Equivalence of Definitions of Normal Subset/3 iff 4

From ProofWiki
Jump to: navigation, search

Theorem

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

Let $S \subseteq G$.


Then:

$S$ is a normal subset of $G$ by Definition 3

iff:

$S$ is a normal subset of $G$ by Definition 4.


That is, the following conditions are equivalent:

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


Proof

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:

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

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 $(1)$.



$\blacksquare$