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.
Best Answer
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:
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:
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.