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4.2: Normal Groups and Factor Groups

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    Example \(\PageIndex{1}\)

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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 \(H \unlhd G.\)Screen Shot 2023-07-06 at 4.04.44 PM.png

    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 \(A_3=\{\langle (123) \rangle\} \unlhd S_3\)?


    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 \(G\).


    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 \(S_n, \; \forall \; n \in \mathbb{N}\).


    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 \(G\).


    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 \(ghg^{-1}=gaba^{-1}b^{-1}g^{-1}\)


             \(=(gag^{-1})(gbg^{-1})(ga^{-1}g^{-1})(gb^{-1}g^{-1})\) 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 \(G\).


    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 \(G\).◻


    Screen Shot 2023-07-06 at 4.06.14 PM.pngNote:  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.


    4.2: Normal Groups and Factor Groups is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.

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