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.