Conjugate of Set with Inverse is Closed

Theorem
Let $G$ be a group.

Let $S \subseteq G$.

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

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

Let $W \left({\tilde S}\right)$ be the set of words of $\tilde S$.

Then $\forall w \in W \left({\tilde S}\right): \forall a \in G: a w a^{-1} \in W \left({\tilde S}\right)$.

Proof
Let $w \in W \left({\tilde S}\right)$.

From the definition of $W \left({\tilde S}\right)$, we have:

$w = \left({a_1 s_1 a_1^{-1}}\right) \left({a_2 s_2 a_2^{-1}}\right) \cdots \left({a_n s_n a_n^{-1}}\right), n \in \N^*, a_i \in G, s_1 \in \hat S, 1 \le i \le n$.

Thus:

As $G$ is a group, all of the $a a_i \in G$.

The result follows.