Equivalence of Definitions of Normal Subset

Theorem

The following definitions of the concept of Normal Subset are equivalent:

Definition 1

$\forall g \in G: g \circ S = S \circ g$

Definition 2

$\forall g \in G: g \circ S \circ g^{-1} = S$

or, equivalently:

$\forall g \in G: g^{-1} \circ S \circ g = S$

Definition 3

$\forall g \in G: g \circ S \circ g^{-1} \subseteq S$

or, equivalently:

$\forall g \in G: g^{-1} \circ S \circ g \subseteq S$

Definition 4

$\forall g \in G: S \subseteq g \circ S \circ g^{-1}$

or, equivalently:

$\forall g \in G: S \subseteq g^{-1} \circ S \circ g$

Definition 5

$\forall x, y \in G: x \circ y \in S \implies y \circ x \in S$

Definition 6

$\map {N_G} S = G$

where $\map {N_G} S$ denotes the normalizer of $S$ in $G$.

Definition 7

$\forall g \in G: g \circ S \subseteq S \circ g$

or:

$\forall g \in G: S \circ g \subseteq g \circ S$

Proof

Definition 1 is Equivalent to Definition 2

Let $e$ be the identity of $G$.

First note that:

$(4): \quad \paren {\forall g \in G: g \circ S \circ g^{-1} = S} \iff \paren {\forall g \in G: g^{-1} \circ S \circ g = S}$

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

Necessary Condition

Suppose that $S$ satisfies $(1)$.

Let $g \in G$.

Then:

 $\ds g \circ S$ $=$ $\ds S \circ g$ $(1)$ $\ds \leadsto \ \$ $\ds \paren {g \circ S} \circ g^{-1}$ $=$ $\ds \paren {S \circ g} \circ g^{-1}$ $\ds \leadsto \ \$ $\ds g \circ S \circ g^{-1}$ $=$ $\ds S \circ \paren {g \circ g^{-1} }$ Subset Product within Semigroup is Associative: Corollary $\ds \leadsto \ \$ $\ds g \circ S \circ g^{-1}$ $=$ $\ds S \circ e$ Definition of Inverse Element $\ds \leadsto \ \$ $\ds g \circ S \circ g^{-1}$ $=$ $\ds S$ Subset Product by Identity Singleton $\ds \leadsto \ \$ $\ds g^{-1} \circ S \circ g$ $=$ $\ds S$ $(4)$

$\Box$

Sufficient Condition

Suppose that $S$ satisfies $(2)$ or $(3)$.

By $(4)$, as long as one of these statements holds, the other one holds as well.

Let $g \in G$.

Then:

 $\ds g \circ S \circ g^{-1}$ $=$ $\ds S$ $\ds \leadsto \ \$ $\ds \paren {g \circ S \circ g^{-1} } \circ g$ $=$ $\ds S \circ g$ $\ds \leadsto \ \$ $\ds \paren {g \circ S} \circ \paren {g^{-1} \circ g}$ $=$ $\ds S \circ g$ Subset Product within Semigroup is Associative: Corollary $\ds \leadsto \ \$ $\ds \paren {g \circ S} \circ e$ $=$ $\ds S \circ g$ Definition of Inverse Element $\ds \leadsto \ \$ $\ds g \circ S$ $=$ $\ds S \circ g$ Subset Product by Identity Singleton

$\blacksquare$

Definition 2 Implies Definition 3

We have that:

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

The result follows by definition of set equality.

$\blacksquare$

Definition 3 is Equivalent to Definition 4

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 $(1)$.

$\blacksquare$

Definitions 3 and 4 imply Definition 2

By Equivalence of Definitions of Normal Subset: 3 iff 4, $S$ being a normal subset of $G$ by Definition 3 and Definition 4 implies that the following hold:

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

By $(1)$ and $(3)$ and definition of set equality:

$\forall g \in G: g \circ S \circ g^{-1} = S$

By $(2)$ and $(4)$ and definition of set equality:

$\forall g \in G: g^{-1} \circ S \circ g = S$

$\blacksquare$

Definition 3 is Equivalent to Definition 5

3 implies 5

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

That is:

$\forall g \in G: g^{-1} \circ S \circ g \subseteq S$.

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

Then:

 $\ds y \circ x$ $=$ $\ds e \circ \paren {y \circ x}$ Group Axiom $\text G 2$: Existence of Identity Element $\ds$ $=$ $\ds \paren {x^{-1} \circ x} \circ \paren {y \circ x}$ Group Axiom $\text G 3$: Existence of Inverse Element $\ds$ $=$ $\ds x^{-1} \circ \paren {x \circ y} \circ x$ Group Axiom $\text G 1$: Associativity $\ds \leadsto \ \$ $\ds y \circ x$ $\in$ $\ds x^{-1} \circ S \circ x$ $x \circ y \in S$ $\ds \leadsto \ \$ $\ds y \circ x$ $\in$ $\ds S$ by hypothesis: Definition 3 of Normal Subset

$\Box$

5 implies 3

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

That is:

$\forall x, y \in G: x \circ y \in S \implies y \circ x \in S$

Let $g \in G$.

Then:

 $\ds \forall x \in S: \,$ $\ds e \circ x \circ e$ $\in$ $\ds S$ Group Axiom $\text G 2$: Existence of Identity Element $\ds \leadsto \ \$ $\ds \forall x \in S: \,$ $\ds \paren {g \circ g^{-1} } \circ x \circ \paren {g \circ g^{-1} }$ $\in$ $\ds S$ Group Axiom $\text G 3$: Existence of Inverse Element $\ds \leadsto \ \$ $\ds \forall x \in S: \,$ $\ds g \circ \paren {g^{-1} \circ x \circ g \circ g^{-1} }$ $\in$ $\ds S$ Group Axiom $\text G 1$: Associativity $\ds \leadsto \ \$ $\ds \forall x \in S: \,$ $\ds \paren {g^{-1} \circ x \circ g \circ g^{-1} } \circ g$ $\in$ $\ds S$ by hypothesis: Definition 5 of Normal Subset $\ds \leadsto \ \$ $\ds \forall x \in S: \,$ $\ds \paren {g^{-1} \circ x \circ g} \circ \paren {g^{-1} \circ g}$ $\in$ $\ds S$ Group Axiom $\text G 1$: Associativity $\ds \leadsto \ \$ $\ds \forall x \in S: \,$ $\ds \paren {g^{-1} \circ x \circ g} \circ e$ $\in$ $\ds S$ Definition of Inverse Element $\ds \leadsto \ \$ $\ds \forall x \in S: \,$ $\ds g^{-1} \circ x \circ g$ $\in$ $\ds S$ Definition of Identity Element $\ds \leadsto \ \$ $\ds g^{-1} \circ S \circ g$ $\subseteq$ $\ds S$ Definition of Subset Product

$\blacksquare$