Subgroups of product of two finite groups with coprime order I want to explain the following proposition (if true).

Let $G_1$ and $G_2$ be finite groups with coprime orders and let $H$ be a subgroup of $G_1 \times G_2$. Then there exists subgroups $H_1$ and $H_2$ of $G_1$ and $G_2$, respectively, so that $H = H_1 \times H_2$.

Of course, if familiar, it can be handled by the the Gourat's lemma. Note that it suffices to prove that 'if the images of $H$ under the natural projections are $G_1$ and $G_2$, then $H=G_1 \times G_2$'. By the lemma the quotients of $G_j$ by the kernels are isomorphic to each other and hence, should be trivial because of the coprime orders. This implies that $H = G_1 \times G_2$.
I want to explain the proposition to the students of undergraduate sophomore level. Can you help to develop a suitable explanation or proof of the proposition? (By the way, they are familiar with Sylow blah, I guess)
PS: Sadly(?) I just found the following question. This post may be treated as duplicate.
$ |G_1 |$ and $|G_2 | $ are coprime. Show that $K = H_1 \times H_2$
 A: Here is a completely elementary proof, which does not use nine-lemma, Goursat's lemma or Sylow theorems.
Let $H_1$ the set of elements $h_1\in G_1$ such that $H$ contains an element of the form $(h_1,g_2)$ for some $g_2\in G_2$, and similarly, let $H_2$ the set of elements $h_2\in G_2$ such that $H$ contains an element of the form $(g_1,h_2)$ for some $g_1\in G_1$.
Clearly, $H_i$ is a subgroup of $G_i$ and $H\subset H_1\times H_2$.
Let us prove the reverse inclusion. Let $(h_1,h_2)\in H_1\times H_2$.
By definition, $(h_1,g_2)\in H$ and $(g_1,h_2)\in H$ for some $g_i\in G_i$.
Let $m_i$ the order of $g_i$. Since $H$ is a subgroup of $G_1\times G_2$, we get that $(h_1^{am_2},1), (1,h_2^{bm_1})\in H$ for all $a,b\in\mathbb{Z}$.
Since $m_2$ divides the order of $G_2$, it is coprime to the order of $G_1$, hence to the order of $h_1$. Hence $uo(h_1)+vm_2=1$ for some $u,v\in\mathbb{Z}$.
Thus $h_1^{vm_2}=h_1$ , so taking $a=v$ above yields $(h_1,1)\in H$. Similarly, $(1,h_2)\in H$, and finally $(h_1,h_2)\in H$, that is $H_1\times H_2\subset H$. a
Consequently, $H=H_1\times H_2$.
