Show that $ G $ is isomorphic to the direct product of $ H $ and $ K $. Let $ G $ be a group of order $ 20 $. Suppose that $ G $ has a subgroup $ H $ of order $ 4 $ and a subgroup $ K $ of order $ 5 $ such that $ hk = kh $ for all $ h \in H $ and $ k \in K $. Show that $ G $ is isomorphic to the direct product of $ H $ and $ K $.
Idea: If I could prove that $ G $ is cyclical then immediately $ H $ and $ K $ are cyclical and therefore $ H \cong \mathbb {Z}_4 $ and $ K \cong \mathbb {Z}_5$. Then $ G \cong \mathbb{Z}_{20} \cong \mathbb {Z}_4\times\mathbb {Z}_5 \cong H \times K.$ But if it's true, I don't see how to prove it. Can you help me please? Thanks so much for reading.

Hint:  It suffices to check that the following three properties are satisfied:  $1)H\cap K=\emptyset \quad 2)G=HK$ and $3)$the elements of $H$ and $K$ commute.
These properties are easy to verify:  $1)$ follows from Lagrange.  $2)$ does also, after noting that $HK\le G$ since both $H$ and $K$ are normal. For $H$ and $K$ are proper subgroups of $HK$.  $3)$ is given.
Btw:  the group need not be cyclic, as $G\cong V_4\times C_5$ shows.
