Subgroup equals 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:

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


Proof 1

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:

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

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


Necessary Condition

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

Then:

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


Similarly:

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

$\Box$


Sufficient Condition

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

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

and so from $(1)$ above:

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


Then:

\(\ds \forall g \in G: \, \) \(\ds g \circ N \circ g^{-1}\) \(=\) \(\ds N\)
\(\ds \leadsto \ \ \) \(\ds \paren {g \circ N \circ g^{-1} } \circ g\) \(=\) \(\ds N \circ g\) Definition of Subset Product
\(\ds \leadsto \ \ \) \(\ds \paren {g \circ N} \circ \paren {g^{-1} \circ g}\) \(=\) \(\ds N \circ g\) Subset Product within Semigroup is Associative: Corollary
\(\ds \leadsto \ \ \) \(\ds \paren {g \circ N} \circ e\) \(=\) \(\ds N \circ g\) Definition of Inverse Element
\(\ds \leadsto \ \ \) \(\ds g \circ N\) \(=\) \(\ds N \circ g\) Coset by Identity

$\blacksquare$


Proof 2

From Subgroup is Superset of Conjugate iff Normal, $N$ is normal in $G$ if and only if:

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

From Subgroup is Subset of Conjugate iff Normal, $N$ is normal in $G$ if and only if:

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

The result follows by definition of set equality.

$\blacksquare$


Also see


Sources

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