Recipe

Build the quotient group G/N

Use when: you're given a group G and a confirmed normal subgroup N, and need G/N's elements, its operation, or a specific identity or inverse inside it

Goal: every coset of N, the operation between any two of them, and G/N's identity and inverses

Taught in The quotient group G/N

1. 1

   Write down every distinct left coset gN. These are exactly the elements of G/N. There are [G:N]=|G|/|N| of them.

   Example

   In \mathbb{Z}_6/\{0,3\}, the three cosets are N=\{0,3\}, 1+N=\{1,4\}, 2+N=\{2,5\}.

   Why: G/N's elements are, by definition, exactly the cosets of N, nothing more and nothing less.

2. 2

   To combine g_1N and g_2N, pick any representative from each. Multiply them in G. Report which listed coset the product falls into.

   Example

   (1+N)+(2+N): representatives 1 and 2 give 1+2=3, and 3\in N, so (1+N)+(2+N)=N.

   • the product already matches one of the cosets listed in Step 1 — read the answer off directly.
   • the product doesn't obviously match a listed coset — reduce it, for example mod n in \mathbb{Z}_n, until it matches one of them.

   Why: the previous section proved this result never depends on which representatives were chosen, precisely because N is normal.

3. 3

   The identity of G/N is always eN=N itself. The inverse of gN is always g^{-1}N, the coset of g's own inverse in G.

   Example

   In \mathbb{Z}_6/\{0,3\}, the inverse of 2+N is the coset of -2\equiv4\pmod6. That is 4+N, which equals 1+N since 4\in1+N.

   Why: both facts fall directly out of G's own identity and inverse axioms, carried over coset by coset.

4. 4

   Repeat Step 2 for every remaining pair of cosets if a full table is wanted. Use N as the identity row and column to check the work.

   Example

   Filling in the rest of \mathbb{Z}_6/\{0,3\}'s 3\times3 table this way reproduces the same table found by listing every sum directly.

   Why: every entry follows the identical rule, so no entry ever needs a different method than the one already used.