Criterion for a subset to be a subgroup of a group

Lemma I. A non-empty suset H of a group G is a subgroup iff
                     (i) ab ∈ H,      ∀ a, b ∈ H
                    (ii) a^-1 ∈ H,      ∀ a ∈ H.

Proof: Necessary part: Suppose that a non-empty subset H of a group G is its subgroup.
Therefore H itself forms a group.
               ∴ The conditions (i) and (ii) hold in H, by the definition of a group.

Sufficient Part: Suppose that H is a non-empty subset of a group G such that the conditions (i) and (ii) hold in H.
               ∴ The closure property and the existence of inverse holds in H.
               Now, let a, b, c ∈ H
               ∴ a, b, c ∈ G                                                    [∵ H ⊆ G]
               ∴ (ab)c = a(bc)      Since G is a group.
               So, the associative law holds in H.
               Since H is a non-empty subset of G, so ∃  a ∈ H.
               ∴ By (ii), a^-1 ∈ H
               ∴ a, a^-1 ∈ H
               ∴ a a^-1 ∈ H, from (i)
               ⇒ e ∈ H, where e is identity element of G.
               ⇒ The identity element exists in H.
                    Therefore, H itself is a group.
               Thus H is a subgroup of G.

Lemma II: A non-empty subset H of a group G is a subgroup iff
                       ab^-1 ∈ H,      ∀ a, b ∈ H

Proof: Necessary Part: Suppose that a non-empty subset H of a group G is its subgroup.
              ∴ H itself forms a group.
              ⇒ ∀ b ∈ H              ⇒     b^-1 ∈ H.
              Also ∀ a, b ∈ H      ⇒    a, b^-1 ∈ H.
                                                ⇒    ab^-1 ∈ H.       (∵ H is a group)

Sufficient Part: Suppose that H is a non-empty subset of a group G such that
                             ∀ a, b ∈ H      ⇒    ab^-1 ∈ H.
               To prove H is a subgroup of G.
               Clearly associative law holds in H, as it holds in G and elements of H are also elements of G.
Further,
                ∀ a ∈ H  ⇒ a, a^-1 ∈ H  ⇒ aa^-1 ∈ H  ⇒ e ∈ H
                i.e., identity element exists in H.
                Now e, a ∈ H   ⇒ ea^-1 ∈ H     ⇒ a^-1 ∈ H
                ∴ ∀ a ∈ H ⇒ a^-1 ∈ H
                i.e., inverses are there in H.
                Also ∀ a, b ∈ H      
                ⇒    a, b^-1 ∈ H.                  (∵ inverses are there in H)
                ⇒  a (b^-1)^-1 ∈ H
                ⇒  ab ∈ H.
                So, closure property holds in H.
                ∴ H is a group in itself under the operation of G.
                Hence H is a subgroup of G.

Lemma III: A non-empty finite subset H of a group is a subgroup of G iff
                  ab ∈ H, ∀ a, b ∈ H

Proof: Necessary Part: Let a non-empty finite subset H of a group G be its subgroup.
              ∴ H itself is a group.
              ∴ ab ∈ H, ∀ a, b ∈ H                  (By closure property)

Sufficient Part: Suppose that H is a non-empty finite subset of a group G such that
              ab ∈ H, ∀ a, b ∈ H   
              ∴ The operations of multplication is a binary operation on H.
              Let a, b, c ∈ H
              ∴ a, b, c ∈ G                                                     [∵ H ⊆ G]
              ∴ (ab)c = a(bc)      Since G is a group.
               So, the associative law holds in H under multiplication.
              Firstly we prove that cancellation laws hold in H.
              Let a, b, c ∈ H      such that ab = ac.
              Since a ∈ H,        so a ∈ G.
              ∴ a^-1 ∈ G       such that      aa^-1 = e = a^-1a
              Now ab = ac
              ⇒ a^-1(ab) = a^-1(ac)
              ⇒ (a^-1a)b = (a^-1a)c
              ⇒ eb = ec               ⇒ b = c
              ⇒ ab = ac              ⇒ b = c
Similarly, ba = ca               ⇒ b = c
              So. The cancellation laws hold in H.
              ∴ H is a non-empty finite set with an associative binary operation in H and the cancellation laws hold in H.
              ∴ H itself is a group.
              Thus H is a subgroup of G.