Conjugate of Set with Inverse Closed for Inverses

Theorem

Let $G$ be a group.

Let $S \subseteq G$.

Let $\hat S = S \cup S^{-1}$, where $S^{-1}$ is the set of all the inverses of all the elements of $S$.

Let $\tilde S = \set {a s a^{-1}: s \in \hat S, a \in G}$.

That is, $\tilde S$ is the set containing all the conjugates of the elements of $S$ and all their inverses.

Then:

$\forall x \in \tilde S: x^{-1} \in \tilde S$

Proof

Let $x \in \tilde S$.

That is:

$\exists s \in \hat S: x = a s a^{-1}$

Then:

\(\ds x^{-1}\)

\(=\)

\(\ds \paren {a s a^{-1} }^{-1}\)

\(\ds \)

\(=\)

\(\ds a s^{-1} a^{-1}\)

Power of Conjugate equals Conjugate of Power

Since $s^{-1} \in \hat S$, it follows that $x^{-1} \in \tilde S$.

$\blacksquare$