Cosets are Equivalent/Proof 1

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


 * $\forall u \in H: \theta \left({u}\right) = u x$
 * $\forall v \in H x: \phi \left({v}\right) = v x^{-1}$

Note that $v \in H x \implies v x^{-1} \in H$, from Element in Right Coset iff Product with Inverse in Subgroup.

Now:


 * $\forall v \in H x: \theta \circ \phi \left({v}\right) = v x^{-1} x = v$
 * $\forall u \in H: \phi \circ \theta \left({u}\right) = u x x^{-1} = u$

Thus $\theta \circ \phi = I_{Hx}$ 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 $Hx \simeq H$ and $Hy \simeq H$ 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:


 * $\forall u \in H: \alpha \left({u}\right) = x u$
 * $\forall v \in x H: \beta \left({v}\right) = x^{-1} v$

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.