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.
$\blacksquare$
Sources
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 43.2$: Lagrange's theorem