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

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

Definition 4

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

Definition 5

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

Definition 6

$\forall g \in G: \paren {n \in N \iff g \circ n \circ g^{-1} \in N}$
$\forall g \in G: \paren {n \in N \iff g^{-1} \circ n \circ g \in N}$

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:

\(\ds \forall g \in G: \, \) \(\ds N \circ g\) \(=\) \(\ds g \circ N\) 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:

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

\(\ds \forall g \in G: \, \) \(\ds N\) \(\subseteq\) \(\ds g^{-1} \circ N \circ g\)
\(\ds \leadsto \ \ \) \(\ds g \circ N\) \(\subseteq\) \(\ds g \circ \paren {g^{-1} \circ N \circ g}\) 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
\(\ds \leadstoandfrom \ \ \) \(\ds g \circ n \circ g^{-1}\) \(=\) \(\ds n_1 \circ g \circ g^{-1}\)
\(\ds \) \(=\) \(\ds n_1 \circ e\) Definition of Inverse Element
\(\ds \) \(=\) \(\ds n_1\) Definition of Identity Element
\(\ds \leadstoandfrom \ \ \) \(\ds g \circ n \circ g^{-1}\) \(\in\) \(\ds N\) 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 $g \circ n \circ g^{-1} \in N$.


\(\ds \exists n_1 \in N: \, \) \(\ds g \circ n \circ g^{-1}\) \(=\) \(\ds n_1\)
\(\ds \leadsto \ \ \) \(\ds g \circ n\) \(=\) \(\ds n_1 \circ g\) Group Axioms
\(\ds \leadsto \ \ \) \(\ds g \circ n\) \(\in\) \(\ds N \circ g\) Definition of Coset
\(\ds \leadsto \ \ \) \(\ds g \circ N\) \(\subseteq\) \(\ds N \circ g\) Definition of Subset


Similarly:

$n \circ g \in N \circ g \implies N \circ G = g \circ N$
\(\ds \exists n_1 \in N: \, \) \(\ds g \circ n \circ g^{-1}\) \(=\) \(\ds n_2\)
\(\ds \leadsto \ \ \) \(\ds n \circ g^{-1}\) \(=\) \(\ds g^{-1} \circ n_2\) Group Axioms
\(\ds \leadsto \ \ \) \(\ds n \circ g^{-1}\) \(\in\) \(\ds g^{-1} \circ N\) 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).

$\blacksquare$