Left and right coset spaces are naturally isomorphic

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 $$x^{-1}$$.

Other facts about left and right cosets

 * Left cosets are in bijection via left multiplication
 * Equivalence of definitions of coset: This states that every left coset of a subgroup is the right coset of a (possibly different) subgroup.
 * Left cosets partition a group

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:


 * Every group is naturally isomorphic to its opposite group: This states that the inverse map establishes a natural isomorphism between any group and its opposite group.

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.

Textbook references

 * , Page 5 (informal, inline proof)
 * , Page 47, Exercise 5
 * , 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)
 * , Page 96, Exercise 12, Section 3.2 (More on Cosets and Lagrange's theorem)