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.

Solutions Collecting From Web of "How is the quotient group related to the direct product group?"

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 $G\simeq H\times K$.