How is the quotient group related to the direct product group?

I'm trying to understand what the relation is between the direct product and the quotient group.

If we let $H$ be a normal subgroup of a group $G$, then it is not too difficult to show that the set of all cosets of $H$ in $G$ forms a quotient group $G/H$: \begin{equation} G/H = \{ g H \mid g \in G \} \end{equation}

On the other hand, the Cartesian product of two groups $G$ and $H$ is defined as: \begin{equation} G \times H = \{ (g,h) \mid g \in G \text{ and } h \in H \} \end{equation} where $(g,h)$ denotes the set of ordered pairs. The direct product operation on this set is defined as: \begin{equation} (g_1,h_1)(g_2,h_2) = (g_1g_2,h_1h_2) \in G \times H \end{equation} and it is easy to see that the direct product forms a group.

Is the following statement true: \begin{equation} K = G \times H \implies G \simeq K / (\{e_G \} \times H) \end{equation} If so, under what conditions is it true? And how can we see it is true (or false)?

Edit 26/03:

Up to this point, I believe I have found a method (see below) of showing the isomorphism relation. I would be really grateful if someone could tell me whether this proof is correct or not.

Let us identity the elements of $h \in H$ with element of $K$ by setting $h \equiv (e_G,h)$. The elements of $K/H$ are as usual defined by: \begin{equation} K/H = \{ (g,h) H \mid g \in G \text{ and } h \in H \} \end{equation} Since $h_1H=H$ for some $h_1 \in H$, we have: \begin{equation} (g,h)H = (g',h')H \iff g=g' \text{ and } h' = h h_1 \tag{1} \end{equation} and so without loss of generality we can write every element of $K/H$ in the form $(g,e_H)H$. Now, let the map: \begin{equation} f : G \to K/H \end{equation} be defined by: \begin{equation} f(g) = (g,e_H) H \tag{2} \end{equation} The map is one-to-one. This can be seen by equation $(1)$, because if: \begin{equation} f(g) = f(g') \end{equation} then: \begin{equation} (g,e_H) H = (g',e_H) H \implies g=g' \end{equation} Furthermore, the map is trivially onto: \begin{equation} \forall (g,h) H \in K/H \; \exists \; g \in G \; , \; f(g)=(g,h) H \end{equation} and thus the map is bijective. Finally, the map is also a homomorphism, because: \begin{equation} f(gg') = (gg',e_H) H = (g,e_H)(g',e_H) H = (g,e_H)(g',e_H) HH = (g,e_H) H (g',e_H) H = f(g) f(g') \end{equation} and so $f$ is a isomorphism. Thus, by definition of equation $(2)$, we have shown that $G \simeq K/H$.

Any input is much appreciated.

Your solution is not wrong but it has unnecassary steps. You can simply use following arguments.

Let $\pi:G\times H\to G$ be projection map .i.e. $\pi(g,h)=g$. It is clear that map is onto.

Claim$1:$ $\pi$ is an homomorphism;

$$\pi((g_1,h_1)(g_2,h_2))=\pi((g_1g_2,h_1h_2))=g_1g_2=\pi((g_1,h_1))\pi((g_2,h_2))$$ and since it is onto map it is an epimorphism.

Claim$2$: $Ker(\pi)=e_G\times H$

Let $\pi(x,y)=e_g$ then you can say that $x=e_g$ and $y$ is any element in $H$ so result follows. By the way it also show that $e_G\times H$ is normal in $G\times H$.

Result: By first isomorphim theorem; $$(G\times H)/ker(\pi)=(G\times H)/(e_G\times H)\cong G$$

Notes: By using the other projection you can show smiliar argument for $G\times e_H$ and $H$. I hope you are familiar with isomorphism theorems.

This statement is always true, since $H$ is a normal subgroup of $G \times H$ (i.e. $xH = Hx$ for all $x \in K$). To see why, consider two cases:

1. The case $x \in H$

2. The case $x \notin H$

Then, the statement follows from the fundamental theorem of homomorphisms. By the way, you should replace the second $=$ sign by the $\simeq$ sign, in your statement.

Suppose $G=H\times K$. We want to show $G/K\simeq H$. Consider the map $H\times K\to H\times 0\simeq H$ given by $(h,k)\to (h,0)$. What is the kernel of this? What is it's image?

In general, the whole point of the direct product is given groups $H$ and $K$ to form a group $G=H\times K$ with copies $\hat K=1\times K$ and $\hat H=H\times 1$ such that

$(\rm i)$ $\hat H,\hat K\lhd G$

$(\rm ii)$ $G=\hat H\hat K$

$(\rm iii)$ $\hat K\cap \hat H=1$

Moreover, $G/\hat K\simeq H$ and $G/\hat H\simeq K$.

Conversely, if $G$ contains subgroups $H,K$ with $H,K\lhd G$, $H\cap K=1$ and $HK=G$, then $G\simeq H\times K$.