Tour:Left cosets are in bijection via left multiplication

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This article adapts material from the main article: left cosets are in bijection via left multiplication

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The different left cosets of a subgroup are in bijection with each other, under the action of the group via left multiplication. This means that all left cosets of a subgroup are of the same size.
PREREQUISITES: Definition of group, subgroup and coset, invertible implies cancellative, and Tour:Mind's eye test two (beginners)#Left and right multiplication maps
WHAT YOU NEED TO DO: Read and understand the statement and proof below. If you find it hard, refer back to the suggested prerequisites pages.

Statement

Statement with symbols

Let H be a subgroup of a group G and let xH and yH be two left cosets of H. Then, there is a bijection between xH and yH as subsets of G, given by the left multiplication by yx^{-1}.


Facts used

  1. Invertible implies cancellative in monoid: In particular, we can cancel elements in a group: if ab = ac, then b = c.

Proof

Given: A group G, a subgroup H, and two left cosets xH, yH of H

To prove: Left multiplication by yx^{-1} establishes a bijection between xH and yH.

Proof: We prove that left multiplication by yx^{-1} sends xH to yH, is surjective, and is injective.

  1. Well-defined as a map from xH to yH: First, note that if g = xh then yx^{-1}g = yh. Thus, any element in xH gets mapped to an element in yH.
  2. Surjective: Every element of the form yh with h in H arises as yx^{-1}(xh), hence, it arises as the image of left multiplication by yx^{-1}. Thus, the map from xH to yH is surjective.
  3. Injective: Given two distinct elements xh_1, xh_2 \in xH, the elements (yx^{-1})xh_1 = yh_1 and (yx^{-1})xh_2 = yh_2 are also distinct, because by fact (1), if they were equal, then canceling yx^{-1} from both sides would give xh_1 = xh_2. Thus, left multiplication by yx^{-1} sends distinct elements to distinct elements, so the map is injective.

Thus, left multiplication by yx^{-1} is a bijection from xH to yH.

Sidenote

In general, there is no natural bijection between two left cosets -- the bijection depends on a choice of element in both cosets (the elements x and y in the above description).

This page is part of the Groupprops guided tour for beginners (Jump to beginning of tour)
PREVIOUS: Left cosets partition a group| UP: Introduction three (beginners)| NEXT: Right coset of a subgroup
General instructions for the tour | Pedagogical notes for the tour | Pedagogical notes for this part