User:Dfeuer/Coset stuff in progress

Theorem
Let $\left({G, \circ}\right)$ be a group and let $H$ be a subgroup of $G$.

Let $x, y \in G$.

Let:
 * $y \circ H$ denote the left coset of $H$ by $y$
 * $H \circ y$ denote the right coset of $H$ by $y$.

Then:

Theorem
Let $\left({G, \circ}\right)$ be a group and let $H$ be a subgroup of $G$.

Let $x, y \in G$.

Let:
 * $y \circ H$ denote the left coset of $H$ by $y$
 * $H \circ y$ denote the right coset of $H$ by $y$.

Then:

Theorem
Let $\left({G, \circ}\right)$ be a group and let $H$ be a subgroup of $G$.

Let $x, y \in G$.

Let:
 * $y \circ H$ denote the left coset of $H$ by $y$
 * $H \circ y$ denote the right coset of $H$ by $y$.

Then:

Proof
Let $\left({G,*}\right)$ be the opposite group of $\left({G,\circ}\right)$.

Then:
 * $x \in H \circ y \iff x \in y * H$
 * $x \circ y^{-1} \in H \iff y^{-1} * x \in H$

Since $H$ is closed under inverses:
 * $x \circ y^{-1} \in H \iff x^{-1} * y \in H$

By Element in Left Coset iff Product with Inverse in Subgroup:
 * $x \in y * H \iff x^{-1} * y \in H$

Thus $(2)$ holds.