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A group homomorphism is a function f : G → G such that f (u + v) = f (u) · f (v) . Some examples follow. 7 Example. Let G = R3 and G = R with the usual sum. We define f : G → G by the rule f (x, y, z) = 8x − 4y + 4z . We will show that f is a homomorphism. As f ((x1 , y1 , z1 ) + (x2 , y2, z2 )) = f (x1 + x2 , y1 + y2 , z1 + z2 ) = 8(x1 + x2 ) − 4(y1 + y2 ) + 4(z1 + z2 ) y f (x1 , y1, z1 ) + f (x2 , y2 , z2 ) = (8x1 , −4y1 + 4z1 ) + (8x2 − 4y2 + 4z2 ), f is a homomorphism. 8 Proposition. Let f : G → G be a homomorphism of groups.

Consider x x = e for any element x ∈ G . Consider the left inverse element of x−1 , that is, (x−1 )−1 x−1 = e . Then xx−1 = e(xx−1 ) = ((x−1 )−1 x−1 )(xx−1 ) = (x−1 )−1 ex−1 = (x−1 )−1 x−1 = e. Hence x−1 is a right inverse of x . Now, for any element x , consider the equalities xe = x(x−1 x) = (xx−1 )x = ex = x. Thus e is a right identity. We say that e is the identity element of a group G if e is a left or right identity element and we talk about the inverse of an element if its left or right inverse exist.

Then g(xhx−1 ) = g(x)g(h)g(x−1) = g(x)eg(x−1 ) = g(x)(g(x))−1 = e. Therefore, xhx−1 ∈ ker(g : G −→ G ) = H . By the previous corollary and proposition, the normality condition is necessary, and sufficient, for the concept of quotient group. 9 Theorem. (Lagrange) If G is a group of order n and H < G , then o(H)|o(G) . com 64 An Introduction to Group Theory Quotient Groups Proof. 2) and they are either disjoint or equal. Thus, n = rm , that is, o(H)|o(G) . The number of left (or right) cosets of a subgroup H < G will be denoted (G : H) and we will call it the índex of H in G , that is, (G : H) = o(G/H) .

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