# Equivalence of Definitions of Normal Subset/1 iff 2

## Theorem

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

Let $S$ be a subset of $G$.

Then Normal Subset/Definition 1 is equivalent to Normal Subset/Definition 2.

That is, the following three statements are equivalent:

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

## Proof

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

$\blacksquare$