Equivalence of Definitions of Normal Subset/3 iff 4
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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:
\(\, \ds \forall g \in G: \, \) | \(\ds g \circ S \circ g^{-1}\) | \(\subseteq\) | \(\ds S\) | |||||||||||
\(\ds g^{-1} \circ \left({ g \circ S \circ g^{-1} }\right)\) | \(\subseteq\) | \(\ds g^{-1} \circ S\) | Subset Relation is Compatible with Subset Product/Corollary 2 | |||||||||||
\(\ds S \circ g^{-1}\) | \(\subseteq\) | \(\ds g^{-1} \circ S\) | Subset Product within Semigroup is Associative/Corollary and the definition of inverse | |||||||||||
\(\ds \left({ S \circ g^{-1} }\right) \circ g\) | \(\subseteq\) | \(\ds \left({ g^{-1} \circ S }\right) \circ g\) | Subset Relation is Compatible with Subset Product/Corollary 2 | |||||||||||
\(\ds S\) | \(\subseteq\) | \(\ds g^{-1} \circ S \circ g\) | 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 $(1)$.
$\blacksquare$