# Left and right coset spaces are naturally isomorphic

This article gives the statement, and possibly proof, of a basic fact in group theory.
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This article describes a natural isomorphism between two structures or between a family of structures

## Statement

### Statement with symbols

Let $G$ be a group and $H$ a subgroup of $G$. Then, if $L$ denotes the set of left cosets of $H$ (viz, sets of the form $xH$), and $R$ denotes the set of right cosets of $H$ (viz, sets of the form $Hx$), then there is a natural bijection between $L$ and $R$.

This bijection is induced by the set-theoretic map sending $x$ to $x^{-1}$.

## Related facts

### More general equivalence of left and right

This result is part of a more general result, which states that left and right play equivalent roles in the theory of groups:

### Breakdown for other algebraic structures

The result has no analogue for monoids, semigroups, and other structures. It is true for involutive monoids (monoids equipped with an involution) when the submonoid is closed under that involution. Of course, in the case of monoids, other facts about cosets break down -- for instance, the cosets are not necessarily pairwise disjoint, and they do not necessarily have equal size.

## Facts used

• Inverse map is involutive: we use the fact that $(ab)^{-1} = b^{-1}a^{-1}$, and also that $(a^{-1})^{-1} = a$.

## Proof

Given: A group $G$, subgroup $H$

To prove: The map $x \mapsto x^{-1}$ establishes a bijection between the left cosets of $H$ in $G$ and the right cosets of $H$ in $G$.

Proof: Note that by fact (1), the map $x \mapsto x^{-1}$ is bijective, so every element occurs as the image of exactly one element.

1. The map sends each left coset bijectively to exactly one right coset, and every right coset occurs as the image of exactly one left coset: Any element of the form $xh$ with $h$ in $H$ gets mapped to $h^{-1}x^{-1}$ (we're using fact (1) here), which lies in $Hx^{-1}$. Hence, the image of $xH$ under the inverse map is inside $Hx^{-1}$. Further, any element in $Hx^{-1}$ arises as the image of a unique element under the inverse map and that element is in $xH$: if $hx^{-1}$ is such an element, it arises as the image of the element $xh^{-1} \in xH$ (again, using fact (1)).
2. The map sends each right coset bijectively to exactly one left coset, and every left coset occurs as the image of exactly one right coset: Any element of the form $hx$ with $h$ in $H$ gets mapped to $x^{-1}h^{-1}$, which lies in $x^{-1}H$ (we're using fact (1) here). Hence, the image of $Hx$ under the inverse map is inside $x^{-1}H$. Further, any element in $x^{-1}H$ arises as the image of a unique element under the inverse map under the inverse map, and that element is in $Hx$: if $x^{-1}h$ is such an element, it arises as the image of the element $h^{-1}x \in Hx$ (again using fact (1)).

Thus, the map $x \mapsto x^{-1}$ gives a well-defined bijective map from left cosets to right cosets, as well as from right cosets to left cosets.

## References

### Textbook references

• Groups and representations by Jonathan Lazare Alperin and Rowen B. Bell, ISBN 0387945261, More info, Page 5 (informal, inline proof)
• Topics in Algebra by I. N. Herstein, More info, Page 47, Exercise 5
• Algebra by Michael Artin, ISBN 0130047635, 13-digit ISBN 978-0130047632, More info, Exercise 9 (b) of Section 6 (Cosets) (this doesn't ask for a natural isomorphism; it simply asks for a proof that the number of left cosets equals the number of right cosets)
• Abstract Algebra by David S. Dummit and Richard M. Foote, 10-digit ISBN 0471433349, 13-digit ISBN 978-0471433347, More info, Page 96, Exercise 12, Section 3.2 (More on Cosets and Lagrange's theorem)