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.
 A: The assumption $hk = kh$ means that you can define a group homomorphism $$\phi: H\times K \rightarrow G, \phi(h, k) = h\cdot k,$$ where $\cdot$ is the group operation in $G$. (In general, one can always define such a map, but it is a group homomorphism only if we assume that elements of $H$ and $K$ commute.)
We want to show that $\phi$ is an isomorphism. Since both sides have order $20$, it suffices to show that $\phi$ is injective.
The kernel of $\phi$ consists of elements of the form $(t, t^{-1})$ with $t\in H\cap K$. But the intersection $H\cap K$ is the trivial group, as its order must divide both the order of $H$ and the order of $K$.
Therefore the kernel of $\phi$ is trivial and $\phi$ is injective, hence an isomorphism.
A: 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.
