Cosets are Equivalent

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Theorem

All left cosets of a group $G$ with respect to a subgroup $H$ are equivalent.

That is, any two left cosets are in one-to-one correspondence.

The same applies to right cosets.


As a special case of this:

$\forall x \in G: \order {x H} = \order H = \order {H x}$

where $H$ is a subgroup of $G$.


Proof 1

Let us set up mappings $\theta: H \to H x$ and $\phi: H x \to H$ as follows:

$\forall u \in H: \map \theta u = u x$
$\forall v \in H x: \map \phi v = v x^{-1}$

From Element in Right Coset iff Product with Inverse in Subgroup:

$v \in H x \implies v x^{-1} \in H$


Now:

$\forall v \in H x: \theta \circ \map \phi v = v x^{-1} x = v$
$\forall u \in H: \phi \circ \map \theta u = u x x^{-1} = u$

Thus $\theta \circ \phi = I_{H x}$ and $\phi \circ \theta = I_H$ are identity mappings.

So $\theta = \phi^{-1}$: both are bijections and one is the inverse of the other.

This establishes, for each $x \in G$, the set equivalence between $H$ and $H x$:

$H \simeq H x$

In particular, for any $x, y \in G$, it follows from $H x \simeq H$ and $H y \simeq H$ that:

$H x \simeq H y$

by Set Equivalence is Equivalence Relation.


Similarly, we can set up mappings $\alpha: H \to x H$ and $\beta: x H \to H$ as follows:

$\forall u \in H: \map \alpha u = x u$
$\forall v \in x H: \map \beta v = x^{-1} v$

Analogous to above reasoning gives $\alpha = \beta^{-1}$ which establishes $H \simeq x H$.

Also similarly, $x H \simeq y H$ for all $x, y \in G$.


Hence the result.

$\blacksquare$


Proof 2

Follows directly from Set Equivalence of Regular Representations.

$\blacksquare$


Sources