Left and Right Coset Spaces are Equivalent
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:
- $\order {\set {x H: x \in G} } = \order {\set {H y: y \in G} }$
That is:
- The number of right cosets is the same as the number of left cosets of $G$ with respect to $H$.
- The left and right coset spaces are equivalent.
Proof 1
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.
$\blacksquare$
Proof 2
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 $a H = b H$.
\(\ds a H = b H\) | \(\iff\) | \(\ds a^{-1} b \in H\) | Left Cosets are Equal iff Product with Inverse in Subgroup‎ | |||||||||||
\(\ds H a^{-1} = H b^{-1}\) | \(\iff\) | \(\ds a^{-1} \paren {b^{-1} }^{-1} \in H\) | 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.
$\blacksquare$
Also see
- Index of Subgroup, which is the number of left (or right) cosets of a subgroup.
Sources
- 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $2$: Subgroups and Cosets: $\S 37 \beta$