Equivalence of Definitions of Normal Subset/3 iff 5

Theorem
Let $\left({G,\circ}\right)$ be a group.

Let $S \subseteq G$.

Then:
 * $S$ is a normal subset of $G$ by Definition 2


 * $S$ is a normal subset of $G$ by Definition 5.
 * $S$ is a normal subset of $G$ by Definition 5.

3 implies 5
Suppose that for each $g \in G$:
 * $g^{-1} \circ S \circ g \subseteq S$

Let $x, y \in G$ such that $x \circ y \in S$.

Then:
 * $x^{-1} \circ \left({ x \circ y }\right) \circ x \in S$

Since $\circ$ is associative and by the definition of inverse:
 * $y \circ x \in S$

5 implies 3
Suppose that $S$ is a normal subset of $G$ by Definition 5.

Then for each $x, y \in G$:
 * $x \circ y \in S \implies y \circ x \in S$

Let $g \in G$ and let $n \in S$.

Then by the definition of inverse and the definition of identity:
 * $\left({ g \circ g^{-1} }\right) \circ x \circ \left({ g \circ g^{-1} }\right) \in S$

By group axiom $G1$: associativity:
 * $g \circ \left({ g^{-1} \circ x \circ g \circ g^{-1} }\right) \in S$

By the premise:
 * $\left({ g^{-1} \circ x \circ g \circ g^{-1} }\right) \circ g \in S$

By group axiom $G1$: associativity, group axiom $G2$: identity and group axiom $G3$: inverses:
 * $g^{-1} \circ x \circ g \in S$

Since this holds for all $x$, $S$ is a normal subset by Definition 3.