Left and Right Coset Spaces are Equivalent/Proof 1

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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 there be exactly $r$ different left cosets of $H$ in $G$.

Let a complete repetition-free list of these left cosets be:

$a_1 H, a_2 H, a_3 H \ldots, a_r H: a_1, a_2, \ldots, a_r \in G$


From Left Coset Space forms Partition, every element of $G$ is contained in exactly one of the left cosets.

Let $x \in G$.

Then, for $1 \le i \le r$:

\(\ds x\) \(\in\) \(\ds H a_i^{-1}\)
\(\ds \leadstoandfrom \ \ \) \(\ds x \paren {a_i^{-1} }^{-1}\) \(\in\) \(\ds H\) Element in Right Coset iff Product with Inverse in Subgroup
\(\ds \leadstoandfrom \ \ \) \(\ds x a_i\) \(\in\) \(\ds H\) Inverse of Group Inverse
\(\ds \leadstoandfrom \ \ \) \(\ds \paren {x^{-1} }^{-1} a_i\) \(\in\) \(\ds H\) Inverse of Group Inverse
\(\ds \leadstoandfrom \ \ \) \(\ds x^{-1}\) \(\in\) \(\ds a_i H\) Element in Left Coset iff Product with Inverse in Subgroup


Since $x^{-1} \in a_i H$ is true for precisely one value of $i$, it follows that $x \in H a_i^{-1}$ is also true for precisely that value of $i$.


So there are exactly $r$ different right cosets of $H$ in $G$, and a complete repetition-free list of these is:

$H a_1^{-1}, H a_2^{-1}, H a_3^{-1}, \ldots, H a_r^{-1}$


The result follows.

$\blacksquare$


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