Cosets are Equivalent/Proof 1

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


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

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.