Algebraic Structures [Lecture notes] by Thomas Markwig Keilen

By Thomas Markwig Keilen

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D. The set S1 = {(x, y) ∈ R2 | x2 + y2 = 1} is the circle of radius one whose centre is the origin (0, 0). Show that the map is surjective. Φ : S1 −→ P1R : (x, y) → (x, y) e. Suppose we allow in the definition of ∼ all elements v, w ∈ R2, not only the non-zero ones. Is ∼ then an equivalence relation on R2? If so, what is the equivalence class of (0, 0)? 13 (The integers) Let M = N × N and let m = (a, b) ∈ M and m ′ = (a ′ , b ′ ) ∈ M be two elements in M. We define m ∼ m ′ ←→ a + b ′ = a ′ + b.

A) The Theorem of Lagrange In this section we consider a particular type of equivalence relation. The underlying set will be a group, and for the definition of the equivalence relation we will use the following notion. 1 Let (G, ·) be a group and A, B ⊆ G be two subsets of G. We define A · B = {a · b | a ∈ A, b ∈ B}. Sometimes we write short AB for A · B, and if A = {g} consists only of one element we write gB instead of {g}B and Bg instead of B{g}. e. for A, B, C ⊆ G we have (A · B) · C = {(a · b) · c | a ∈ A, b ∈ B, c ∈ C} = {a · (b · c) | a ∈ A, b ∈ B, c ∈ C} = A · (B · C).

Then there is a z ∈ x ∩ y, and we have z ∼ x and z ∼ y. By symmetry we also get x ∼ z and then x ∼ y by transitivity. Let now u ∈ x be any element. Then u ∼ x and again by transitivity we have u ∼ y. Hence, u ∈ y and therefore x ⊆ y. Exchanging the roles of x and y the same argument shows that x = y. 9 Let M be a finite set, ∼ be an equivalence relation on M, and M1, . . , Ms be the pairwise different equivalence classes of ∼. Then: s |M| = |Mi|. 8. 10 Let M = {(an)n∈N | an ∈ Q} be the set of all sequences of rational numbers.

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