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 $(1)$.
the necessary corollary is currently only available in subset form, but these arguments are all the same. Alternatively, the exact same method can be used to move the gs from the right to the left as was used to move them from the left to the right.
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