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

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \phi \left({y_1}\right)\)	\(=\)	\(\displaystyle \phi \left({y_2}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle x \circ y_1\)	\(=\)	\(\displaystyle x \circ y_2\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle y_1\)	\(=\)	\(\displaystyle y_2\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	(by 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 $Y \circ X$.

$\blacksquare$