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- https://math.libretexts.org/Bookshelves/Abstract_and_Geometric_Algebra/An_Inquiry-Based_Approach_to_Abstract_Algebra_(Ernst)/05%3A_Cosets_Lagranges_Theorem_and_Normal_Subgroups/5.01%3A_CosetsIf a∈G, then the “left" and “right" equivalence classes containing a are given by [a]∼L={g∈G∣a∼Lg} and [a]∼R={g∈G∣a∼Rg}. The next theorem t...If a∈G, then the “left" and “right" equivalence classes containing a are given by [a]∼L={g∈G∣a∼Lg} and [a]∼R={g∈G∣a∼Rg}. The next theorem tells us that the equivalence classes determined by ∼L and ∼R are indeed the left and right cosets of H≤G, respectively.
- https://math.libretexts.org/Bookshelves/Abstract_and_Geometric_Algebra/Abstract_Algebra%3A_Theory_and_Applications_(Judson)/06%3A_Cosets_and_Lagrange's_Theorem/6.01%3A_CosetsLet g1H and g2H be two cosets of H in G. We must show that either g1H∩g2H=∅ or g1H=g2H. Suppose that \(g_1 H \cap g_2 H \neq \emptyse...Let g1H and g2H be two cosets of H in G. We must show that either g1H∩g2H=∅ or g1H=g2H. Suppose that g1H∩g2H≠∅ and a∈g1H∩g2H. Then by the definition of a left coset, a=g1h1=g2h2 for some elements h1 and h2 in H. Hence, g1=g2h2h−11 or g1∈g2H. By Lemma 6.3, g1H=g2H.
- https://math.libretexts.org/Bookshelves/Abstract_and_Geometric_Algebra/Elementary_Abstract_Algebra_(Clark)/01%3A_Chapters/1.08%3A_Cosets_and_Lagrange's_TheoremNote that each corresponds to a way of factoring 72 as a product of prime powers. \[\begin{array} {ll} \mathbb{Z}_9 \times \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2 & \qquad 72 = 9 \cdot 2 ...Note that each corresponds to a way of factoring 72 as a product of prime powers. \[\begin{array} {ll} \mathbb{Z}_9 \times \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2 & \qquad 72 = 9 \cdot 2 \cdot 2 \cdot 2 \\ \mathbb{Z}_9 \times \mathbb{Z}_4 \times \mathbb{Z}_2 & \qquad 72 = 9 \cdot 4 \cdot 2 \\ \mathbb{Z}_9 \times \mathbb{Z}_8 & \qquad 72 = 9 \cdot 8 \\ \mathbb{Z}_3 \times \mathbb{Z}_3 \times \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2 & \qquad 72 = 3 \cdot 3 \cdot 2 \cdot 2…
