# 4.2: Normal Groups and Factor Groups

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##### Definition: Term

Let $$G$$ be a group, and let $$H \leq G$$. Then $$H$$ is called a normal subgroup of $$G$$ if $$g H g^{-1} =H (\,or\, gH=Hg), \forall g \in G ,$$ denoted  as $$H \unlhd G.$$

Suppose $$H \le G$$.

\)H \) is a normal subgroup of $$G$$, if $$gH=Hg, \;\forall g \in G$$.  Means left=right cosets.

Note if $$G$$ is abelian, every subgroup is normal to $$G$$.  For any group $$G$$, the trivial subgroups $$\{e\}$$ and $$G$$ are normal subgroups.

##### Example $$\PageIndex{1}$$

Consider $$S_3=\{e,(123),(321),(12),(13),(23)\}$$.

1. Is $$H=\{e, (12)\}\unlhd S_3$$?

Since $$(123)H=\{(123),(13)\} \ne H(123)=\{(123),(23\}$$, thus $$H \not \unlhd S_3$$.

1. Is $$A_3=\{\langle (123) \rangle\} \unlhd S_3$$?

$$A_3=\{e,(123),(321)\}$$.

There will be 2 cosets of $$A_3$$ in $$S_3$$ since $$\frac{|S_3|}{|A_3|}=\frac{6}{3}=2$$.

The first will be the set itself since $$eH=He=H$$.  To find the second, take any element

in $$S_3$$, not in $$S_4$$ and determine its coset.  Having done so, the cosets are: $$A_3, \{(12),(13),(23)\}$$.

##### Theorem $$\PageIndex{1}$$

Let $$G$$ be a group, and $$H$$ be a subgroup of $$G$$.  If $$[G:H]=2$$, then show that $$H$$ is a normal subgroup of $$G$$.

Proof:

Let $$G$$ be a group and $$H \le G$$ s.t. $$[G:H] =2$$.

\)H\) partitions $$G$$ into 2 cosets.  The left cosets being $$\{ H, xH\}$$.  Similarly, the right cosets are $$\{H, Hx\}$$ where $$x \in G$$ and $$x \not \in H$$.

Case 1:

If $$x \in H$$ then $$xH=H=Hx$$.

Since $$xH=Hx$$, $$H \unlhd G$$.

Case 2:

If $$x \in G$$, but not in $$H$$.  We will show this as $$x \in G-H$$.

Then $$xH = G-H =Hx$$.

Since $$xH=Hx$$, $$H \unlhd G$$.

Thus $$xH=Hx, \; \forall x \in G$$, thus $$H \unlhd G$$.◻

##### Example $$\PageIndex{1}$$

Show that $$A_n$$ is a normal subgroup of $$S_n, \; \forall \; n \in \mathbb{N}$$.

###### Solution

Consider $$\frac{|S_n|}{|A_n|}=\frac{n!}{\frac{n!}{2}}=2$$.

From previous theorem, given $$H \le G$$ then if $$[G:H]=2$$ then $$H \unlhd G$$.

Since $$A_n \le S_n$$ and $$[S_n:A_n]=2$$, $$H \unlhd G$$.◻

##### Example $$\PageIndex{1}\ Let \(G$$ be a group and let $$G^{'} = \langle aba^{-1}b^{-1} \rangle$$, that is, $$G^{'}$$ is the subgroup of all infinite products of elements in $$G$$ of the form $$aba^{-1}b^{-1}$$.  The subgroup $$G^{'}$$ is called the commutator subgroup of $$G$$.

1. Show that $$G^{'}$$ is a normal subgroup of $$G$$.

Proof:

Let $$g \in G$$ and $$h \in G^{'}$$.

We will show that $$G^{'} \unlhd G$$.

Consider $$h=aba^{-1}b^{-1}, a,b \in G$$.

Then $$ghg^{-1}=gaba^{-1}b^{-1}g^{-1}$$

$$=geaebea^{-1}eb^{-1}g^{-1}$$

$$=(gag^{-1})(gbg^{-1})(ga^{-1}g^{-1})(gb^{-1}g^{-1})$$ Note: $$g^{-1}g=e$$

$$=(gag^{-1})(gbg^{-1})(gag^{-1})^{-1}(gbg^{-1})^{-1} \in G'$$.

Thus $$G^{'} \unlhd G$$.◻

2. Let $$N$$ be a normal subgroup of $$G$$.  Prove that $$G/N$$ is abelian iff $$N$$ contains the commutator subgroup of $$G$$.

Proof:

Let $$N \unlhd G$$.

We shall show that $$G/N$$ is abelian if and only if $$N$$ contains the commutator subgroup of $$G$$.

Let $$G/N$$ be abelian.

Let $$a,b \in G$$.

We shall show that the commutator subgroup of $$G \subseteq N$$

Consider $$(aN)(bN)=(bN)(aN)$$ since $$G/N$$ is abelian.

Thus, $$abN=baN$$.

Thus $$ab(ba)^{-1} \in N$$ Note: due to cosets.

Thus $$ab(ba)^{-1}=aba^{-1}b^{-1} \in N$$.

Therefore the commutator subgroup of $$G \subseteq N$$.

Conversely, let the commutator subgroup of $$G \subseteq N$$ and $$a,b \in G$$.

We shall show that $$G/N$$ is abelian.

Since the commutator subgroup of $$G \subseteq N$$, $$aba^{-1}b^{-1} \in N$$.

Thus, $$ab(ba)^{-1}N=N$$.

Thus $$(aN)(bN)=(bN)(aN)$$.

Therefore $$G/N$$ is abelian.

Therefore $$G/N$$ is abelian iff $$N$$ contains the commutator subgroup of $$G$$.◻

Note:  Let $$G$$ be a group.

If $$H \le G$$ and $$K \le H$$ then $$K \le G$$.  However this does not work for normal sub groups.  Thus given $$H\unlhd G$$ and $$K \unlhd H$$, it does not follow that $$K \unlhd G$$, see the following example.

##### Example $$\PageIndex{1}$$

Let $$G=S_4, H= A_4, \( and \( K= \{e, (1,2)(3,4),(1,3)(2,4), (2,3)(1,4) \} .$$

Then $$H \unlhd G$$ and $$K \unlhd H, \( but \(K$$ is not a normal subgroup of $$S_4.$$ That is $$K \not\trianglelefteq G.$$

##### Theorem $$\PageIndex{1}$$

let $$G$$ be a group and $$H \le G$$.  Then the following statements are equivalent:

1. \)H \unlhd G \).

2. \)\forall \; g\in G, \; gHg^{-1} \subseteq H \).

3. \)\forall \; g \in G, \; gHg^{-1} = H \).

#### Simple Subgroups

##### Definition: Term

A group $$G$$ is called simple if $$G$$ has no nontrivial normal subgroups.

##### Example $$\PageIndex{1}$$

$$\mathbb{Z}_2$$  is simple since the normal subgroups are  $$\{0\}, \mathbb{Z}_2$$.  $$\mathbb{Z}_p$$, for prime $$p$$ and $$A_n$$ for $$n\geq 5$$ are simple.

#### Factor Groups

##### Definition: Term

Let $$G$$ be a group and $$N \unlhd G$$.

Thus $$\{gN|g \in G\}$$ are all the cosets (ie, the set of sets) and this is defined as $$G/N= \{gN|g \in G\}$$, which is a group with the operation of $$(g_1N)(g_2N)=g_1g_2N$$.  If $$G$$ is finite, the order $$|G/N|=[G:N]$$.

We shall show  that $$G/N$$ is a group $$gN \star hN=ghN$$ with $$\star$$ being the operation in $$G$$ .

Note:  this is the process used for combining groups.

Let $$N=$$ the identity of $$G$$ .

Consider $$gN\star N=gN\star N=gN=N\star gN$$ .  Thus $$e_N$$ exists.

The inverse of $$gN\star g^{-1}N=(gg^{-1})N=eN=N$$ .  Thus the inverses exist.

$$G /N$$ is associative since $$G$$ was associative.

Since $$G/N$$ has an identity, has inverses for all elements, and is associative, $$G/N \le G$$, which is called a factor group

##### Example $$\PageIndex{1}$$

$$S_3/A_3=\{A_3, (12)A_3\}$$.

##### Example $$\PageIndex{1}$$

Let $$(\mathbb{Z}, +)$$ s.t. $$n\mathbb{Z}=\{\ldots, -2n,-n, 0, n, 2n, \ldots\}$$.

Is the subset a group?  Yes since it is a non-empty set that contains the identity and $$gh^{-1}\in \mathbb{Z}$$.

Therefore the set is a subgroup of  $$(\mathbb{Z}$$ and because of addition, it is a normal subgroup of  $$(\mathbb{Z}$$, algebraically, .

$$\mathbb{Z} / n \mathbb{Z}=\{\ldots , -1+n \mathbb{Z}, 0+n\mathbb{Z}, 1+ n \mathbb{Z}, \ldots \}$$.

##### Theorem $$\PageIndex{1}$$

Let $$G$$ be a group and $$Z(G)$$ be the centre of $$G$$.  If $$G/Z(G)$$ is cyclic, then $$G$$ is abelian.

Ex

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