4.2: Normal Supgroups and Factor Groups

Normal subgroups 

Definition: Normal subgroup

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

That is, normal subgroups are those that are invariant under conjugation by any element of the group.

Suppose \(H \le G \).

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

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 \(H \not \unlhd S_3\).

2. 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 \(\PageIndex{2}\)

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 \(\PageIndex{3}\)

List all normal subgroups in \(D_4.\)

Solution

\(D_4= \langle r,s \mid r^4=e, s^2=e, srs=r^{-1} \rangle =\{e, r, r^2, r^3, s, sr^2, sr^3\}\). Note that \(\{1\}\) and \(D_4\) are normal subgroups of \(D_4\).  

Since \(|D_4|=8\), by Lagranges thereom, nontrivial subgroups of  \(D_4\)  has order \(2\) or \(4\).

The subgroups of order \(4\) are:

\(H_1=\langle r\rangle=\{e, r, r^2, r^3\}, H_2=\langle s,r^2\rangle =\{e, s, r^2, sr^2\}, H_3=\langle sr, r^2\rangle =\{e, sr, r^2, sr^3\}.\)

Since \([D_4: H_i]=\dfrac{8}{4}=2, \forall i=1,2,3 \), \(H_i \unlhd D_4 ,\, \forall i=1,2,3\).

We know that \(Z(D_4)=\{e,r^2\}\), \(\{e,r^2\} \unlhd D_4\). (see below Theorem 3)

Verify that the subgroup of order \(2, \langle s\rangle \not \unlhd D_4\).

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 subgroups.  Thus, given \(H\unlhd G \) and \(K \unlhd H \), it does not follow that \(K \unlhd G \); see the following example. 

Example \(\PageIndex{4}\)

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

Then \( H \unlhd G \) and \( K \unlhd H, \) but \(K \) is not a normal subgroup of \(A_4. \) That is \( K \not\trianglelefteq G. \)

Answer

Let \(G = A_4, H= K_4 = \{e,(12)(34), (13)(24), (14)(23)\}\) and \(K=\{e, (12)(34)\}.\) Consider the Cayley table for \(H\) and \(K\),

 

Then clearly, \(H\leq G \text{, } K\leq G.\) We shall show that \(H \trianglelefteq G\) and \(K \trianglelefteq H\).

By Lagranges Theorem. \([H:K] = \frac{|H|}{|K|} = 2, gK = Kg, \forall\ g\in H\). Thus, \(K \trianglelefteq H\).    Now, by Lagranges Theorem. \([G:H] = \frac{|G|}{|H|} = \dfrac{12}{4}=3\), Hence \(H\) has \(3\) cosets in \(G\).

 Now, consider the left cosets of \( H\) in \(G\):\[ eH = \{ ee, e(12)(34), e(13)(24), e(14)(23) \}  = \{ e, (12)(34), (13)(24), (14)(23) \} \] \[ = ((1, 2)(3, 4))H = ((1, 3)(2, 4))H  = ((1, 4)(2, 3))H \] \[ (1, 2, 3)H = \{ (1, 2, 3)e, (1, 2, 3)(12)(34), (1, 2, 3)(13)(24), (1, 2, 3)(14)(23) \} \] \[ = \{ (1, 2, 3), (1, 3, 4), (2, 4, 3), (4, 2, 1) \} = (1, 3, 4)H  = (2, 4, 3)H  = (4, 2, 1)H \] \[ (3, 2, 1)H= \{ (3, 2, 1)e, (3, 2, 1)(12)(34), (3, 2, 1)(13)(24), (3, 2, 1)(14)(23) \} \] \[ = \{ (3, 2, 1), (2, 3, 4), (1, 2, 4), (1, 4, 3) \} = (2, 3, 4)H = (1, 2, 4)H = (1, 4, 3)H. \] And the right cosets of \( H \) are: \[ He = \{ ee, (12)(34)e, (13)(24)e, (14)(23)e \}  = \{ e, (12)(34), (13)(24), (14)(23) \}  = H((1, 2)(3, 4))  = H((1, 3)(2, 4)) = H((1, 4)(2, 3)) \] \[ H(1, 2, 3) = \{ e(1, 2, 3), (12)(34)(1, 2, 3), (13)(24)(1, 2, 3), (14)(23)(1, 2, 3) \}  = \{ (1, 2, 3), (2, 4, 3), (1, 4, 2), (1, 3, 4) \} = H(1, 3, 4)= H(2, 4, 3)= H(1, 4, 2) \] \[ H(3, 2, 1) = \{ e(3, 2, 1), (12)(34)(3, 2, 1), (13)(24)(3, 2, 1), (14)(23)(3, 2, 1) \}  = \{ (3, 2, 1), (1, 4, 3), (2, 3, 4), (1, 2, 4) \} = H(2, 3, 4)  = H(1, 2, 4)  = H(1, 4, 3) \] Since the right cosets are the same as the left cosets, we know that \( K \) is normal in \( H \).

Since the right cosets are the same as the left cosets, we know that \( H \) is normal in \( G \).

We shall show that \(K\) is not a normal subgroup of \(A_4.\) Since \((1, 2, 3)(12)(34)(3,2,1)=(1,3)(2,4)\notin K\), \(K\) is not a normal subgroup of \(A_4.\)

Theorem \(\PageIndex{2}\)

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

Theorem \(\PageIndex{3}\)

Let \(G \) be a group and \(H \le G \). Then

1. if \(G\) is abelian, then \(H \unlhd G \).

2. \(Z(G)  \unlhd G \).

 

Example \(\PageIndex{5}\)

Prove or disprove the statements:

1. If all the subgroups of a group \(G\) are normal subgroups of \(G\) then \(G\) is abelian.

2. If \(H \unlhd G \) then \(H=Z(G).\)

3. If  \(H \unlhd G \) then \([G:H]=2.\)

4. If  \(H \le G \) and \(H\) is abelian then  \(H \unlhd G \).

Solution

1. Counterexample: \(G=Q_8=\{ \pm 1, \pm i, \pm j, \pm k \mid i^2=j^2=k^2, ij=k, jk=i, ki=j\}\)

2. Counterexample: \(G=S_3\). Then \(Z(S_3)=\{e\}\) and \(A_3 \unlhd G \).  Normal subgroups are not necessarily the center of the group.

3.  Counterexample: \(G=A_4\) and \( H= \{e, (1,2)(3,4),(1,3)(2,4), (2,3)(1,4) \} . \)

4. Counterexample: \(G=A_4\) and \(K=\{e, (12)(34)\}.\)

Simple Subgroups

Definition: Simple

A group \(G\) is called simple if \(G\) has no nontrivial normal subgroups.

Example \(\PageIndex{5}\)

\(\mathbb{Z}_2\)  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: Factor Groups

Let \(G \) be a group and \(N \unlhd G \).

Thus \( \{gN|g \in G\} \) are all the cosets (i.e., the set of sets), and the factor group is defined as \(G/N= \{gN|g \in G\} \), which is a group with the operation of \((g_1N)(g_2N)=g_1g_2N \) for \(g_1g_2 \in G \).  If \(G \) is finite, the order \(|G/N|=[G:N] \).

 We shall show  that \(G/N\) is a group \(gN \, hN=(g \star h)N\) with \(\star\) being the operation in \(G\) .

Note: This is the process used to combine groups.

First, we shall show that \(N=\) is the identity of \(G/N\).

Consider \(gN\, N=(g\star e)N=N\, gN\) .  Thus, the identity  \(N\) exists.

 Suppose \(g \in G\). Then  \(gN\star g^{-1}N=(gg^{-1})N=eN=N\) .  Thus, the inverse exists  \(g^{-1}N\) for each element \(gN\).

\(G /N\) is associative since \(G\) was associative.

Since \(G/N\) has an identity, has an inverse for each element, and is associative, \(G/N\) is a group, which is called a factor group (or quotient group). 

 

Example \(\PageIndex{6}\)

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

The Cayley Table for \(S_3/A_3\):

	\(A_3\)	\((12)A_3\)
\(A_3\)	\(A_3\)	\((12)A_3\)
\((12)A_3\)	\((12)A_3\)	\(A_3\)

 

Example \(\PageIndex{7}\)

Let \((\mathbb{Z}, +) \). Consider \(n\mathbb{Z}=\{\ldots, -2n,-n, 0, n, 2n, \ldots\} \). Is this a group? 

Yes, since it is a non-empty set that contains the identity and \(gh^{-1}\in n\mathbb{Z}, \forall g,h \in n\mathbb{Z}\).

Therefore, the set is a subgroup of  \(\mathbb{Z}\). Since \(\mathbb{Z}\) is abelian, it is a normal subgroup of  \(\mathbb{Z}\). Hence,

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

Example \(\PageIndex{8}\)

Let \(G=<a>\) where   \(|a|=12,\) and let \(H=<a^3>.\) Find all cosets in \(G/H\) and  write down the Cayley table. Is \(G/H\) cyclic? Why or why not?

Solution

By Theorem 2.4.8, \(H \le G.\) Since \(G\) is abelian, \(H \unlhd G\). The factor group \(G/H=\{H, aH, a^2H\}.\) 

Cayley table for \(G/H\)
	\(H\)	\(aH\)	\(a^2H\)
\(H\)	\(H\)	\(aH\)	\(a^2H\)
\(aH\)	\(aH\)	\(a^2H\)	\(H\)
\(a^2H\)	\(a^2H\)	\(H\)	\(aH\)

Notice that \(G/H=\{H, aH, a^2H\}=\{\langle  aH \rangle \mid a^3H=H \}.\)  Thus \(G/H\)  is cyclic.

 

Theorem \(\PageIndex{4}\)

1. A factor group of a cyclic group is cyclic.

2. A factor group of an abelian group is abelian.

3. A factor group of a finite group is finite.

 

Note

A factor group of a non-abelian group can be abelian or cyclic. ( see example \(\PageIndex{6}\))

A factor group of an infinite group can be finite.

Theorem \(\PageIndex{5}\)

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

Corollary\(\PageIndex{1}\)

 

 Let \(G\) be a non-abelian group of order \(pq\), where \(p\) and \(q\) are different prime numbers. Then the center of \(G\),  \(Z(G)=\{e\}.\)

Proof

Let \(G\) be a non-abelian group of order \(pq\), where \(p\) and \(q\) are prime. 

By Lagranges thereom, the subgroups of  \(G\)  has order \(1\) , \(p\), \(q\) or \(pq\).

 Since \(G\) is non-abelian, \(Z(G) \ne G\).

Assume that \(|Z(G)|\) is \(p\) or \(q\), then \(G/Z(G)\) is of prime order. Therefore, it is cyclic. Hence by Theorem \(\PageIndex{3}\), \(G\) is abelian.

This contradicts the fact that \(G\) is non-abelian. Hence \(|Z(G)|=1\). Thus \(Z(G)=\{e\}.\)

 

Example \(\PageIndex{9}\)

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 \(G\).◻