Theorems

Theorem 1: A semi-group in which both the equations ax = b and ya = b have a unique solution, is a group. 
(It is also called a definition of  a group)

Proof: Let G be a semi-group under an operation denoted multiplicatively in which both the equations 
ax = b                     ... (1)        and ya = b                            ... (2)
have a unique solution.
To show that G be a group. For this we show that
(i) identity element exists in G. and 
(ii) inverse of each element exists in  G.
For (i) By condition (1). For any element a ∈ G, we have
         ax = a, has a unique solution in G.
         ∴ ∃ an element e ∈ G such that ae = a
         Let b ∈ G be an element of G. Then by condition (2)
                              ya = b     i.e.,    b = ya
         Now be = (ya)e = y(ae) = ya = b
         ⇒                            be = b
         ∴ e is the right identity of G.
         Similarly, by condition (2), for any element a ∈ G, we have
         ya = a, has a unique solution in G.
         ∴ ∃ an element f ∈ G such that fa = a and fb = b
         i.e., f is the left identity of G.
         Now              fe = f                        [∵ e is the right identity]
         and               fe = e                        [∵ e is the right identity]
         ⇒                   e = f
         ∴ e is the identity element of G.
For (ii) Let a ∈ G be any element and e be the identity element of G.
         Then by condition (1) and (2) ∃ a' , a'' ∈ G such that
                          aa' = e        and          a''a = e
         Now         a'' = a'' e = a'' (aa') = (a'' a)a' = ea' = a'
         Thus inverse of each element in G exists and is unique.
         Hence G is a group.

Theorem 2: Prove that any finite semi-group iff both the cancellation laws hold.
         (It is also called a definition of a group, but for finite sets)

Proof: Let G be a semi-group under an operation denoted multiplicatively.
         Let G be a group, then both the cancellation laws hold.  (by III property in last article)

        Conversely: Let both the cancellation laws hold.
        To prove G is a group.
        Since G is finite.
        Let G = {a₁, a₂, ... , a_n} be different elemnets of G.
        ∴ O(G) = n.
        ∀ a ∈ G, consider S = {a₁a, a₂a, ... , a_na}.
        Due to closed property in G, S ⊆ G.
        Further all the elements of S are different.
        For it, let                a_ia = a_ja,  i ≠ j           i.e.,  a_i ≠  a_j ∈ G.
        Using right cancellation law, we get
                                          a_i = a_j , which is absurd.
        ∴ all the elements of S are different.
        ∴ O(S) = n = O(G)
        ⇒ S = G.
        ∀ a, b ∈ G, b ∈ G but G = S ⇒ b ∈ S
         let                            b = a_la 
         i.e., a_l is a solution of the equation ya = b, ∀ a, b ∈ G.
         Consider another set T = {aa₁, aa₂, ... , aa_n}.
         T ⊆ G and all the elements of T are different.
         For it, let              aa_i = aa_j,  i ≠ j           i.e.,  a_i ≠  a_j ∈ G.
         Using left cancellation law, we get
                                          a_i = a_j , which is absurd.
        ∴ all the elements of T are different.
        ∴ O(T) = n = O(G)
        ⇒ T = G.
        ∀ a, b ∈ G, b ∈ G but G = T ⇒ b ∈ T
         let                            b = aa_k 
         i.e., a_k is a solution of the equation ax = b, ∀ a, b ∈ G.
         Thus both the equations ax = b and ya = b ∀ a, b ∈ G have solutions in G.
         Hence G is a group.