Prove that $G$ is the internal direct product of $H$ and $K$ Let $G$ be a group of order $20$. If $G$ has a subgroups $H$ and $K$ of orders $4$ and $5$ respectively such that $hk=kh$ for all $h \in H$ and $k \in K$, prove that $G$ is the internal direct product of $H$ and $K$.
What I have so far:
claim: $G = H \times K$ internal direct product of $H$ and $K$
(i)
$(H\cap K = \phi)$
Since $K$ is 5-cycle so each element of $K$ is of order $5$, same for the elements of $H$ since $|H| = 4$
(ii) Calculating order of $HK$
We have $hk = kh$ for all $h \in H$ and $k \in K$ ---- (i)
=> $HK = KH$
Suppose for $h$ and $h'$ in $H$ and $k$ in $K$ if   $hk = h'k$, then
$(hk = h'k \rightarrow h^{-1}hk = h^{-1}h'k \Rightarrow e.h = h^{-1}h'k \Rightarrow h^{-1}h'=e)$
Thus, using (i) $h = h'$ (note that $(h^{-1}h' \in H))$
Thus all the elements shown in $HK$ are distinct
$(HK=\{hk :h \in H, k \in K \}$
and hence $|HK| = |H| \times |K| = 4 \times 5 = 20$
Since, $HK$ is subgroup of G and $|HK|=|G| => G = HK$
$HK$ is subgroup since $hk$ and $h'k'$ belonging to
$HK => (hk)(h'k') = h (kh') k' = hh' kk'$ belongs to $HK$, and hence is closed
This is an exam review problem. Any help would be appreciated!
 A: Only $(1)$.
To prove $H\cap K=\{e\} (i.e.|H\cap K|=1) $, you may also think,
Suppose there is an element $x \in H\cap K $, it follows that $x \in H$ and $x \in K$
obviously , $|x| $ must divides $|H|=4$ and $|K|=5$,
therefore $|x|=1$ implies $x=e$
A: Hints:
== Using Sylow Theorems prove the subgroup of order $\;5\;$ is normal in any case
== The above already gives us $\;G=K\rtimes H\;$ , the semidirect product. This will be a direct product iff $\;H\lhd G\;$ , as obviously  $\;H\cap K=1\;$ (why?) .
Now use the remaining piece of information to deduce that indeed $\;H\lhd G\;$ and you shall be done.
Added on reequest:
You already know that $\;G=HK\;$ , so that $\;\forall h'\in H\;,\;\;g=hk\in G\;,\;\;h\in H\,,\,\,k\in K:\:$
$$[h',g]=h'^{-1}g^{-1}h'g=h'^{-1}k^{-1}h^{-1}h'hk=\left[h'h^{-1}h'h\right]k^{-1}k=h'h^{-1}h'h\in H$$
and thus we get that
$$[H,G]\le H\iff H\lhd G$$ 
Likewise $\;K\lhd G\;$ ,which is what was left to show in order to have $\;HK\cong H\times K\;$ , without semidirect products and without Sylow theorems.
