Equivalence of Definitions of Normal Subset/3 iff 4

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

Let $N \subseteq G$.

Then:
 * $N$ is a normal subset of $G$ by Definition 3

iff:
 * $N$ is a normal subset of $G$ by Definition 4.

That is, the following conditions are equivalent:
 * $(1)\quad \forall g \in G: g \circ N \circ g^{-1} \subseteq N$
 * $(2)\quad \forall g \in G: g^{-1} \circ N \circ g \subseteq N$
 * $(3)\quad \forall g \in G: N \subseteq g \circ N \circ g^{-1}$
 * $(4)\quad \forall g \in G: N \subseteq g^{-1} \circ N \circ g$

Proof
First note that:


 * $(5): \quad \left({\forall g \in G: g \circ N \circ g^{-1} \subseteq N}\right) \iff \left({\forall g \in G: g^{-1} \circ N \circ g \subseteq N}\right)$


 * $(6): \quad \left({\forall g \in G: N \subseteq g \circ N \circ g^{-1}}\right) \iff \left({\forall g \in G: N \subseteq g^{-1} \circ N \circ g}\right)$

which is shown by, for example, setting $h := g^{-1}$ and substituting.

Therefore conditions $(1)$ and $(2)$ are equivalent and conditions $(3)$ and $(4)$ are equivalent, so we need only show that condition $(1)$ is equivalent to condition $(3)$.

Suppose that $(1)$ holds. Then:

Thus condition $(1)$ implies condition $(3)$.

The exact same argument, substituting $\supseteq$ for $\subseteq$ and using Superset Relation is Compatible with Subset Product instead of Subset Relation is Compatible with Subset Product proves that $(3)$ implies $(1)$.