# 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 $G/H$: $$G/H = \{ g H \mid g \in G \}$$

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

Is the following statement true: $$K = G \times H \implies G \simeq K / (\{e_G \} \times H)$$ 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: $$K/H = \{ (g,h) H \mid g \in G \text{ and } h \in H \}$$ Since $h_1H=H$ for some $h_1 \in H$, we have: $$(g,h)H = (g',h')H \iff g=g' \text{ and } h' = h h_1 \tag{1}$$ 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: $$f : G \to K/H$$ be defined by: $$f(g) = (g,e_H) H \tag{2}$$ The map is one-to-one. This can be seen by equation $(1)$, because if: $$f(g) = f(g')$$ then: $$(g,e_H) H = (g',e_H) H \implies g=g'$$ Furthermore, the map is trivially onto: $$\forall (g,h) H \in K/H \; \exists \; g \in G \; , \; f(g)=(g,h) H$$ and thus the map is bijective. Finally, the map is also a homomorphism, because: $$f(gg') = (gg',e_H) H = (g,e_H)(g',e_H) H = (g,e_H)(g',e_H) HH = (g,e_H) H (g',e_H) H = f(g) f(g')$$ 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.

• Note that $H$ isn't even a subgroup of $G\times H$ strictly speaking. What you want to show is $G \cong K\big/ \left(\{e_G\}\times H\right)$. – Christoph Mar 23 '14 at 19:56
• @Christoph thanks, I've updated my question. – Hunter Mar 23 '14 at 20:02

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.

• Thanks for your reply! No, I am not familiar with isomorphism theorems, but I'm always willing to learn. However, this is why I wanted to be sure that my "proof" is correct. – Hunter Mar 28 '14 at 20:19
• I'm not sure how it works now: I want to get as much attention as possible to this question, in order to see if anybody can find a mistake. But if by the end of the 7 day period (5 days left) no one has found one, then I want to give the bounty to you for confirming my proof. – Hunter Mar 28 '14 at 20:21
• @Hunter: Actually, your proof includes it that is why your solution is long; first isomorphism theorem says that $\phi:G\to H$ is a epimorphim then $G/ker(\phi)\cong H$. I suggest you to learn all isomorphism theorems: en.wikipedia.org/wiki/Isomorphism_theorem – mesel Mar 28 '14 at 20:29
• @Hunter: people.reed.edu/~jerry/332/09isom.pdf this source seems to good. – mesel Mar 28 '14 at 20:35

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.

• Thanks for your reply. I understand that $(e_G,H)$ is a normal subgroup of $G \times H$ and that $(e_G,H)$ is isomorphic to $H$, but I don't see why that would make the statement true. Can you elaborate? – Hunter Mar 23 '14 at 19:32
• I have tried to answer the question myself and would be really interested to have feedback from you if you have time. – Hunter Mar 24 '14 at 16:28

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 $G\simeq H\times K$.

• Shouldn't we look at $G\times H\to G$ to obtain the desired isomorphism? – Christoph Mar 23 '14 at 19:50
• @Christoph I usually write $H\times K=G$, so I slightly skipped over the OPs notation. – Pedro Tamaroff Mar 23 '14 at 19:56
• So are we looking for: $$\mathrm{Ker} = \{h \in H \text{ and } k \in K \mid f((h,k)) =(h,e_K) \}$$ where $f$ is a homomorphic function? If so, if have no idea what this information tells us. (I'm sorry, I'm studying physics and not mathematics.) – Hunter Mar 23 '14 at 19:58
• @PedroTamaroff I have tried to answer the question myself and would be really interested to have feedback from you if you have time. – Hunter Mar 24 '14 at 16:28
• @Hunter I don't understand what you're asking. – Pedro Tamaroff Mar 24 '14 at 16:51