Conjugate of a Set by Product

Theorem

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

Let $S \subseteq G$.

Let $S^a$ denote the $G$-conjugate of $S$ by $a$ as:

$S^a := \left\{{y \in G: \exists x \in S: y = a \circ x \circ a^{-1}}\right\} = a \circ S \circ a^{-1}$

Then:

$\left({S^a}\right)^b = S^{b \circ a}$

Theorem for Alternative Definition

The concept of set conjugate can be defined in a different way:

Let $S^a$ denote the $G$-conjugate of $S$ by $a$ as:

$S^a := \left\{{y \in G: \exists x \in S: y = a^{-1} \circ x \circ a}\right\} = a^{-1} \circ S \circ a$

Then:

$\left({S^a}\right)^b = S^{a \circ b}$

Proof

$S^a$ is defined as $a \circ S \circ a^{-1}$ from the definition of the conjugate of a set.

From the definition of subset product with a singleton, this can be seen to be the same thing as:

$S^a = \left\{{a}\right\} \circ S \circ \left\{{a^{-1}}\right\}$.

Thus we can express $\left({S^a}\right)^b$ as $b \circ \left({a \circ S \circ a^{-1}}\right) \circ b^{-1}$, and understand that the RHS refers to subset products.

From Subset Product of Associative is Associative (which applies because $\circ$ is associative), it then follows directly that:

	\(\displaystyle \)	\(\displaystyle \left({S^a}\right)^b\)	\(=\)	\(\displaystyle b \circ \left({a \circ S \circ a^{-1} }\right) \circ b^{-1}\)	\(\displaystyle \)	from above
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left({b \circ a}\right) \circ S \circ \left({a^{-1} \circ b^{-1} }\right)\)	\(\displaystyle \)	Subset Product of Associative is Associative
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left({b \circ a}\right) \circ S \circ \left({b \circ a}\right)^{-1}\)	\(\displaystyle \)	Inverse of Group Product
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle S^{b \circ a}\)	\(\displaystyle \)	Definition of Conjugate of a Set

$\blacksquare$

Proof for Alternative Definition

Using the same preliminary argument as above, we then follow:

	\(\displaystyle \)	\(\displaystyle \left({S^a}\right)^b\)	\(=\)	\(\displaystyle b^{-1} \circ \left({a^{-1} \circ S \circ a}\right) \circ b\)	\(\displaystyle \)	from above
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left({b^{-1} \circ a^{-1} }\right) \circ S \circ \left({a \circ b}\right)\)	\(\displaystyle \)	Subset Product of Associative is Associative
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left({a \circ b}\right)^{-1} \circ S \circ \left({a \circ b}\right)\)	\(\displaystyle \)	Inverse of Group Product
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle S^{a \circ b}\)	\(\displaystyle \)	Definition of Conjugate of a Set

$\blacksquare$

Comment

This is not always correct in the literature.

For example, Allan Clark: Elements of Abstract Algebra (1971) defines set conjugate as:

$S^a := a \circ S \circ a^{-1}$

but then states (without proof) the assertion:

$\left({S^a}\right)^b = S^{a \circ b}$

Sources

• Allan Clark: Elements of Abstract Algebra (1971)... (previous)... (next): $\S 45$