Left and Right Coset Spaces are Equivalent/Proof 2

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Theorem

Let $\struct {G, \circ}$ be a group.

Let $H$ be a subgroup of $G$.


Let:

$x H$ denote the left coset of $H$ by $x$
$H y$ denote the right coset of $H$ by $y$.


Then:

$\order {\set {x H: x \in G} } = \order {\set {H y: y \in G} }$


Proof

Let $G$ be a group and let $H \le G$.

Consider the mapping $\phi$ from the left coset space to the right coset space defined as:

$\forall g \in G: \map \phi {g H} = H g^{-1}$


We need to show that $\phi$ is a bijection.


First we need to show that $\phi$ is well-defined.

That is, that $a H = b H \implies \map \phi {a H} = \map \phi {b H}$.

Suppose $a H = b H$.

\(\displaystyle a H = b H\) \(\iff\) \(\displaystyle a^{-1} b \in H\) Left Cosets are Equal iff Product with Inverse in Subgroup‎
\(\displaystyle H a^{-1} = H b^{-1}\) \(\iff\) \(\displaystyle a^{-1} \paren {b^{-1} }^{-1} \in H\) Right Cosets are Equal iff Product with Inverse in Subgroup


But $a^{-1} \paren {b^{-1} }^{-1} = a^{-1} b \in H$ as $a H = b H$.

So $H a^{-1} = H b^{-1}$ and $\phi$ is well-defined.


Next we show that $\phi$ is injective:

Suppose $\exists x, y \in G: \map \phi {x H} = \map \phi {y H}$.

Then $H x^{-1} = H y^{-1}$, so $x^{-1} = e_G x^{-1} = h y^{-1}$ for some $h \in H$.

Thus $h = x^{-1} y \implies h^{-1} = y^{-1} x$.

As $H$ is a subgroup, $h^{-1} \in H$.

Thus:

$y^{-1} x \in H$

So by Left Cosets are Equal iff Product with Inverse in Subgroup:

$x H = y H$

Thus $\phi$ is injective.


Next we show that $\phi$ is surjective:

Let $H x$ be a right coset of $H$ in $G$.

Since $x = \paren {x^{-1} }^{-1}$, $H x = \map \phi {x^{-1} H}$ and so $\phi$ is surjective.


Thus $\phi$ constitutes a bijection from the left coset space to the right coset space, and the result follows.

$\blacksquare$


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