# Tour: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).