Equivalence of Definitions of Normal Subgroup
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Theorem
Let $G$ be a group.
Let $N$ be a subgroup of $G$.
The following definitions of the concept of Normal Subgroup are equivalent:
Definition 1
- $\forall g \in G: g \circ N = N \circ g$
Definition 2
- Every right coset of $N$ in $G$ is a left coset
that is:
- The right coset space of $N$ in $G$ equals its left coset space.
Definition 3
| \(\ds \forall g \in G: \, \) | \(\ds g \circ N \circ g^{-1}\) | \(\subseteq\) | \(\ds N\) | |||||||||||
| \(\ds \forall g \in G: \, \) | \(\ds g^{-1} \circ N \circ g\) | \(\subseteq\) | \(\ds N\) |
Definition 4
| \(\ds \forall g \in G: \, \) | \(\ds N\) | \(\subseteq\) | \(\ds g \circ N \circ g^{-1}\) | |||||||||||
| \(\ds \forall g \in G: \, \) | \(\ds N\) | \(\subseteq\) | \(\ds g^{-1} \circ N \circ g\) |
Definition 5
| \(\ds \forall g \in G: \, \) | \(\ds N\) | \(=\) | \(\ds g \circ N \circ g^{-1}\) | |||||||||||
| \(\ds \forall g \in G: \, \) | \(\ds N\) | \(=\) | \(\ds g^{-1} \circ N \circ g\) |
Definition 6
| \(\ds \forall g \in G: \, \) | \(\ds \leftparen {n \in N}\) | \(\iff\) | \(\ds \rightparen {g \circ n \circ g^{-1} \in N}\) | |||||||||||
| \(\ds \forall g \in G: \, \) | \(\ds \leftparen {n \in N}\) | \(\iff\) | \(\ds \rightparen {g^{-1} \circ n \circ g \in N}\) |
Definition 7
- $N$ is a normal subset of $G$.
Proof
Definition 1 iff Definition 2: Subgroup is Normal iff Left Cosets are Right Cosets
Necessary Condition
Let $N$ be a normal subgroup of $G$ by Definition 1.
Then the equality of the coset spaces follows directly from definition of normal subgroup and coset.
$\Box$
Sufficient Condition
Suppose that every right coset of $N$ in $G$ is a left coset of $N$ in $G$.
Let $g \in G$.
Since every right coset of $N$ in $G$ is a left coset, there exists an $h \in G$ such that $N \circ g = h \circ N$.
By Element of Group is in its own Coset:
- $g \in N \circ g = h \circ N$
From Element in Left Coset iff Product with Inverse in Subgroup:
- $g^{-1} \circ h \in N$
Then:
| \(\ds N \circ g\) | \(=\) | \(\ds \paren {g \circ g^{-1} } \circ \paren {h \circ N}\) | Definition of Inverse Element and Coset by Identity | |||||||||||
| \(\ds \) | \(=\) | \(\ds g \circ \paren {\paren { g^{-1} \circ h } \circ N}\) | Subset Product within Semigroup is Associative: Corollary | |||||||||||
| \(\ds \) | \(=\) | \(\ds g \circ N\) | $g^{-1} \circ h \in N$ and Left Coset Equals Subgroup iff Element in Subgroup |
Since this holds for all $g \in G$, $N$ is normal in $G$ (by definition 1).
$\blacksquare$
Definition 1 iff Definition 3: Subgroup is Superset of Conjugate iff Normal
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.
$\blacksquare$
Definition 1 iff Definition 4: Subgroup is Subset of Conjugate iff Normal
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.
$\blacksquare$
Definitions 3 and 4 iff Definition 5: Subgroup equals Conjugate iff Normal
From Subgroup is Superset of Conjugate iff Normal, $N$ is normal in $G$ if and only if:
- $\forall g \in G: N \supseteq g \circ N \circ g^{-1}$
- $\forall g \in G: N \supseteq g^{-1} \circ N \circ g$
From Subgroup is Subset of Conjugate iff Normal, $N$ is normal in $G$ if and only if:
- $\forall g \in G: N \subseteq g \circ N \circ g^{-1}$
- $\forall g \in G: N \subseteq g^{-1} \circ N \circ g$
The result follows by definition of set equality.
$\blacksquare$
Definition 1 iff Definition 6: Subgroup is Normal iff Contains Conjugate Elements
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:
- $g \circ N = N \circ g$
Definition 1 iff Definition 7: Subgroup is Normal iff Normal Subset
Necessary Condition
Let $N$ be normal in $G$ (by definition 1):
Thus for each $g \in G$:
- $\forall g \in G: g \circ N = N \circ g$
where $g \circ N$ denotes the subset product of $g$ with $N$.
Thus $N$ is a normal subset of $G$ (by definition 1):
- $\forall g \in G: g \circ N = N \circ g$
Sufficient Condition
Let $N$ be a normal subset of $G$ (by definition 1):
- $\forall g \in G: g \circ N = N \circ g$
Since $N$ is a subgroup, $N$ is a normal subgroup of $G$ (by definition 1).
$\blacksquare$