Generator of Normal Subgroup

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 $$\tilde S$$.

Let $$N$$ be the smallest normal subgroup of $$G$$ that contains $$S$$.

Then $$N = \left \langle {S} \right \rangle = W \left({\tilde S}\right)$$.

Proof
Let $$N$$ be the smallest normal subgroup of $$G$$ that contains $$S$$, where $$S \subseteq G$$.


 * $$N$$ must certainly include $$\hat S$$, because any group containing $$s \in S$$ must also contain $$s^{-1}$$.

Therefore, $$N$$ must be the smallest normal subgroup containing $$\hat S$$.

Since $$N \triangleleft G$$, it follows that $$\forall a \in G: \forall s \in \hat S: a s a^{-1} \in N$$.

Thus, $$\forall x \in \tilde S: x \in N$$.

Thus $$\tilde S \subseteq N$$.


 * By the closure axiom, $$N$$ must also contain all products of any finite number of elements of $$\tilde S$$. Thus $$W \left({\tilde S}\right) \subseteq N$$.


 * Now we prove that $$W \left({\tilde S}\right) \triangleleft G$$.

$$\tilde S \ne \varnothing$$, as $$e \left({s s^{-1}}\right) e^{-1} = e \in \tilde S$$.

By Conjugate of Set with Inverse Closed for Inverses, $$\tilde S$$ is closed under taking inverses.

So from Set of Words Generates Group: Corollary, $$W \left({\tilde S}\right) \le G$$.

From Conjugate of Set with Inverse is Closed, $$\forall w \in W \left({\tilde S}\right): \forall a \in G: a w a^{-1} \in W \left({\tilde S}\right)$$.


 * From Normal Subgroup Equivalent Definitions: 3, it follows that $$W \left({\tilde S}\right) \triangleleft G$$.

The result follows by the definition of the minimality of $$N$$.