1. Properties of Groups Proposition 1: Let (G,  ) be a group. i.The inverse element of any element of G is unique. Remark: In view of i., we may use the notation a –1 for the inverse of a. ii.Cancellation Law: for a,b,c  G, if a  b = a  c, then b = c. iii.The equations a  x = b and x  a = b, where a, b  G, have the solutions x = a –1  b and x = b  a –1 respectively. iv.For a  G, and n a positive integer, define a n = a  a  …  a (n times), a –n = a –1  a –1  …  a –1 (n times) and a 0 = e (identity element of the group G). Then the usual exponentiation laws hold, i.e. a m  a n = a m + n, a m  a – n = a m – n and (a m ) n = a m n. v.For a,b  G, (a  b) –1 = b –1  a –1

2. Isomorphism of Groups Definition: Let (G,  ) and (H,  ) be groups, and let  : G  H be a bijection on G to H with the further condition that  (a  b ) =  (a)   (b) for all a, b  G. Then  is said to be an isomorphism between G and H, and G and H are said to be isomorphic groups. Example of isomorphism: Consider the group (H,  ), where H = {(0,0),(1,0),(0,1),(1,1)} and the operation  is defined as component-wise addition mod 2. Then the mapping   : K 4  H defined by: e  (0,0), a  (1,0), b  (0,1), c  (1,1) is an isomorphism.

3. Subgroups Definition: Let (G,  ) be a group and let H be a non- empty subset of G such that H with the operation  is itself a group, i.e. (H,  ) satisfies all the four group axioms. Then H is said to be a subgroup of G. We use the notation H  G to indicate that H is a subgroup of G. Examples: 1.( Z, +) is a subgroup of ( Q, +), which is in turn a subgroup of ( R, +), which is in turn a subgroup of ( C, +). 2.( Q  {0},  ) is a subgroup of ( R  {0},  ). However, ( Q  {0},  ) is not a subgroup of ( Q, +), since the operation is not the same for the two sets.

4. Properties of Subgroups Proposition 2: Let (G,  ) be a finite group. Then a non- empty subset H of G is a subgroup if and only if H satisfies the closure property. Proof: [  ] Obvious from definition of subgroup. [  ] Suppose H satisfies the closure property. Let H = {x 1,x 2,…,x n }. Now consider the set J = {x 1 x 1, x 1 x 2,…, x 1 x n }. 1.Because of the closure property, J  H, and because of the Cancellation Law for G, all elements of J are distinct. Hence, |J| = n, and so J = H. In particular, x 1 x k = x 1 for some k. Hence, x 1 x k = x 1 e, and so by Cancellation Law for G, x k = e, i.e. e  H. This shows that H satisfies the identity property. 2.Again, since J = H, we must have x 1 x j = e = x 1 x 1 -1 for some j. Applying Cancellation Law, we get that x j = x 1 -1, so again we have that x 1 -1  H. Repeating this argument, we see that x i -1  H for i = 1,2,..,n. This shows that H satisfies the inverse property. Hence, H is a subgroup of G, as was required to be proved.