# Subgroup is Superset of Conjugate iff Normal

Jump to navigation
Jump to search

It has been suggested that this page or section be merged into Normal Subgroup Test.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 `{{Mergeto}}` from the code. |

It has been suggested that this page or section be merged into Definition:Normal Subgroup/Definition 3.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 `{{Mergeto}}` from the code. |

## Theorem

Let $\struct {G, \circ}$ be a group.

Let $N$ be a subgroup of $G$.

Then $N$ is normal in $G$ (by definition 1) if and only if:

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

## 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:

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

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 | ||||||||||

\(\ds \leadsto \ \ \) | \(\ds g \circ N\) | \(\subseteq\) | \(\ds N \circ g\) | Definition of Set Equality | ||||||||||

\(\ds \leadsto \ \ \) | \(\ds \paren {g \circ N} \circ g^{-1}\) | \(\subseteq\) | \(\ds \paren {N \circ g} \circ g^{-1}\) | Definition of Subset Product | ||||||||||

\(\ds \leadsto \ \ \) | \(\ds g \circ N \circ g^{-1}\) | \(\subseteq\) | \(\ds N \circ \paren {g \circ g^{-1} }\) | Subset Product within Semigroup is Associative: Corollary | ||||||||||

\(\ds \leadsto \ \ \) | \(\ds g \circ N \circ g^{-1}\) | \(\subseteq\) | \(\ds N \circ e\) | Definition of Inverse Element | ||||||||||

\(\ds \leadsto \ \ \) | \(\ds g \circ N \circ g^{-1}\) | \(\subseteq\) | \(\ds N\) | 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. If it is true for all $g$, of course true for all $g^{-1}$.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. |

\(\ds \forall g \in G: \, \) | \(\ds N \circ g\) | \(=\) | \(\ds g \circ N\) | Definition of Normal Subgroup | ||||||||||

\(\ds \leadsto \ \ \) | \(\ds N \circ g\) | \(\subseteq\) | \(\ds g \circ N\) | Definition of Set Equality | ||||||||||

\(\ds \leadsto \ \ \) | \(\ds g^{-1} \circ \paren {N \circ g}\) | \(\subseteq\) | \(\ds g^{-1} \circ \paren {g \circ N}\) | Definition of Subset Product | ||||||||||

\(\ds \leadsto \ \ \) | \(\ds g^{-1} \circ N \circ g\) | \(\subseteq\) | \(\ds \paren {g^{-1} \circ g} \circ N\) | Subset Product within Semigroup is Associative: Corollary | ||||||||||

\(\ds \leadsto \ \ \) | \(\ds g^{-1} \circ N \circ g\) | \(\subseteq\) | \(\ds e \circ N\) | Definition of Inverse Element | ||||||||||

\(\ds \leadsto \ \ \) | \(\ds g^{-1} \circ N \circ g\) | \(\subseteq\) | \(\ds N\) | Coset by Identity |

$\Box$

### Sufficient Condition

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

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

and so from $(1)$ above:

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

Then:

\(\ds \forall g \in G: \, \) | \(\ds g \circ N \circ g^{-1}\) | \(\subseteq\) | \(\ds N\) | |||||||||||

\(\ds \leadsto \ \ \) | \(\ds \paren {g \circ N \circ g^{-1} } \circ g\) | \(\subseteq\) | \(\ds N \circ g\) | Definition of Subset Product | ||||||||||

\(\ds \leadsto \ \ \) | \(\ds \paren {g \circ N} \circ \paren {g^{-1} \circ g}\) | \(\subseteq\) | \(\ds N \circ g\) | Subset Product within Semigroup is Associative: Corollary | ||||||||||

\(\ds \leadsto \ \ \) | \(\ds \paren {g \circ N} \circ e\) | \(\subseteq\) | \(\ds N \circ g\) | Definition of Inverse Element | ||||||||||

\(\ds \leadsto \ \ \) | \(\ds g \circ N\) | \(\subseteq\) | \(\ds N \circ g\) | 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. |

\(\ds \forall g \in G: \, \) | \(\ds g^{-1} \circ N \circ g\) | \(\subseteq\) | \(\ds N\) | |||||||||||

\(\ds \leadsto \ \ \) | \(\ds g \circ \paren {g^{-1} \circ N \circ g}\) | \(\subseteq\) | \(\ds g \circ N\) | Definition of Subset Product | ||||||||||

\(\ds \leadsto \ \ \) | \(\ds \paren {g \circ g^{-1} } \circ \paren {N \circ g}\) | \(\subseteq\) | \(\ds g \circ N\) | Subset Product within Semigroup is Associative: Corollary | ||||||||||

\(\ds \leadsto \ \ \) | \(\ds e \circ \paren {N \circ g}\) | \(\subseteq\) | \(\ds g \circ N\) | Definition of Inverse Element | ||||||||||

\(\ds \leadsto \ \ \) | \(\ds N \circ g\) | \(\subseteq\) | \(\ds g \circ N\) | 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 known as

Because of its convenience as a technique for determining whether a subgroup is normal or not, this result is often seen referred to as the Normal Subgroup Test.

## Also see

## Sources

- 1965: J.A. Green:
*Sets and Groups*... (previous) ... (next): $\S 6.6$. Normal subgroups - 1965: Seth Warner:
*Modern Algebra*... (previous) ... (next): Chapter $\text {II}$: New Structures from Old: $\S 11$: Quotient Structures: Theorem $11.2: \ 2^\circ, 3^\circ$ - 1996: John F. Humphreys:
*A Course in Group Theory*... (previous) ... (next): Chapter $7$: Normal subgroups and quotient groups: Proposition $7.4 \ \text{(b)}$