Proof: To establish one to one correspondence between two sets X and Y we have to exhibit a mapping f : X → Y such that ∀ a ₁ , a ₂ ∈X , a ₁ = a ₂ if and only if f( a ₁ )=f( a ₂ ) where f( a ₁ ), f( a ₂ ) ∈Y . Also f has to be onto. As in our case we have if h ₁⊕ a = h ₂ ⊕ a then h ₁ = h ₂ and thus h ₁ ⊕ b = h ₂ ⊕ b . The function is also onto because for every element h ⊕ b in the range there is an inverse element h ⊕ a in the domain.

Theorem 2 [Lagrange]: Let (G, ⊕ ) be a finite group and H is a subgroup of G. Then o(H) | o(G).

Proof: Notation o(S) = |S|

Let k be the number of right cosets. Thus k * o(H) = o(G) and o(H) | o(G).

Order of an element

Let (G, ⊕ ) be a finite group. Let a ∈ G. Then order (a) is defined as the smallest positive integer t such that a ^{( t )} = e. [ a ^t = a ⊕ a ⊕ a ... ⊕ a t times].

Theorem 3 : For any finite group (G,*) and any a ∈ G the order of the element is equal to the size of the subgroup it generates i.e., ord (a) =|<a>|.

Proof: < a > = e , a , a ² , a ³ ..........

Let t = ord ( a ). So a ^(t) = e .

⇒  a^{( t ) *} a ^{( k )}

⇒  e ^*a^{( k )} = a ^{( k )}

If j > t ∃ i < j such that a^{( i )} = a^{( j )}

Thus no new elements are generated beyond a^{( t )} . Hence |< a >| ≤ t .

Now we have to show that |< a >| ≥ t by proving all elements in < a > = { a¹ , a² ,….., a^t } are distinct. Assume otherwise ⇒∃1≤ i < j ≤ t such that a ⁱ = a ^j . Let t be j + k . Hence a ^i+k = a ^{j + k} = a^{( t )} = e ⇒ a^{i +( t - j )} = e and we know that i + t - j < t . Thus we arrive at a contradiction since t = ord ( a ) is the smallest power to which a has to be raised to become identity. Thus our assumption ∃1 ≤ i < j ≤ t such that a ⁱ = a ^j is incorrect . Therefore, each element of the sequence a⁽¹⁾ , a⁽²⁾ , ..., a^{( t )} is distinct, and |< a > | ≥ t . Thus we conclude that ord ( a ) = |< a > |

Corollary 1 :

Let (G, ⊕ ) be a finite group with identity e then for all a ∈ G we have a^ord(G) = e .

Proof : Consider the subgroup < a > of G. From Theorem 3 |< a >| = ord( a ). From Lagrange's theorem ord ( a ) | ord(G). Let ord (G) be k * ord( a ). Thus a ^ord(G) = a ^{k *ord(a)} = e ^k = e .

Consider the group ( Z^*_n, ^* n ). We already know that | Z_n^* | = Φ( n ).

Euler's Theorem :

For any integer n > 1

a^{Φ( n )} ≡ 1 mod n for all a ∈ Z_n^* . [Corollary 1]

Fermat's Theorem :

If p is a prime then | Z_p^* | = Φ ( p ) = p -1.

a^{( p -1)} ≡ 1 mod p for all a ∈ Z_p^*.[Corollary 1]

Reference:

1. Introduction to Algorithms , Second Edition, T. H. Cormen, C. E. Leiserson, R. Rivest and C. Stein, Prentice Hall India .

2. Topics in Algebra , Second Edition, I. N.Herstein, John Wiley .