Order of Subset Product with Singleton

Theorem
Let $$\left({G, \circ}\right)$$ be a group.

Let $$X, Y \subseteq \left({G, \circ}\right)$$ such that $$X$$ is a singleton, i.e. $$X = \left\{{x}\right\}$$.

Then $$\left|{X \circ Y}\right| = \left|{Y}\right| = \left|{Y \circ X}\right|$$, where $$\left|{S}\right|$$ is defined as the order of $S$.

Proof
From the definition of regular representations, we have that the left regular representation of $$\left ({S, \circ}\right)$$ with respect to $$a$$ is $$\lambda_x \left({S}\right) = \left \{{x}\right\} \circ S = x \circ S$$.

The result then follows directly from Regular Representations of Invertible Elements are Permutations.

Alternative Proof
Let $$\left|{Y}\right| = k$$.

We define the mapping $$\phi: Y \to X \circ Y$$ such that $$\forall y \in Y: \phi \left({y}\right) = x \circ y$$.


 * First we show that $$\phi$$ is injective.

Let $$y_1, y_2 \in Y$$.

Let $$\phi \left({y_1}\right) = \phi \left({y_2}\right)$$.

$$ $$ $$

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 $$Y \circ X$$.