Cosets are Equivalent/Proof 1

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:
 * $\forall x \in G: \left|{x H}\right| = \left|{H}\right| = \left|{H x}\right|$

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

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 Elements in Coset iff Product with Inverse in Coset.

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 $H \simeq Hx$. Because $x$ was chosen arbitrarily $H \simeq Hx$ holds for every $x in G$ which in particular means that $Hx \simeq H$ and $Hy \simeq H$ for every two cosets $Hx$ and $Hy$, so $Hx \simeq Hy$ for every $x$ and $y$ in $G$.

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 $H \simeq Hx$. Also similarly $xH \simeq yH$ for $x$ and $y$ in $G$.

Hence the result: $Hy \simeq Hx \simeq H \simeq Hx \simeq Hy$.

If order $|H|$ is finite, then bijections above give $|Hy| = |Hx| = |H| = |Hx| = |Hy|$, which might be not true if $|H|$ is infinite (this result holds if $|H| = \infty$ for cardinality $\# H \geq \aleph_0$).