Conjugate of Set with Inverse is Closed

Theorem
Let $G$ be a group.

Let $S \subseteq G$.

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

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

Let $\map W {\tilde S}$ be the set of words of $\tilde S$.

Then:
 * $\forall w \in \map W {\tilde S}: \forall a \in G: a w a^{-1} \in \map W {\tilde S}$

Proof
Let $w \in \map W {\tilde S}$.

From the definition of $\map W {\tilde S}$, we have:


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

Thus:

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

The result follows.