Normal subgroup

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In abstract algebra, a normal subgroup is a subgroup which is invariant under conjugation by members of the group. Normal subgroups can be used to construct quotient groups from a given group. In other words, a subgroup H of a group G is normal in G if and only if aH = Ha for all a in G.

Évariste Galois was the first to realize the importance of the existence of normal subgroups.

Definitions

A subgroup, N, of a group, G, is called a normal subgroup if it is invariant under conjugation; that is, for each element n in N and each g in G, the element gng⁻¹ is still in N. We write

$N \triangleleft G\,\,\Leftrightarrow\,\forall\,n\in{N},\forall\,g\in{G}\ , gng^{-1}\in{N}.$

For any subgroup, the following conditions are equivalent to normality. Therefore any one of them may be taken as the definition:

• For all g in G, gNg⁻¹ ⊆ N.
• For all g in G, gNg⁻¹ = N.
• The sets of left and right cosets of N in G coincide.
• For all g in G, gN = Ng.
• N is a union of conjugacy classes of G.
• There is some homomorphism on G for which N is the kernel.

The last condition accounts for some of the importance of normal subgroups; they are a way to internally classify all homomorphisms defined on a group. For example, a non-identity finite group is simple if and only if it is isomorphic to all of its non-identity homomorphic images, a finite group is perfect if and only if it has no normal subgroups of prime index, and a group is imperfect if and only if the derived subgroup is not supplemented by any proper normal subgroup.

Examples

• The subgroup {e} consisting of just the identity element of G and G itself are always normal subgroups of G. The former is called the trivial subgroup, and if these are the only normal subgroups, then G is said to be simple.
• The center of a group is a normal subgroup.
• The commutator subgroup is a normal subgroup.
• More generally, any characteristic subgroup is normal, since conjugation is always an automorphism.
• All subgroups N of an abelian group G are normal, because gN = Ng. A group that is not abelian but for which every subgroup is normal is called a Hamiltonian group.
• The translation group in any dimension is a normal subgroup of the Euclidean group; for example in 3D rotating, translating, and rotating back results in only translation; also reflecting, translating, and reflecting again results in only translation (a translation seen in a mirror looks like a translation, with a reflected translation vector). The translations by a given distance in any direction form a conjugacy class; the translation group is the union of those for all distances.
• In the Rubik's Cube group, the subgroup consisting of operations which only affect the corner pieces is normal, because no conjugate transformation can make such an operation affect an edge piece instead of a corner. By contrast, the subgroup consisting of turns of the top face only is not normal, because a conjugate transformation can move parts of the top face to the bottom and hence not all conjugates of elements of this subgroup are contained in the subgroup.

However ,evenif H is a normal subgroup of G ,this does not mean that ah=ha for all h€H and for all a€G.As the following example shows : Consider the group S3. Let H={e,(1 2 3),(1 3 2)}. Then H is a subgroup of S3. Since e,(1 2 3) and (1 3 2) are elements of H , it follows that eH=He ,(1 2 3)H=H(1 2 3) and (1 3 2)H=H(1 3 2) .Now (1 2)H={(1 2),(1 2)(1 2 3),(1 2)(1 3 2)} H(1 2)={(1 2),(1 3 2)(1 2),(1 2 3)(1 2)} Hence (1 2)H=H(1 2) (2 3)H={(2 3),(2 3)(1 2 3),(2 3)(1 3 2)}={(2 3),(1 3),(1 2)} H(2 3)={(2 3),(1 2 3)(2 3),(1 3 2)(2 3)}={(2 3),(1 2),(1 3)} Hence (2 3)H=H(2 3) Also in the same way this can be shown that (1 3)H=H(1 3) Consequently H is a normal subgroup.However we point out that for (1 2 3)€H and (2 3)€G(=S3) (2 3)(1 2 3)=(1 3)≠(1 2)=(1 2 3)(2 3)

Properties

• Normality is preserved upon surjective homomorphisms, and is also preserved upon taking inverse images.
• Normality is preserved on taking direct products
• A normal subgroup of a normal subgroup of a group need not be normal in the group. That is, normality is not a transitive relation. However, a characteristic subgroup of a normal subgroup is normal. Also, a normal subgroup of a central factor is normal. In particular, a normal subgroup of a direct factor is normal.
• Every subgroup of index 2 is normal. More generally, a subgroup H of finite index n in G contains a subgroup K normal in G and of index dividing n! called the normal core. In particular, if p is the smallest prime dividing the order of G, then every subgroup of index p is normal.

Lattice of normal subgroups

The normal subgroups of a group G form a lattice under subset inclusion with least element {e} and greatest element G. Given two normal subgroups N and M in G, meet is defined as

$N \wedge M := N \cap M$

and join is defined as

$N \vee M := N M = \{nm \,|\, n \in N \text{, and } m \in M\}.$

The lattice is complete and modular.

Normal subgroups and homomorphisms

If N is normal subgroup, we can define a multiplication on cosets by

(a₁N)(a₂N) := (a₁a₂)N.

This turns the set of cosets into a group called the quotient group G/N. There is a natural homomorphism f : G → G/N given by f(a) = aN. The image f(N) consists only of the identity element of G/N, the coset eN = N.

In general, a group homomorphism f: G → H sends subgroups of G to subgroups of H. Also, the preimage of any subgroup of H is a subgroup of G. We call the preimage of the trivial group {e} in H the kernel of the homomorphism and denote it by ker(f). As it turns out, the kernel is always normal and the image f(G) of G is always isomorphic to G/ker(f) (the first isomorphism theorem). In fact, this correspondence is a bijection between the set of all quotient groups G/N of G and the set of all homomorphic images of G (up to isomorphism). It is also easy to see that the kernel of the quotient map, f: G → G/N, is N itself, so we have shown that the normal subgroups are precisely the kernels of homomorphisms with domain G.

Theorem: If H and K are two subgroups of a group G, then

 (i) if H is a normal subgroup of G, then HK=KH is a subgroup of G.
 (ii) if H and K are both normal subgroups of G then HK=KH is a normal subgroup of G.
 (iii) if H and K are both normal subgroups of G then HດK is a normal subgroup of G.

Proof :  
        (i) Let b€K. Then Hb=bH implies Hb≤KH. This is true for any b€K. Hence HK≤KH. Similarly it can be shown that KH≤HK. So HK=KH. So HK is a subgroup of G. (Note: it is a theorem that if H and K are two subgroups of G and HK=KH then HK is a subgroup of G.)
        (ii) Suppose that H and K are both normal subgroups of G. Now from (i) HK=KH is a subgroup of G. Let g€G. Then gHKg(-1)=gHg(-1)gKg(-1)=(gHg(-1))(gKg(-1))≤HK. Hence  HK  is  a  normal  subgroup  of  G.
        (iii) Since H and K are subgroups, HດK is also a subgroup. Let g€G and a€HດK. Then gag(-1)€gHg(-1)≤HK and gag(-1) €gKg(-1)≤K. Since gag(-1)€HດK for all a€HດK, we find that g(HດK)g(-1)≤HດK for all g€G. Hence HດK is a normal subgroup of G.

(Q.E.D)

Theorem:

    Let H  be  a  normal  subgroup  of  a  group  G .Denote  the  set  of  all 
cosets {aH : a€G}  by G/H  and and  define  * on  G/H  by 
 for  all  aH , bH € G/H ,(aH)*(bH)=abH.
Then (G/H,*) is  a  group.

 Proof  :  First  of  all  we  have  to  show  that  the  operation  *  is  well
defined  binary  operation  on  G/H. In  other  words  we  have  to  show  that  
if  aH=a1H  and  bH=b1H  then  (ab)H=(a1b1)H ,
 as  this  will show   that  aH*bH=(ab)H=(a1b1)H=a1H * b1H.

Now let aH=a₁H and bH=b₁H is given. This imply that

a=a1h1  and  b=  b1h2  for  some  h1,h2 €H. 

Then (a₁b₁)^-1ab=(b₁)^-1(a₁)^-1a₁h₁b₁h₂ (1.1) Since H is a normal subgroup , Hb₁ = b₁H and hence h₁b₁ =b₁h₃ for some h₃€H . Hence from (1.1) (a₁b₁)^-1 = b₁^-1b₁h₃h₂ = h₃h₂ € H. This implies that (a₁b₁)H= (ab)H.Hence aH*bH=abH=a₁b₁H=a₁H*b₁H and so * is a

well  defined  binary  operation  on  G/H.
    Next  we  show  the  associativitty  of  *  on  G/H.  Let aH ,bH, cH € G/H.Now 
aH *(bH*cH)=aH*bcH=a(bc)H=(ab)cH=abH*cH=(aH*bH)*cH .  Hence  *  is  associative .
     Now  eH € G/H  and  aH*eH=aeH=aH=eaH=eH*aH
 for  all  aH€ G/H .Hence  eH  is  the  identity  of  (G/H,*).Also  for  all  aH€(G/H),

a^-1H€ G/H and

  aH*a-1H=aa-1H=eH=a-1aH=a-1H* aH.
 Here  for  all  aH€ G/H,a-1H  is  the  inverse  of  aH .Thus  (G/H,*)  is  a  group.

(Q.E.D)

//SOME  RESULTS  RELATED  TO  NORMAL  SUBGROUPS//
 * Let H  be  a  subgroup  of  G  such  that  [G:H]=2.Then  H  is  a  normal  subgroup.
   Proof::  Since [G:H]=2 ,the  group  G  has only  two  distinct  left 
cosets  and  only  two  distinct right  cosets.Now  H  itself  is  a  left
 as  well  as  roght  coset  in  G. Let  a€G.If a€H , then  aH=H=Ha. Suppose
 a  does  not  belong  to  H.Then  aH≠H .Hence  G=HυaH  and  HດaH=Φ .Then
aH=G-H.Again  since  a  does  not  belong  to  H  and  G  has  only  two
 right   cosets , we  find  that  G=HυHa  where HດHa=Φ . Thus  Ha=G-H.
Hence  Ha=aH.Thus  we  find  that  aH=Ha  for  all  a€G  and  so  H is
 a  normal  subgroup  of  G.(Q.E.D.)

• The center of a group G , given by Z(G)={a€G:ag=ga for all g€G} is a normal

subgroup of G.

  Proof :: We  know  that  the  center  of  a  group  is a  subgroup  of  that  group.
Now  for  any  g€G  

and any a€Z(G).gag^-1=agg^-1=a€Z(G) and hence , gZ(G)g^-1≤Z(G).Consequently Z(G) is a normal subgroup.

* Let /h  be  subgroup  of  the  group  G.If x2€H  for  all  x€G , then H  is  a  normal  subgroup  of  G  and  G/H  is  commutative.
  Proof :: Let g€G  and  h€H . Consider  ghg-1  and  note  that 
          ghg-1=ghghh-1g-2=(gh)2h-1g-2.
    Now  h-1€H  and  by  our  hypothesis  (gh)2,g-2€H .
This  implies   that  ghg-1€H 
  which  in  turn  shows  that  gHg-1⊆H .Hence  H  is a  normal  subgrpup  of  G . 
Toshow  G/H is   commutative , let  xH,yH ∈G/H,.We  show  that  xHyH=yHxH  or  xyH-yxH  or 
(yx)-1(xy)∈H. Now ,
 (yx)-1(xy)=(x-1y-1)(xy)=(x-1y-1)2(yxy-1)2y2. Since  a2∈H  for  all  a∈G, it  follows  that  (x-1y-1)2(yxy-1)2y2∈H  and   so  (yx)-1(xy)∈H .

Hence G/H is commutative.

* Let  G  be  a group  such  that  every  cyclic  subgroup  of  G  is  a  normal  

subgroup of G .

 Then every  subgroup  of  G  is  a  normal  subgroup  of  G.
 Proof :: Let  H  be  a  subgroup  of  G . Let  g∈G  and  a∈H.Then  gag-1∈<a>⊆H.

Hence H is normal in G.

 * Let  H  is a proper  subgroup  of  a  group  G . such  that  for  all  x,y∈G-H ,xy∈H.

Then H is a normal subgroup of G.

   Proof :: Let  x∈G-H .Then x-1∈G-H .Let  y∈H .Then  xy∈G-H,

(for otherwise ,x=xyy^-1∈H).Thus xy,x^-1∈G-H.

  Hence  xyx-1∈H.Also  for  any  x∈H, we  have  xyx-1∈H. 
Thus  H  is a  normal  subgroup  of  G.

  * Let   H  be  a  subgroup  of  the  group  G .Suppose  that  the  product  of  two 
left  cosets  of  of  H  in  G is  again  a  left  coset  of  H  in G .Then  H  is  a 
Normal  subgroup  of  G .
  Proof ::  Let  g∈G ,Then  gHg-1H=tH  for    some  t∈G.. Thus  e=geg-1e∈tH.

Hence e=th for some h∈H.Thus t=h^-1∈H so that tH=H. Now gHg^-1⊆gHg^-1H=H.

Hence  H  is  a  normal  subgroup.

See also

Operations taking subgroups to subgroups normalizer conjugate closure normal core Subgroup properties complementary (or opposite) to normality malnormal subgroup contranormal subgroup abnormal subgroup self-normalizing subgroup Subgroup properties stronger than normality characteristic subgroup fully characteristic subgroup

Subgroup properties weaker than normality subnormal subgroup ascendant subgroup descendant subgroup quasinormal subgroup seminormal subgroup conjugate permutable subgroup modular subgroup pronormal subgroup paranormal subgroup polynormal subgroup c normal subgroup Related notions in algebra ideal (ring theory)

References

• I. N. Herstein, Topics in algebra. Second edition. Xerox College Publishing, Lexington, Mass.-Toronto, Ont., 1975. xi+388 pp.
• David S. Dummit; Richard M. Foote, Abstract algebra. Prentice Hall, Inc., Englewood Cliffs, NJ, 1991. pp. xiv, 658 ISBN 0-13-004771-6
• Sen,Ghosh,Mokhopadhyay,"Topics in Abstract Algebra".Second edition. Universities Press(india) Private Limited . ISBN 978-81-7371-551-8

External links

• Weisstein, Eric W., "normal subgroup" from MathWorld.
• Normal subgroup in Springer's Encyclopedia of Mathematics
• Robert Ash: Group Fundamentals in Abstract Algebra. The Basic Graduate Year
• John Baez, What's a Normal Subgroup?