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**Example text**

2 Proposition. The left and right congruence relations are equivalence relations. Proof. As x ≡i x (mod H) ⇐⇒ x−1 x = e ∈ H , the relation is reflexive. As x ≡i y (mod H) ⇐⇒ x−1 y ∈ H ⇐⇒ (x−1 y)−1 ∈ H ⇐⇒ y −1 x ∈ H ⇐⇒ y ≡i x (mod H) the relation is symmetric. Finally, if x ≡i y (mod H) and y ≡i z (mod H) then x−1 y ∈ H and y −1 z ∈ H . Hence (x−1 y)(y −1z) ∈ H ⇐⇒ x−1 ez = x−1 z ∈ H. Thus x ≡i z (mod H) and the relation is transitive. Analogously for the right congruence. 3 Proposition. The left and right equivalence clases [x] of the relation defined can be expressed as xH = {xh|h ∈ H} and Hx = {hx|h ∈ H} respectively.

In line with the corevalue to be ‘First’, the company intends to expand its market position. Dedicated Analytical Solutions FOSS Slangerupgade 69 3400 Hillerød Tel. com 40 Click on the ad to read more An Introduction to Group Theory Elementary Properties Proof. Let f : G → G and g : G → G be group homomorphisms. Then (g ◦ f )(x + y) = g(f (x + y)) = g(f (x) + f (y)) = g(f (x)) + g(f (y)) = (g ◦ f )(x) + (g ◦ f )(y) Hence (g ◦ f ) is a homomorphism. 11 Definition. Let f : G → G be a group homomorphism.

Proof. If xy = xz , then x−1 (xy) = x−1 (xz) . By associativity, (x−1 x)y = (x−1 x)z . Hence, ey = ez and, finally, y = z . If yx = zx , then y = z , which can be proved in the same way. 4 Proposition. In an arbitrary group G , the inverse of any element is unique. Proof. Let x be another inverse element of the element x . Then, x x = e . We also know that x−1 x = e . Thus, x x = x−1 x = e . By the previous proposition, x = x−1 . 5 Proposition. In an arbitrary group G , if x, y ∈ G, the equations xa = y and bx = y have a unique solution in G .