2.5 Normal subgroups and factor groups

Cosets

For any pair of subsets A and B of a group G, define AB to be the set

$$AB=\{ab\mid a\in A\text{ and }b\in B\}.$$

If H is a subgroup of G then $HH=H$

When A is a singleton set, say $A=\{a\}$ , we usually write aB instead of $\{a\}B$ . If H is a subgroup of G, then each subset aH where $a\in G$ is called a (left) coset of H. Two cosets of a given subgroup H are either identical or non-intersecting. For, suppose there exists an element $g\in aH\cap bH$ . Setting

$$g=a{h}_1=b{h}_2(h_1,h_2\in H)$$

we have for any $h\in H$

$$ah=b{h}_2{h}_1^{-1}h\in bH,$$

so that a $H\subseteq bH$ . Equally, it can be argued that $bH\subseteq aH$ , whence either $aH\cap bH=\varnothing$ or a $H=bH$ . Since $g=ge$ and $e\in H,$ , any element $g\in G$ always belongs to the coset $gH$ . Thus the cosets of H form a family of disjoint subsets covering all of G. There is an alternative way of demonstrating this partitioning property. The relation $a\equiv b$ on $G,$ defined by

$$a\equiv b\text{ iff }{b}^{-1}a\in H,$$

is an equivalence relation since it is (i) reflexive, $a^{-1}a=e\in H;(\mathrm{ii})$ symmetric, $a^{-1}b=$ $(b^{-1}a{)}^{-1}\in H$ if $b^{-1}a\in H$ ; and (iii) transitive, $a^{-1}b\in H,b^{-1}c\in H$ implies ${\mathbf{a}}^{-1}\mathbf{c}=$ $a^{-1}b{b}^{-1}c\in H$ . The equivalence classes defined by this relation are precisely the left cosets of the subgroup H, for $b\equiv a$ if and only if $b\in aH$

Theorem 2.3

(Lagrange) IfG is afinite group oforder n, then the order ofevery subgroup H is a divisor ofn.

Proof

Every coset gH is in one-to-one correspondence with H, for if $g{h}_1=g{h}_2$ then $h_1=g^{-1}g{h}_2=h_2$ . Hence every coset $gH$ must have exactly H elements, and since the cosets partition the group G it follows that n is a multiple of $|H|$ -

Corollary 2.4

The order ofany element is a divisor of G .

Proof

Let g be any element of G and let m be its order. As shown in Example 2.5 the elements $\{g,g^2,...,g^m=e\}$ are then all unequal to each other and form a cyclic subgroup of order m. By Lagrange’s theorem m divides the order of the group, G .

Exercise

If G has prime order p all subgroups are trivial – they are either the identity subgroup e or G itself. Show that G is a cyclic group.

Normal subgroups

The right cosets Hg of a subgroup H are defined in a completely analogous way to the lef cosets. While in general there is no obvious relationship between right and left cosets, there is an important class of subgroups for which they coincide. A subgroup N of a group G is called normal if

$$gN{g}^{-1}=N,\forall g\in G.$$

Such subgroups are invariant under inner automorphisms; they are sometimes referred to as invariant or self-conjugate subgroups. The key feature of normal subgroups is that the systems of left and right cosets are identical, for

$$gN=gN{g}^{-1}g=Ng,\forall g\in G.$$

This argument may give the misleading impression that every element of N commutes with every element of $G,$ , but what it actually demonstrates is that for every $n\in N$ and every $g\in G$ there exists an element $n^{\prime }\in N$ such that $gn=n^{\prime }g$ . There is no reason, in general, to expect that $n^{\prime }=n$

For any group G the trivial subgroups e and G are always normal. A group is called simple if it has no normal subgroups other than these trivial subgroups.

Example 2.20

The centre Z of a group G is defined as the set of elements that commute with all elements of $G_{:}$

$$Z=\{z\in G|zg=gz\text{ for all }g\in G\}.$$

This set forms a subgroup of G since the three essential requirements hold:

Closure: if $z,z^{\prime }\in Z$ then $z{z}^{\prime }\in Z$ since

$$(z{z}^{\prime })g=z(z^{\prime }g)=z(g{z}^{\prime })=(zg)z^{\prime }=(gz)z^{\prime }=g(z{z}^{\prime }).$$

Identity: $e\in Z$ , as $eg=ge=g$ for all $g\in G$

Inverse: i $z\in Z$ then $z^{-1}\in Z$ since

$$z^{-1}g=z^{-1}ge=z^{-1}gz{z}^{-1}=z^{-1}zg{z}^{-1}=g{z}^{-1}.$$

This subgroup is clearly normal since $gZ=Zg$ for all $g\in G$

Factor groups

When we multiply left cosets of a subgroup H together, for example

$$gH{g}^{\prime }H=\{gh{g}^{\prime }{h}^{\prime }\mid h,h^{\prime }\in H\},$$

the result is not in general another coset. On the other hand, the product ofcosets ofa norma subgroup N is always another coset,

$$gN{g}^{\prime }N=g{g}^{\prime }NN=(g{g}^{\prime })N,$$

and satisfies the associative law,

$$(gN{g}^{\prime }N)g^{\prime \prime }N=(g{g}^{\prime }{g}^{\prime \prime })N=gN(g^{\prime }N{g}^{\prime \prime }N).$$

Furthermore, the coset $eN=N$ plays the role of an identity element, while every coset has an inverse $(gN{)}^{-1}=g^{-1}N$ . Hence the cosets of a normal subgroup N form a group called the factor group of $G$ by N, denoted $G/N$

Example 2.21

The even integers 2Z form a normal subgroup of the additive group of integers $\mathbb{Z},$ since this is an abelian group. The factor group $\mathbb{Z}/2\mathbb{Z}$ has just two cosets $[0]=0+2\mathbb{Z}$ and $[1]=1+2\mathbb{Z}.$ , and is isomorphic to the additive group ofintegers modulo 2, denoted by ${\mathbb{Z}}_2$ (see Example 2.4).

Kernel of a homomorphism

Let $\phi :G\to G^{\prime }$ be a homomorphism between two groups $G$ and $G^{\prime }$ . The kernel of ${\phi }_{:}$ denoted $\ker (\phi )$ , is the subset of $G$ consisting of those elements that map onto the identity $e^{\prime }$ of $G^{\prime }$ ,

$$\ker (\phi )={\phi }^{-1}(e^{\prime })=\{k\in G|\phi (k)=e^{\prime }\}.$$

The kernel $K=\ker (\phi )$ of any homomorphism $\phi$ is a subgroup of $G$ :

Closure: $\mathrm{if}{k}_1$ and $k_2$ belong to K then so does $k_1{k}_2$ , since

$$\phi (k_1{k}_2)=\phi (k_1)\phi (k_2)=e^{\prime }{e}^{\prime }=e^{\prime }.$$

Identity: $e\in K$ as $\phi (e)=e^{\prime }$

Inverse: if $k\in K$ then $k^{-1}\in K$ , for

$$\phi (k^{-1})=(\phi (k){)}^{-1}=(e^{\prime }{)}^{-1}=e^{\prime }.$$

Furthermore, K is a normal subgroup since, for all $k\in K$ and $g\in G$

$$\phi (gk{g}^{-1})=\phi (g)\phi (k)\phi (g^{-1})=\phi (g)e^{\prime }(\phi (g){)}^{-1}=e^{\prime }.$$

The following theorem will show that the converse of this result also holds, namely that every normal subgroup is the kernel of a homomorphism.

Theorem 2.5

Let G be a group. Then thefollowing two properties hold:

1. IfN is a normal subgroup ofG then there is a homomorphism $\mu :G\to G/N.$

2. If $\dot{\phi }:G\to G^{\prime }$ is a homomorphism then thefactor group $G/\ker (\phi )$ is isomorphic with the image subgroup im $(\phi )\subseteq G^{\prime }$ defined in Eq. (2.13),

$$\mathrm{im}(\phi )\cong G/\ker (\phi ).$$

Proof

1. The map $\mu :G\to G/N$ defined by $\mu (g)=gN$ is a homomorphism, since

$$\mu (g)\mu (h)=gNhN=ghNN=ghN=\mu (gh).$$

2. Let $K=\ker (\phi )$ and $H^{\prime }=\mathrm{im}(\phi )$ . The map $\phi$ is constant on each coset $gK$ , for

$$k\in K⟹\phi (gk)=\phi (g)\phi (k)=\phi (g)e^{\prime }=\phi (g).$$

Hence the map ϕ defines a map $\psi :G/K\to H^{\prime }$ by setting

$$\psi (gK)=\phi (g),$$

and this map is a homomorphism since

$$\psi (gKhK)=\psi (ghK)=\phi (gh)=\phi (g)\phi (h)=\psi (gK)\psi (hK).$$

Furthermore $\psi$ is one-to-one, for

$$\begin{array}{rl}\psi (gK)=\psi (hK)& ⟹\phi (g)=\phi (h)\\ & ⟹\phi (g{h}^{-1})=\phi (g)(\phi (h){)}^{-1}=e^{\prime }\\ & ⟹g{h}^{-1}\in K\\ & ⟹g\in hK.\end{array}$$

Since every element $h^{\prime }$ of the image set H is of the form $h^{\prime }=\phi (g)=\psi (gK)$ , the map ψ is an isomorphism between the groups $G/K$ and $H^{\prime }$ -

Example 2.22

Let $G$ and H be two groups with respective identity elements $e_G$ and $e_H$ . A law of composition can be defined on the cartesian product $G\times H$ by

$$(g,h)(g^{\prime },h^{\prime })=(g{g}^{\prime },h{h}^{\prime }).$$

This product clearly satisfies the associative law and has identity element $(e_G,e_H)$ . Furthermore, every element has a unique inverse $(g,h{)}^{-1}=(g^{-1},h^{-1})$ ). Hence, with this law of composition, $G\times H$ is a group called the direct product of G and H. The group G is clearly isomorphic to the subgroup $(G,e_H)=\{(g,e_H)|g\in G\}$ . The latter is a norma subgroup, since

$$(a,b)(G,e_H)(a^{-1},b^{-1})=(aG{a}^{-1},b{e}_H{b}^{-1})=(G,e_H).$$

It is common to identify the subgroup of elements $(G,e_H)$ with the group $G$ . In a similar way H is identified with the normal subgroup $(e_G,H)$ ).

Problems

Problem 2.13

(a) Show that if H and K are subgroups of G then their intersection $H\cap K$ is always a subgroup of G.

(b) Show that the product $HK=\{hk\mid h\in H,k\in K\}$ is a subgroup if and only if $HK=KH$

Problem 2.14

Find all the normal subgroups ofthe group ofsymmetries ofthe square $D_4$ described in Example 2.7.

Problem 2.15

The quaternion group Q consists of eight elements denoted

$$\{1,-1,i,-i,j,-j,k,-k\},$$

subject to the following law of composition:

$$1g=g1=g,\text{for all }g\in Q,$$ $$-1g=-g,\mathrm{for}g=i,j,k,$$ $$i^2=j^2=k^2=-1,$$ $$ij=k,jk=i,ki=j.$$

(a) Write down the full multiplication table for Q, justifying all products not included in the above list.

(b) Find all subgroups of Q and show that all subgroups of Q are normal.

(c) Show that the subgroup consisting of $\{1,-1,i,-i\}$ is the kernel of a homomorphism $Q\to \{1,-1\}$.

(d) Find a subgroup H of $S_4$, the symmetric group of degree 4, such that there is a homomorphism $Q\to H$ whose kernel is the subgroup $\{1,-1\}$.

Problem 2.16

A M¨obius transformation is a complex map,

$$z\mapsto z^{\prime }=\frac{az+b}{cz+d}\text{ where }a,b,c,d\in \mathbb{C},ad-bc=1.$$

(a) Show that these are one-to-one and onto transformations of the extended complex plane, which includes the point $z=\infty ,$ , and write out the composition of an arbitrary pair of transformations given by constants $(a,b,c,d)$ and $(a^{\prime },b^{\prime },c^{\prime },d^{\prime })$

(b) Show that they form a group, called the M¨obius group.

(c) Show that the map $\mu$ from $SL(2,\mathbb{C})$ to the M¨obius group, which takes the unimodular matrix $\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ to the above M¨obius transformation, is a homomorphism, and that the kernel of this homomorphism is $\{I,-I\}$; i.e. the M¨obius group is isomorphic to $SL(2,\mathbb{C})/{\mathbb{Z}}_2$

Problem 2.17

Assuming the identification of G with $(G,e_H)$ and H with $(e_G,H)$ , show that $G\cong$ $(G\times H)/H$ and $H\cong (G\times H)/G$

Problem 2.18

Show that the conjugacy classes of the direct produc $G\times H$ of two groups G and H consist precisely of products of conjugacy classes from the groups

$$\left(C_i,D_j\right)=\left\{\left(g_i,h_j\right)\mid g_i\in C_i,h_j\in D_j\right\}$$

where $C_i$ is a conjugacy class of G and $D_j$ a conjugacy class of H.