Subgroup is Subset of Conjugate iff Normal

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Theorem

Let $\struct {G, \circ}$ be a group whose identity is $e$.

Let $N$ be a subgroup of $G$.


Then $N$ is normal in $G$ (by definition 1) if and only if:

$\forall g \in G: N \subseteq g \circ N \circ g^{-1}$
$\forall g \in G: N \subseteq g^{-1} \circ N \circ g$


Proof

By definition, a subgroup is normal in $G$ if and only if:

$\forall g \in G: g \circ N = N \circ g$


First note that:

$(1): \quad \paren {\forall g \in G: N \subseteq g \circ N \circ g^{-1} } \iff \paren {\forall g \in G: N \subseteq g^{-1} \circ N \circ g}$

which is shown by, for example, setting $h := g^{-1}$ and substituting.


Necessary Condition

Suppose that $N$ is normal in $G$.

Then:

\(\, \displaystyle \forall g \in G: \, \) \(\displaystyle g \circ N\) \(=\) \(\displaystyle N \circ g\) $\quad$ Definition of Normal Subgroup $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle g^{-1} \circ \paren {g \circ N}\) \(=\) \(\displaystyle g^{-1} \circ \paren {N \circ g}\) $\quad$ Definition of Subset Product $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle \paren {g^{-1} \circ g} \circ N\) \(=\) \(\displaystyle g^{-1} \circ N \circ g\) $\quad$ Subset Product within Semigroup is Associative: Corollary $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle e \circ N\) \(=\) \(\displaystyle g^{-1} \circ N \circ g\) $\quad$ Definition of Inverse Element $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle N\) \(=\) \(\displaystyle g^{-1} \circ N \circ g\) $\quad$ Coset by Identity $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle N\) \(\subseteq\) \(\displaystyle g^{-1} \circ N \circ g\) $\quad$ Definition 2 of Set Equality $\quad$

$\Box$


Then by $(1)$ above:

$\paren {\forall g \in G: N \subseteq g \circ N \circ g^{-1} } \implies \paren {\forall g \in G: N \subseteq g^{-1} \circ N \circ g}$

$\Box$


Sufficient Condition

Let $N$ be a subgroup of $G$ such that:

$\forall g \in G: N \subseteq g \circ N \circ g^{-1}$

and so from $(1)$ above:

$\forall g \in G: N \subseteq g^{-1} \circ N \circ g$


Then:

\(\, \displaystyle \forall g \in G: \, \) \(\displaystyle N\) \(\subseteq\) \(\displaystyle g \circ N \circ g^{-1}\) $\quad$ $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle N \circ g\) \(\subseteq\) \(\displaystyle \paren {g \circ N \circ g^{-1} } \circ g\) $\quad$ Subset Relation is Compatible with Subset Product: Corollary 2 $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle N \circ g\) \(\subseteq\) \(\displaystyle \paren {g \circ N} \circ \paren {g^{-1} \circ g}\) $\quad$ Subset Product within Semigroup is Associative: Corollary $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle N \circ g\) \(\subseteq\) \(\displaystyle \paren {g \circ N} \circ e\) $\quad$ Definition of Inverse Element $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle N \circ g\) \(\subseteq\) \(\displaystyle g \circ N\) $\quad$ Coset by Identity $\quad$


Similarly:

\(\, \displaystyle \forall g \in G: \, \) \(\displaystyle N\) \(\subseteq\) \(\displaystyle g^{-1} \circ N \circ g\) $\quad$ $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle g \circ N\) \(\subseteq\) \(\displaystyle g \circ \paren {g^{-1} \circ N \circ g}\) $\quad$ Subset Relation is Compatible with Subset Product: Corollary 2 $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle g \circ N\) \(\subseteq\) \(\displaystyle \paren {g \circ g^{-1} } \circ \paren {N \circ g}\) $\quad$ Subset Product within Semigroup is Associative: Corollary $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle g \circ N\) \(\subseteq\) \(\displaystyle e \circ \paren {N \circ g}\) $\quad$ Definition of Inverse Element $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle g \circ N\) \(\subseteq\) \(\displaystyle N \circ g\) $\quad$ Coset by Identity $\quad$


Thus we have:

$N \circ g \subseteq g \circ N$
$g \circ N \subseteq N \circ g$

By definition of set equality:

$g \circ N = N \circ g$

Hence the result.

$\blacksquare$


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