# Question about Wikipedia Coset/Quotient Group Example

I am wondering if anyone can elaborate on the following Wikipedia example of a coset.

I am confused how the quotient group is even a group in this example. I will try to explain each step of what the example does and hopefully my mistakes will help anyone to help me learn where my problems are...

We are considering $G=(\mathbb{Z}_6,+)=\{0,1,2,3,4,5\}$ and the subgroup $N=\{0,3\}$. We want to form the set of left cosets (I know N is normal, so its left and right cosets are equivalent).

The Wikipedia article begins to construct the set $G/N=\{gN\mid g\in G\} = \{g\{0,3\}\mid g\in \{0,1,2,3,4,5\}\}$

One particular coset with $g=0$ is $\{\{0+0\equiv 0\pmod{6}\},\{0+3\equiv 3\pmod{6}\}=\{0,3\}$. Once we construct all six cosets, we eliminate the three equivalent sets and we're left with a set of 3 unique cosets: $G/N=\{\{0,3\},\{1,4\},\{2,5\}\}$. To me this doesn't seem like a group. The elements of this "group" are sets like $\{0,3\}\in G/N$. Another element is $\{1,4\}$. How do I show that $\{0,3\}+\{1,4\}$ is closed and is an element of $G/N$?

Thanks for your help. I hope you don't mind clearing up my misunderstandings.

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The best way to see the group operation on the cosets is to keep them factored in terms of the normal subgroup, i.e. $\{0N,1N,2N\}$, so that $(aN)+(bN)=(a+b)N$. –  anon Sep 16 '11 at 2:58
@aengle: Do you know what the definition of $+$ in the quotient group is though? $\{0, 3\} + \{1, 4\} = \{1, 4\}$, for example. –  Zhen Lin Sep 16 '11 at 3:00

First, remember that a group is just a set together with a binary operation that satisfies certain conditions. Certainly, $G/N$ is a set; so the only question is whether the operation defined on it makes it a group.

The definition of the operation that we use on $G/N$ is the following:

if $gN$ and $hN$ are cosets, then we define the "product" of $gN$ and $hN$ to be the coset $ghN$.

So here, to add $\{0,3\}$ and $\{1,4\}$, we pick one element from each set, add them, and then look for the coset which contains the result. So we can take $0$ and $1$, and look for the coset that contains $0+1=1$, namely $\{1,4\}$. So the result of adding the coset $\{0,3\}$ with $\{1,4\}$ is $\{1,4\}$.

What if you picked a different element from each set? It doesn't matter: you get the same answer in the end:

• If you pick $0$ and $4$, you look for the coset of $0+4=4$, which is $\{1,4\}$;
• If you pick $3$ and $1$, you look for the coset of $1+3 = 4$, which is $\{1,4\}$;
• If you pick $3$ and $4$, you look for the coset of $3+4=1$, which is $\{1,4\}$.

Likewise, $\{0,3\} +\{2,5\} = \text{the coset of }0+2 = \{2,5\}$; and $\{1,4\}+\{2,5\}=\text{the coset of }1+2 = \{0,3\}$. You can check that this makes the set $G/N$ into a group.

But really, you don't want to think of the elements of $G/N$ as the sets. You want to think of them as "equivalence classes", and specifically, as "the equivalence class of an element of $G$".

So you don't want to think of $\{0,3\}$ as "the set whose elements are $0$ and $3$", you want to think of it as $[0]$, the equivalence class of $0$ (also of $3$). And you want to think of $\{1,4\}$ as $[1]$ or as $[4]$; and $\{2,5\}$ as $[2]$ or $[5]$. Then the addition is given by $$[a]+[b] = [a+b].$$ The fact that the subgroup is normal is what guarantees that this is well defined: each element has two "names", but the name you pick does not affect the result you get.

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Thank you. This explained a lot. I appreciate the help with the well defined part too. I was considering asking a question about that too, but now I understand with your example. –  mathmath8128 Sep 16 '11 at 3:09

If $N$ is normal in $G$, and $\overline{x}$, $\overline{y} \in G/N$ (so $\overline{x}$ and $\overline{y}$ are cosets of $N$), we define $\overline{x}+\overline{y} = \overline{x+y}$. The fact that $N$ is normal ensures this definition is independent of the particular choice of $x$ or $y$ that represent the cosets.

Let's take a look at your example. $\{0,3\} + \{1,4\}$ is found by choosing representatives of each coset, say $0$ and $1$, adding them in $G$, $0+1=1$, and so in $G/N$, the sum is the coset $1+N = \{1, 4\}$. Note, if I chose $3$ and $4$ as my representatives, the sum works out the same -- $3+4 = 1$ in $G$.

Hope this helps!

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$G/N = \{gN : g\in G\}$ I will leave $\{0, 3\}$ written as $N$ here as it is more clear.

Then $G/N = \{0N, 1N, 2N\}$ as you have stated. You seem to want to see each particular example so we can exhaust the examples to check if this is a group

add $0N + 0N = (0+0)N = 0N \in Z/G$

add $0N + 1N = (0+1)N = 1N\in Z/G$

add $0N + 2N = (0+2)N = 2N\in Z/G$

add $1N + 1N = (1+1)N = 2N\in Z/G$

add $1N + 2N = (1+2)N = 3N = 0N \in Z/G$ because we can represent the coset with either element that is a member of it and $0N = \{0 , 3\}$.

add $2N + 2N = (2+2)N = 4N = 1N \in Z/G$ for the same reason as above.

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Deven, I thought about the problem a little last night and I have a sentence that is only true in infinite graphs (this is the only way I could reach you). Consider "there is one vertex with only one neighbor, and all other vertices have exactly two distinct neighbors." If you start at the vertex with only one neighbor and move along that path, you can never loop back. So the only graphs that model this are infinite graphs with a subgraph isomorphic to the natural numbers. –  Isaac Solomon Nov 10 '12 at 19:20