# Equivalence of Definitions of Normal Subgroup

## 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 $G$.

## 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 $N \circ g = h \circ N$.

$g \in N \circ g = h \circ N$
$g^{-1} \circ h \in N$

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

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

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).

$\blacksquare$