# Equivalence of Definitions of Normal Subset/3 iff 4

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## Theorem

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

Let $S \subseteq G$.

Then:

- $S$ is a normal subset of $G$ by Definition 3

iff:

- $S$ is a normal subset of $G$ by Definition 4.

That is, the following conditions are equivalent:

- $(1)\quad \forall g \in G: g \circ S \circ g^{-1} \subseteq S$
- $(2)\quad \forall g \in G: g^{-1} \circ S \circ g \subseteq S$
- $(3)\quad \forall g \in G: S \subseteq g \circ S \circ g^{-1}$
- $(4)\quad \forall g \in G: S \subseteq g^{-1} \circ S \circ g$

## Proof

First note that:

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

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

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.

It remains to be shown that condition $(1)$ is equivalent to condition $(3)$.

Suppose that $(1)$ holds.

Then:

\(\, \ds \forall g \in G: \, \) | \(\ds g \circ S \circ g^{-1}\) | \(\subseteq\) | \(\ds S\) | |||||||||||

\(\ds g^{-1} \circ \left({ g \circ S \circ g^{-1} }\right)\) | \(\subseteq\) | \(\ds g^{-1} \circ S\) | Subset Relation is Compatible with Subset Product/Corollary 2 | |||||||||||

\(\ds S \circ g^{-1}\) | \(\subseteq\) | \(\ds g^{-1} \circ S\) | Subset Product within Semigroup is Associative/Corollary and the definition of inverse | |||||||||||

\(\ds \left({ S \circ g^{-1} }\right) \circ g\) | \(\subseteq\) | \(\ds \left({ g^{-1} \circ S }\right) \circ g\) | Subset Relation is Compatible with Subset Product/Corollary 2 | |||||||||||

\(\ds S\) | \(\subseteq\) | \(\ds g^{-1} \circ S \circ g\) | Subset Product within Semigroup is Associative/Corollary and the definition of inverse |

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 $(1)$.

$\blacksquare$