Conjugate of Set by Group Product
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
- 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $2$: Conjugacy, Normal Subgroups, and Quotient Groups: $\S 45$