Equivalence of Definitions of Normal Subset/1 iff 2

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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

$\blacksquare$