Cosets

The following definitions and lemmas are taken from Jean Tignol's Galois' Theory of Algebraic Equations.

Definition 1: g + H

Let g + H = { g + h such that h ∈ H }.

Example 1.1:

Let G be the set { 0, 1, 2, 3 }

Let H be the set { 9, 10, 11 }

Then:

0 + H = { 9, 10, 11 }

1 + H = { 10, 11, 12 }

2 + H = { 11, 12, 13 }

3 + H = { 12, 13, 14 }


Definition 2: coset of H in G

The set g + H is a coset if and only if there exists a group G such that g ∈ G and H is a subgroup of G.

Note: Technically, definition 3 above describes a left coset. A coset can also be defined as H + g which is called a right coset. The notation of g + H defines a coset based on addition. Cosets can also be defined on multiplication and represented as gH or Hg.

Example 2.1:

Let G = the set of integers Z
Let H = 2Z = the set of even integers

1 ∈ Z
1+H = odd integers = { ... -3, -1, 1, 3, ... }

Example 2.2: Cyclic Group Z₄

Z₄ = { 0, 1, 2, 3}

It is a group since:

(1) Closure

Since Z₄ is always modulo 4, it is clear that that operation of addition is closed (for example: 3 + 2 = 1)

(2) Associativity

For any elements a,b,c ∈ Z₄, (a + b) + c = a + (b + c)

(3) Identity Element

0 is the identity element

(4) Inverse Element

Since it is modulo 4, each element has an inverse: 0+0=0, 1+3=0, 2+2=0

Let H = { 0 , 2 }

H is a subset and H is itself a group since:

(1) It has closure: 0+2=2, 0+0=0, 2+2=0

(2) It has associativity.

(3) It has a identity element: 0

(4) From #1, it is clear that each element is its own inverse.

From Z₄ and H, there are 2 distinct cosets:

0 + H = { 0, 2}
1 + H = { 1, 3}
2 + H = { 2, 0} = 0 + H
3 + H = { 3, 1} = 1 + H

Lemma 1:

For the coset H in G:

For all a,b ∈ G:

a + H = b + H if and only if a - b ∈ H

Proof:

(1) Assume that a + H = b + H

(2) So, for all x ∈ (a + H) → x ∈ (b + H)

(3) a ∈ a + H, b ∈ b + H [since H is a group and therefore 0 ∈ H]

(4) a ∈ b + H [follows directly from step #2]

(5) Let c = a - b

(6) b + c ∈ b + H [since b + c = a and a ∈ b + H]

(7) So c ∈ H [from step #6 and Definition 3 above]

(8) Assume that (a-b) ∈ H

(9) Assume that x ∈ a + H

(10) Let y = x - a

(11) y ∈ H [since x = a + y]

(12) y + (a - b) ∈ H since H is a group

(13) x - b ∈ H [since y + (a - b) = (x - a) + (a - b) = x - b]

(14) Then x ∈ b + H since [b + x-b = x]

(15) We can make the same argument if x ∈ b + H so this shows that a + H = b + H.

QED

Lemma 2:

For the coset H in G:

a ∈ H → a + H = H

Proof:

(1) Assume that a ∈ H

(2) Assume that x ∈ a + H

(3) Then there exists h such that x = a + h where h ∈ H [See Definition 1 above]

(4) But if a ∈ H and h ∈ H, then a + h ∈ H. [since H is a group, see Definition 2 above and the Closure property of groups]

(5) So x ∈ H

(6) Assume that x ∈ H

(7) Let b = x - a

(8) Since a ∈ H, it follows that -a ∈ H [See Definition 2 above and the Inverse property of groups]

(9) Since x ∈ H and -a ∈ H, it follows that b ∈ H [from the Closure property of groups]

(10) So then, x ∈ a + H [since b ∈ H and x = a + b from Definition 1 above]

QED

Reference

• Jean-Pierre Tignol, Galois' Theory of Algebraic Equations, World Scientific, 2001