Subgroup equals Conjugate iff Normal/Proof 1
Theorem
- $\forall g \in G: g \circ N \circ g^{-1} = N$
- $\forall g \in G: g^{-1} \circ N \circ g = N$
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:
- $(1): \quad \paren {\forall g \in G: g \circ N \circ g^{-1} = N} \iff \paren {\forall g \in G: g^{-1} \circ N \circ g = N}$
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 |
$\blacksquare$
Sources
- 1965: J.A. Green: Sets and Groups ... (previous) ... (next): $\S 6.6$. Normal subgroups
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): $\S 11$: Theorem $11.2: \ 2^\circ, 3^\circ$
- 1967: George McCarty: Topology: An Introduction with Application to Topological Groups ... (previous) ... (next): $\text{II}$: Morphisms
- 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $7$: Normal subgroups and quotient groups: Proposition $7.4 \ \text{(c)}$