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:
\(\ds \forall g \in G: \, \) | \(\ds g \circ N\) | \(=\) | \(\ds N \circ g\) | 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:
Although this article appears correct, it's inelegant. There has to be a better way of doing it. In particular: Do not repeat this verification again. This part was done at the beginning of the proof. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by redesigning it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Improve}} from the code.If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page. |
\(\ds \forall g \in G: \, \) | \(\ds N\) | \(\subseteq\) | \(\ds g^{-1} \circ N \circ g\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds g \circ N\) | \(\subseteq\) | \(\ds g \circ \paren {g^{-1} \circ N \circ g}\) | 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
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {II}$: New Structures from Old: $\S 11$: Quotient Structures: Theorem $11.2: \ 4^\circ, 5^\circ$