Conjugate of Set with Inverse Closed for Inverses

Theorem
Let $$G$$ be a group.

Let $$S \subseteq G$$.

Let $$\hat S = S \cup S$$.

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

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

$$ $$

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