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:

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

\(=\)

\(\ds a \paren {a_1 s_1 a_1^{-1} } \paren {a_2 s_2 a_2^{-1} } \cdots \paren {a_n s_n a_n^{-1} } a^{-1}\)

\(\ds \)

\(=\)

\(\ds a \paren {a_1 s_1 a_1^{-1} } a^{-1} a \paren {a_2 s_2 a_2^{-1} } a^{-1} \cdots a \paren {a_n s_n a_n^{-1} } a^{-1}\)

\(\ds \)

\(=\)

\(\ds \paren {\paren {a a_1} s_1 \paren {a a_1}^{-1} } \paren {\paren {a a_2} s_2 \paren {a a_2}^{-1} } \cdots \paren {\paren {a a_n} s_n \paren {a a_n}^{-1} }\)

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

The result follows.

$\blacksquare$