Question about Sylow Theorem and normalizer I'm dealing with the following problem. 

Let $G$ be a finite group, $H$ and $K$ Sylow 3- 5- subgroups
  respectively of $G$. Suppose that 3 divides $|N(K)|$, show that 5
  divides $|N(H)|$.

I've looked for this question in the website, but the closest question I've found is this one, the problem is that here we don't have the hypothesis that $|H|=3$ and $|K|=5$.
Someone can give me a hand please? Thank you very much!
 A: The exercise you've been given is incorrect:
Let $G$ be the group of order 75 with a non-normal Sylow 3-subgroup [such a group is uniquely determined, is $O^2(\operatorname{AGL}(1,25))$, and has a normal Sylow 5-subgroup; let me know if you need more structure].
Then $G$ has 25 Sylow 3-subgroups, call one of them $H$, and $[G:N_G(H)]=25$ so $N_G(H)=H$ has order 3 (not divisible by 5). $G$ has only one Sylow 5-subgroup, call it $K$, and $[G:N_G(K)]=1$ so $N_G(K)=G$ has order 75 (divisible by 3).

Here is an explicit description, but I should probably use a slightly bigger group $$\hat G = \operatorname{AGL}(1,25) = \left\{ \begin{bmatrix} a & b \\ 0 & 1 \end{bmatrix} : a \in F^\times, b \in F \right\}$$ where $F$ is a field of size 25. This group has 24(25) = 600 elements. Its Sylow 3-subgroups are all conjugate to the subgroup $$H = \left\{ \begin{bmatrix} \zeta^i & 0 \\ 0 & 1 \end{bmatrix} : i \in \{0,1,2\} \right\}$$
where $\zeta$ is a third root of unity ($F^\times$ is a group of order 24, so Cauchy or Sylow shows it has an element of order 3, say $\zeta$) and $\hat G$'s Sylow 5-subgroup is the normal subgroup $$K = \left\{ \begin{bmatrix} 1 & b \\ 0 & 1 \end{bmatrix} : b \in F \right\}$$
Here $$N_{\hat G}(H) = \left\{ \begin{bmatrix} a & 0 \\ 0 & 1 \end{bmatrix} : a \in F^\times \right\}$$ has order 24 and $N_{\hat G}(K) = \hat G$ has order 600 since $K$ is normal. You can verify these normalizer calculations using the definition of matrix multiplication (they work pretty generally, nothing special about 3 and 5).
