Let $H,K\leq G$ and let $G=H\times K$ Let $H$ and $K$ be groups and let $G=H\times K$
Show $H$(actually$H\times\{e\}$) and $K$(actually $\{e\}\times K$) have the following properties:
$\bullet$ $\forall x\in G, x=hk$ for some $h\in H$ and $k\in K$
$\bullet$ $hk=kh$, $\forall h\in H$, and $\forall k\in K$
$\bullet$ $H\cap K=\{e\}$
We have yet to cover Lagrange's Theorem, and I don't think we yet have the notion of a normal subgroup. My first thought was to break down $G$ in component form of $h, k$. Although, I'm confused as to why it's important we notice $H$(actually$H\times\{e\}$) and $K$(actually $\{e\}\times K$). $H$ and $K$ are already subgroups, so they contain $e$ anyways. Anything from $H\times\{e\}$ is of the form $(h,e)$, and I don't see the significance. Since we're assuming $G=H\times K$, it doesn't matter to show they're isomorphic right?
Does it make sense to just outright start by saying:
Given $G=H\times K$, let $x\in G$. By definition of cartesian product $x=hk$ for some $h\in H$ and $k\in K$?
 A: This looks like nit-picking on definitions. So we need to check carefully what $H \times K$ is defined as. I assume it would be:
$$G = \{(h, k): h ∈ H, k ∈ K\}$$
with the group operation $·_G$:
$$(h_1, k_1) ·_G (h_2, k_2) := (h_1 ·_H h_2, k_1 ·_K k_2)$$
And the identification of $K$ and $H$ is given by
$ι_H: h ↦ (h, e_k)$ and $ι_K: K ↦ (e_h, k)$.
Then the first item becomes $x = (h, k)$ by definition. Then $x = ι_H(h) ·_G ι_K(K)$. Done.
The second item becomes $ι_H(h) ·_G ι_K(k) = (h, e_K) ·_G (e_H, k) = (h ·_H e_H, e_K ·_K k) = (e_H ·_H , k ·_K e_K) = (h, k) = ι_H(h) ·_G ι_K(k)$. Here we use that the identity element in groups $H$ and $K$ commutes with all group elements.
The third item: let $x = (h, k) ∈ ι_H(H) ∩ ι_K(K)$. Then there exist $h' ∈ H$ and $k' ∈ K$ such that $(h, k) = (h, e_K)$ and $(h, k) = (e_H, k)$. From the first we get $k = e_K$ and from the second $h = e_H$ and hence $x = (e_H, e_K) = e_G$. So $ι_H(H) ∩ ι_K(K) = \{e_G\}$.
My impression for this task is that it all boils down to be very rigorous.
PS: What's Lagrange's Theorem?
