Conjugate of Set by Group Product

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Theorem

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

Let $S \subseteq G$.


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

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


Then:

$\paren {S^a}^b = S^{b \circ a}$


Also defined as

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

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

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


Then:

$\paren {S^a}^b = S^{a \circ b}$


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:

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


Thus we can express $\paren {S^a}^b$ as $b \circ \paren {a \circ S \circ a^{-1} } \circ b^{-1}$, and understand that the right hand side refers to subset products.


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

\(\ds \paren {S^a}^b\) \(=\) \(\ds b \circ \paren {a \circ S \circ a^{-1} } \circ b^{-1}\) from above
\(\ds \) \(=\) \(\ds \paren {b \circ a} \circ S \circ \paren {a^{-1} \circ b^{-1} }\) Subset Product within Semigroup is Associative
\(\ds \) \(=\) \(\ds \paren {b \circ a} \circ S \circ \paren {b \circ a}^{-1}\) Inverse of Group Product
\(\ds \) \(=\) \(\ds S^{b \circ a}\) Definition of Conjugate of Group Subset

$\blacksquare$


Proof for Alternative Definition

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

\(\ds \paren {S^a}^b\) \(=\) \(\ds b^{-1} \circ \paren {a^{-1} \circ S \circ a} \circ b\) from above
\(\ds \) \(=\) \(\ds \paren {b^{-1} \circ a^{-1} } \circ S \circ \paren {a \circ b}\) Subset Product within Semigroup is Associative
\(\ds \) \(=\) \(\ds \paren {a \circ b}^{-1} \circ S \circ \paren {a \circ b}\) Inverse of Group Product
\(\ds \) \(=\) \(\ds S^{a \circ b}\) Definition of Conjugate of Group Subset

$\blacksquare$


Comment

This is not always correct in the literature.

For example, 1971: Allan Clark: Elements of Abstract Algebra 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