Symmetric:

Let a , b ∈ G. Since a ≡ b mod H

a ⊕ b ^-1 ∈ H

⇒  ( a ⊕ b ^-1 ) ^-1∈ H [Since H is a subgroup]

⇒  ( b ^-1 ) ^-1 ⊕ a ^-1 ∈ H

⇒  b ≡ a mod H

Transitive:

Let a , b , c ∈ G Λ a = b mod H Λ b ≡ c mod H

                 a ⊕ b ^-1 ∈ H Λ b ⊕ c ^-1∈ H

                 So a ⊕ b ^-1 ⊕ b ⊕ c ^-1∈ H

⇒  a ⊕ c ^-1 ∈ H

⇒  a ≡ c mod H

Cosets: Let (G, ⊕ ) be a group and H is a subgroup of G .Pick an element a belonging to G.

Let H a = { h ⊕ a | h ∈ H} be Right coset

Let a H= { a ⊕ h | h ∈ H} be Left coset

Lemma 2: H a = { x | x ≡ a mod H} ∀a ∈ G

Proof: Let [ a ] = { x | x ≡ a mod H}. We have to show H a =[ a ].

To prove H a ⊆ [ a ] let us pick an element h ⊕ a ∈ H a . Thus ∀ h∈ H a ⊕ ( h ⊕ a ) ^-1 = a ⊕a ^-1 ⊕ h ^-1 = h ^-1 ∈H [Since H is a subgroup].

⇒  h ⊕ a ∈[a]

To prove that [ a ]⊆ H a let us pick an element x ∈[ a ]

⇒  a ⊕ x ^-1 ∈ H

⇒  ( a ⊕ x ^-1 ) ^-1∈ H

⇒  x ⊕ a ^-1 ∈ H

Therefore for some h ∈H we have x = h ⊕ a . Thus x ∈ H a and [a]⊆ Ha.

Lemma 3: There is a one to one correspondence between any 2 right cosets of G.