# 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.

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.

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.