By Thomas Markwig Keilen

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**Additional resources for Algebraic Structures [Lecture notes]**

**Sample text**

What we do not know yet is if this product gives a coset again. If think for a second we might see another obvious way to define the product of gU and hU, namely as (gh) · U. The advantage of the latter definition is that it certainly gives a coset again. However, what we cannot be sure of is that the definition is independent of the chosen representative. In an ideal world these two obvious ways of defining the product of cosets would coincide, solving both of our problems since the first definition is obviously independent of the chosen representative and the second one is obviously a coset.

Analogously we could consider the relation: g∼h ⇐⇒ hg−1 ∈ U. This leads in the same way to an equivalence relation, and the equivalence class corresponding to g is Ug. It is called a right coset. 9 for right cosets holds as well, and thus so does the corresponding version of the Theorem of Lagrange, namely, in a finite group G with subgroup U which has m distinct right cosets we have |G| = |U| · m. In particular, the number m of right cosets of U in G and the number |G : U| of left cosets of U in G must coincide!

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. Show that (an)n∈N ∼ (bn)n∈N :⇐⇒ lim (an − bn) = 0 n→∞ defines an equivalence relation on M. 11 We define for two points (x, y), (x ′ , y ′ ) ∈ R2 (x, y) ∼ (x ′ , y ′ ) :⇐⇒ |x| + |y| = |x ′ | + |y ′ |.

### Algebraic Structures [Lecture notes] by Thomas Markwig Keilen

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