Order of Subset Product with Singleton

Theorem

Let $\struct {G, \circ}$ be a group.

Let $X, Y \subseteq \struct {G, \circ}$ such that $X$ is a singleton:

$X = \set x$

Then:

$\order {X \circ Y} = \order Y = \order {Y \circ X}$

where $\order S$ is defined as the order of $S$.

From Regular Representations of Subset Product, we have that the left regular representation of $\struct {S, \circ}$ with respect to $a$ is:

$\lambda_x \sqbrk S = \set x \circ S = x \circ S$

$\blacksquare$

Let $\order Y = k$.

We define the mapping $\phi: Y \to X \circ Y$ such that:

$\forall y \in Y: \map \phi y = x \circ y$

First we show that $\phi$ is injective.

Let $y_1, y_2 \in Y$.

Let $\map \phi {y_1} = \map \phi {y_2}$.

$\ds \map \phi {y_1}$

$=$

$\ds \map \phi {y_2}$

$\ds \leadsto \ \ $

$\ds x \circ y_1$

$=$

$\ds x \circ y_2$

$\ds \leadsto \ \ $

$\ds y_1$

$=$

$\ds y_2$

left cancellation

Hence $\phi$ is injective.

The fact that $\phi$ is surjective follows from the definition of $X \circ Y$.

Every element of $X \circ Y$ is of the form $x \circ y$ for some $y \in Y$.

Thus $\phi$ is surjective.

The other half of the result follows identically, by defining a similar function for $Y \circ X$.

$\blacksquare$