1. ## Quotient groups

I'm quite confused about the concept of quotient groups. From what I gather from my course so far:

Let $\psi :G \rightarrow H$ and $K = \text{kernel} \, \, \psi$. Then
$G/K =$ the collection of fibres $\{ \psi ^{-1} (a) | a \in H \}$

Let $N \unlhd G$, then
$G/N$ = the collection of left cosets (or right cosets).

But how is that a group? You get a collection of sets, and only one of those sets contain the identity so only one of those sets are a group?

Looking at the Wikipedia example:
Quotient group - Wikipedia, the free encyclopedia
they get G/N = { {0, 3}, {1, 4}, {2, 5} } which is a collection of sets... so how are they a group?

2. ## Re: Quotient groups

{0,3}+{0,3} = {(0+0) (mod 6), (0+3) (mod 6), (3+0) (mod 6), (3+3) (mod 6)} = {0,3,3,0} = {0,3}
{0,3}+{1,4} = {(0+1) (mod 6), (0+4) (mod 6), (3+1) (mod 6), (3+4) (mod 6)} = {1,4,4,1} = {1,4}
{0,3}+{2,5} = {(0+2) (mod 6), (0+5) (mod 6), (3+2) (mod 6), (3+5) (mod 6)} = {2,5,5,2} = {2,5}
Hence {0,3} is the identity.

{1,4}+{1,4} = {(1+1) (mod 6), (1+4) (mod 6), (4+1) (mod 6), (4+4) (mod 6)} = {2,5,5,2} = {2,5}
{1,4}+{2,5} = {(1+2) (mod 6), (1+5) (mod 6), (4+2) (mod 6), (4+5) (mod 6)} = {3,0,0,3} = {0,3}
{2,5}+{2,5} = {(2+2) (mod 6), (2+5) (mod 6), (5+2) (mod 6), (5+5) (mod 6)} = {4,1,1,4} = {1,4}

So, the set G/N is closed under addition, it has an identity element, every element has a unique inverse: -{0,3} = {0,3}, -{1,4} = {2,5}, and -{2,5} = {1,4}. Associativity is inherited from G. Hence, it is a group.

Another way to think about it: for any a in G/N, you can write a=g+N for some g in G. So, {0,3} becomes 0+N=3+N. {1,4} becomes 1+N=4+N. {2,5} becomes 2+N=5+N. Then, the addition becomes:

(0+N) + (0+N) = 0+N
(0+N) + (1+N) = 1+N
(0+N) + (2+N) = 2+N
(1+N) + (1+N) = 1+N
(1+N) + (2+N) = 3+N = 0+N
(2+N) + (2+N) = 4+N = 1+N

This addition table matches the one above.

3. ## Re: Quotient groups

Hi,
I hope the following helps: