectively.
Prove that if $3$ divides the order of $N(K)$ (the normalizer of $K$ in $G$), then $5$ divides the order of $N(H)$.
 A: Hint: apply Sylow theory: in $N_G(K)$ there must be a conjugate of $H$, show that this implies that $N_G(H)$ contains a conjugate of $K$.
Since $3$ divides its order, the subgroup $N_G(K)$ contains a (cyclic) subgroup $S$ of order $3$ (apply Cauchy's Theorem). 
Step 1. $S$ centralizes $K$ (that is, $S \subseteq C_G(K) \subseteq N_G(K)$ (where the second inclusion is always true))

 We will apply the "$N/C$"-theorem: $N_G(K)/C_G(K)$ injects homomorphically in $Aut(K)$, and since $|K|=5$, $K \cong C_5$, whence $Aut(K)\cong C_4$. This shows that index$[N_G(K):C_G(K)]$ is even, and we conclude that $3 | |C_G(K)|$. So $S$ can be found in the centralizer of $K$ in $G$.

Step 2. $S \subseteq C_G(K) \iff K \subseteq C_G(S)$.

 For all $s \in S$ and $k \in K$: $s^{-1}ks=k \iff k^{-1}sk=s$

Step 3. $N_G(H)$ contains a conjugate of $K$.

Since $S$ has order $3$ and also the Sylow $3$-subgroup $H$ has order $3$, there is a $g \in G$ with $S^g=H$. Hence by the previous step, $K \subseteq C_G(H^g)=C_G(H)^{g^{-1}}$. And this gives $K^g \subseteq C_G(H)$. But always $C_G(H) \subseteq N_G(H)$, so we are done.

Step 4. $|N_G(H)|$ is divisible by $5$.

 $|K^g|=|K|=5$ and apply the previous step.

In general the following holds.
Proposition Let $p$ and $q$ be prime numbers such that $q \not \equiv 1$ mod $p$. Assume that the order of the finite group $G$ is divisible by $p$ and $q$, but not by higher powers of these primes. Let $P \in Syl_p(G)$ and $Q \in Syl_q(G)$. Then $p$ divides $|N_G(Q)|$ implies $q$ divides $|N_G(P)|$.
A: Here is an alternative solution.
First, we need to know that all groups of order $15$ are abelian (in fact, cyclic). This is a standard exercise, so I will not give the proof.
Next, we note that since $3$ divides the order of $N_G(K)$, it has a subgroup of order $3$. It also has a normal subgroup of order $5$ (namely $K$), and hence it has a subgroup of order $15$.
This means that $G$ has a subgroup of order $15$, and hence that $G$ has an element of order $5$ which commutes with an element of order $3$. In other words, for some subgroup of order $3$, let's call it $Q$, we get that there is an element of order $5$ in $N_G(Q)$. But now, we also assumed that $3$ was the order of a Sylow subgroup of $G$, and all Sylow subgroups have normalizers of the same order, which proves the claim.
