Subgroup is Superset of Conjugate iff Normal

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It has been suggested that this page or section be merged into Definition:Normal Subgroup/Definition 3.
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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 $g^{-1}$.
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\(\ds \forall g \in G: \, \) \(\ds N \circ g\) \(=\) \(\ds g \circ N\) 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

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