Subgroup is Subset of Conjugate iff Normal

From ProofWiki
Jump to navigation Jump to search

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:

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

$\Box$


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:

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


Similarly:

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


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.

$\blacksquare$


Also see


Sources