Cosets are either identical or disjoint

In the same way that equivalence classes are either identical or disjoint, so are cosets.

Consider a subgroup H of G, and elements a, b &isin G. Then H has right cosets Ha and Hb.

Either:

• Ha &cap Hb = ∅, in which case there's nothing more to show

OR:

• ∃c &isin G s.t. c ∈ Ha &cap Hb

In the second case,

• c = h₁a = h₂b for some h₁, h₂ &isin H

and so

• a = h₁^-1h₂b = h₃b, where h₃ = h₁^-1h₂ &isin H

and

• b = h₂^-1h₁a = h₄a, where h₄ = h₂^-1h₁ &isin H

A coset of a subgroup has the same order as the subgroup.

This one is just for practice at constructing a proof. The result is trivial.

Consider a finite subgroup H of G with order n. Then

• H = {h₁, h₂, ..., h_n}

and ∀g &isin G,

• Hg = {h₁g, h₂g, ..., h_ng}

where h₁g, h₂g, ..., h_ng &isin G.

Assume ∃a,b<n, a &ne b such that

• h_ag = h_bg

Then by the cancellation laws,

• h_a = h_b

a contradiction. Therefore, Hg has n distinct elements. That is, the coset has the same order as the subgroup.

A closed subset of a group is a group

Let (G, ⋅) be a group, and let H ⊂ G. If H is closed under ⋅, then H is a subgroup.

Proof:

Let H have m elements, so that

• H = {a₁, a₂, ..., a_m}

Then

• Ha₁ = {a₁², a₂a₁, ..., a_ma₁}

By the cancellation laws, Ha₁ has m elements as well, and since H is closed and a₁ ∈ H, Ha₁ = H.

Since a₁ &isin H = Ha₁, ∃a_j &isin H such that a₁ = a_ja₁. Therefore, a_j = e (∈H). __(1)

Also, ∃a_k &isin H such that a_ka₁ = a_j. That is, a_k = a₁^-1 &isin H.

Similarly, by looking at Ha_i, for i = 2..m, a_i^-1 &isin H. __(2)

So, H:

• is closed (hypothesis)
• is associative (inherited)
• has identity (1), and
• contains inverses (2)

Therefore, H is a group, and so H is a subgroup of G.

Equivalence classes are either identical or disjoint

Let S be a non-empty set with equivalence relation ~, and ∀a ∈ S let

• [a] = {x∈S | a ~ x}

be the equivalence class of a.

Then two equivalence classes [a] and [b] are either identical or disjoint.

Proof:

Consider [a] ∩ [b]. If [a] ∩ [b] = ∅, well and good. Otherwise, ∃c ∈ S such that c ∈ [a] ∩ [b].

That is, c ~ a and c ~ b. By transitivity of equivalence relations, a ~ b. If x ∈ [a] then x ∈ [b], so [a] ⊆ [b].

Similarly, and by reflexivity of equivalence relations, [b] ⊆ [a].

Since [a] ⊆ [b] and [b] ⊆ [a], [a] = [b].

Thus, equivalence classes are either identical or disjoint.