Equivalence of Definitions of Normal Subset/3 iff 4

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Theorem

Let $\struct {G,\circ}$ be a group.

Let $S \subseteq G$.


Then:

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

if and only if:

$S$ 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:

$(5): \quad \paren {\forall g \in G: g \circ S \circ g^{-1} \subseteq S} \iff \paren {\forall g \in G: g^{-1} \circ S \circ g \subseteq S}$
$(6): \quad \paren {\forall g \in G: S \subseteq g \circ S \circ g^{-1}} \iff \paren {\forall g \in G: S \subseteq g^{-1} \circ S \circ g}$

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:

\(\ds \forall g \in G: \, \) \(\ds g \circ S \circ g^{-1}\) \(\subseteq\) \(\ds S\)
\(\ds \leadsto \ \ \) \(\ds g^{-1} \circ \paren {g \circ S \circ g^{-1} }\) \(\subseteq\) \(\ds g^{-1} \circ S\) Subset Relation is Compatible with Subset Product/Corollary 2
\(\ds \leadsto \ \ \) \(\ds S \circ g^{-1}\) \(\subseteq\) \(\ds g^{-1} \circ S\) Subset Product within Semigroup is Associative/Corollary and the definition of inverse
\(\ds \leadsto \ \ \) \(\ds \paren {S \circ g^{-1} } \circ g\) \(\subseteq\) \(\ds \paren { g^{-1} \circ S} \circ g\) Subset Relation is Compatible with Subset Product/Corollary 2
\(\ds \leadsto \ \ \) \(\ds S\) \(\subseteq\) \(\ds g^{-1} \circ S \circ g\) 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 $(1)$.



$\blacksquare$