# 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:

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

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

### Necessary Condition

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

Then:

 $\, \displaystyle \forall g \in G: \,$ $\displaystyle g \circ S$ $=$ $\displaystyle S \circ g$ $(1)$ $\displaystyle \implies \ \$ $\displaystyle \left({g \circ S}\right) \circ g^{-1}$ $=$ $\displaystyle \left({S \circ g}\right) \circ g^{-1}$ $\displaystyle \implies \ \$ $\displaystyle g \circ S \circ g^{-1}$ $=$ $\displaystyle S \circ \left({g \circ g^{-1} }\right)$ Subset Product within Semigroup is Associative: Corollary $\displaystyle \implies \ \$ $\displaystyle g \circ S \circ g^{-1}$ $=$ $\displaystyle S \circ e$ Definition of Inverse Element $\displaystyle \implies \ \$ $\displaystyle g \circ S \circ g^{-1}$ $=$ $\displaystyle S$ Subset Product by Identity Singleton $\, \displaystyle \forall g \in G: \,$ $\displaystyle g^{-1} \circ S \circ g$ $=$ $\displaystyle S$ $(4)$

$\Box$

### Sufficient Condition

Let $S$ be a subset of $G$ such that:

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

or:

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

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

Then:

 $\, \displaystyle \forall g \in G: \,$ $\displaystyle g \circ S \circ g^{-1}$ $=$ $\displaystyle S$ $\displaystyle \implies \ \$ $\displaystyle \left({g \circ S \circ g^{-1} }\right) \circ g$ $=$ $\displaystyle S \circ g$ $\displaystyle \implies \ \$ $\displaystyle \left({g \circ S}\right) \circ \left({g^{-1} \circ g}\right)$ $=$ $\displaystyle S \circ g$ Subset Product within Semigroup is Associative: Corollary $\displaystyle \implies \ \$ $\displaystyle \left({g \circ S}\right) \circ e$ $=$ $\displaystyle S \circ g$ Definition of Inverse Element $\displaystyle \implies \ \$ $\displaystyle g \circ S$ $=$ $\displaystyle S \circ g$ 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:

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

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:

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

$y \circ x \in S$

$\Box$

#### 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:

$\left({ g \circ g^{-1} }\right) \circ x \circ \left({ g \circ g^{-1} }\right) \in S$
$g \circ \left({ g^{-1} \circ x \circ g \circ g^{-1} }\right) \in S$

By the premise:

$\left({ g^{-1} \circ x \circ g \circ g^{-1} }\right) \circ g \in S$
$g^{-1} \circ x \circ g \in S$

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

$\blacksquare$