Equivalence of Definitions of Normal Subset/1 iff 2
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Theorem
Let $\struct {G, \circ}$ 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:
- $(4): \quad \paren {\forall g \in G: g \circ S \circ g^{-1} = S} \iff \paren {\forall g \in G: g^{-1} \circ S \circ g = S}$
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 |
$\blacksquare$