Left Cosets are Equal iff Element in Other Left Coset
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Theorem
Let $G$ be a group.
Let $H$ be a subgroup of $G$.
Let $x, y \in G$.
Let $x H$ denote the left coset of $H$ by $x$.
Then:
- $x H = y H \iff x \in y H$
Proof 1
| \(\ds x H\) | \(=\) | \(\ds y H\) | ||||||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds x^{-1} y\) | \(\in\) | \(\ds H\) | Left Cosets are Equal iff Product with Inverse in Subgroup | ||||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds x\) | \(\in\) | \(\ds y H\) | Element in Left Coset iff Product with Inverse in Subgroup |
$\blacksquare$
Proof 2
Let $x \in y H$.
Then $x$ is of the form $y h_1$ for some $h_1 \in H$.
Thus every element of the form $x h_2 \in x H$ is of the form $y h_1 h_2$ for some $h_2 \in H$.
But:
- $h_1 h_2 \in H$
and so:
- $x h_2 \in y H$
So by definition of subset:
- $x H \subseteq y H$
Let $x \in y H$ again.
Then $x$ is of the form $y h$ for some $h \in H$.
But then:
- $y = x h^{-1} \in x H$
Thus every element of the form $y h_2 \in y H$ is of the form $x h^{-1} h_2 \in x H$.
Thus by definition of subset:
- $y H \subseteq x H$
By definition of set equality:
- $x H = y H$
$\Box$
Let $x H = y H$.
Then $x h_1 = y h_2$ for some $h_1, h_2 \in H$.
Hence:
- $x = y h_2 h^{-1} \in y H$
The result follows.
$\blacksquare$