Subgroup is Normal iff Contains Conjugate Elements

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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 $G$ if and only if:

\(\text {(1)}: \quad\) \(\ds \forall g \in G: \, \) \(\ds n \in N\) \(\iff\) \(\ds g \circ n \circ g^{-1} \in N\)
\(\text {(2)}: \quad\) \(\ds \forall g \in G: \, \) \(\ds n \in N\) \(\iff\) \(\ds g^{-1} \circ n \circ g \in N\)


Proof

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

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


Necessary Condition

Suppose that $g \circ N = N \circ g$, by definition 1 of normality in $G$.

Let $n \in N$.

Then:

\(\ds g \circ n\) \(\in\) \(\ds N \circ g\) Definition of Coset
\(\ds \leadstoandfrom \ \ \) \(\ds \exists n_1 \in N: \, \) \(\ds g \circ n\) \(=\) \(\ds n_1 \circ g\) Definition of Coset
\(\ds \leadstoandfrom \ \ \) \(\ds g \circ n \circ g^{-1}\) \(=\) \(\ds n_1\) Division Laws for Groups
\(\ds \leadstoandfrom \ \ \) \(\ds g \circ n \circ g^{-1}\) \(\in\) \(\ds N\) Definition of $n_1$

$\Box$


Sufficient Condition

Suppose that:

$\forall g \in G: \paren {n \in N \iff g \circ n \circ g^{-1} \in N}$

Let $g \circ n \circ g^{-1} \in N$.

\(\ds \exists n_1 \in N: \, \) \(\ds g \circ n \circ g^{-1}\) \(=\) \(\ds n_1\)
\(\ds \leadsto \ \ \) \(\ds g \circ n\) \(=\) \(\ds n_1 \circ g\) Division Laws for Groups
\(\ds \leadsto \ \ \) \(\ds g \circ n\) \(\in\) \(\ds N \circ g\) Definition of Coset
\(\ds \leadsto \ \ \) \(\ds g \circ N\) \(\subseteq\) \(\ds N \circ g\) Definition of Subset


Similarly:

\(\ds \exists n_2 \in N: \, \) \(\ds g \circ n \circ g^{-1}\) \(=\) \(\ds n_2\)
\(\ds \leadsto \ \ \) \(\ds n \circ g^{-1}\) \(=\) \(\ds g^{-1} \circ n_2\) Division Laws for Groups
\(\ds \leadsto \ \ \) \(\ds n \circ g^{-1}\) \(\in\) \(\ds g^{-1} \circ N\) Definition of Coset
\(\ds \leadsto \ \ \) \(\ds N \circ g^{-1}\) \(\subseteq\) \(\ds g^{-1} \circ N\) Definition of Subset

As $g$ is arbitrary, then so is $g^{-1}$.

Thus:

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

By definition of set equality:

$g \circ N = N \circ g$


Also see


Sources

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