Quotient groups

One of the most useful facts about normal subgroups is the following result/definition.

Lemma 5.18.1   If $H$ is a normal subgroup of $G$ then the coset space $G/H$ with the binary operation,

$$aH*bH = (ab)H, (aH)^{-1}=a^{-1}H,$$

for all $a,b$ belonging to $G$ , is a group. The identity element of this group is the trivial coset $H$ .

This group $G/H$ is called the quotient group of $G$ by $H$ and is sometimes pronounced ``$G$ mod $H$ ''.

Example 5.18.2   If $f : G_1 \rightarrow G_2$ is a homomorphism between two groups then $G_1/ker(f)$ is a group.

Next we introduce an important idea first emphasized by E. Galois (October 1811-May 1832), the French mathematician mentioned in the introduction. Though Galois attended secondary school, he had trouble entering the French university system. Sadly, by the time he entered the École Normale Súperieure in November 1829, he also began to get caught up in the activities of the French revolution which toppled Charles X. His political activities resulted in him being expelled from schoolin December 1830. Though apparently he had been a rather rebellious teenager, this only frustrated him further. He died in a fight with another Frenchman.

Galois's work was the first to begin to illuminate the ``basic building blocks'' of the collection of finite groups (i.e., the analog of the idea that atoms are the ``basic building blocks'' of molecules). Galois introduced the ideas of solvable groups and normal groups. Roughly speaking, one may say that the basic building blocks of finite groups are those the groups which have no proper non-trivial normal subgroups. Intuitively, this is because a non-trivial quotient group (by a normal subgroup) is closely related to the original group but smaller in size (and hence perhaps subject to analysis by an inductive argument of some type). These basic building blocks are called ``simple'' groups.

Definition 5.18.3   A simple group is a group with no proper normal subgroups other than the trivial subgroup $\{1\}$ .

All finite simple groups have been classified. In fact, the survey book discussing this is free and download able from [GLS].

Example 5.18.4   If $p$ is a prime then $C_p$ (the cyclic group having $p$ elements) is simple. In fact, if $G$ is any group which is both abelian and simple then there is a prime $p$ such that $G\cong C_p$ . If $n>4$ then $A_n$ is simple (as was stated above in Theorem 5.17.7). These facts are proven in [R].

Simple groups are not very abundant. In fact, the first non-abelian simple group is of order $60$ (it's $A_5$ ).

Example 5.18.5   Let $f:\mathbb{Z}\rightarrow \mathbb{Z}/m\mathbb{Z}$ denote the map which sends an integer $n$ to its residue $\overline{n}=n ({\rm mod} m)$ . This map send every multiple of $m$ to $\overline{0}$ . In fact, if $f(n)=\overline{0}$ then $m\vert n$ . Therefore, $ker(f)=m\mathbb{Z}$ .

The following basic result, which generalizes the above example, describes the quotient group $G_1/ker(f)$ .

Theorem 5.18.6 (First Isomorphism Theorem)   If $f : G_1 \rightarrow G_2$ is a homomorphism between two groups then $G_1/ker(f)$ is isomorphic to $f(G_1)$ .

proof: $ker(f)$ is a normal subgroup of $G_1$ , so $G_1/ker(f)$ is a group. We must show that this group is isomorphic to the group $f(G_1)$ . Define $\phi: G_1/ker(f)\rightarrow f(G_2)$ by $\phi(g\cdot ker(f))=f(g)$ , for $g\in G_1$ . We must show

(a) $\phi$ is well-defined,

(b) $\phi$ is a homomorphism,

(c) $\phi$ is a bijection.

If $g\cdot ker(f)=g'\cdot ker(f)$ then $g^{-1}g'\in ker(f)$ , since $ker(f)$ is a group. This implies $f(g^{-1}g')\in f(ker(f))=\{1\}$ , so $f(g)=f(g')$ . This implies $\phi$ is well-defined.

Since $ker(f)$ is normal, $(g\cdot ker(f))(g'\cdot ker(f)) =gg'({g'}^{-1}ker(f)g')ker(f)=gg'\cdot ker(f)$ . Therefore $\phi((g\cdot ker(f))(g'\cdot ker(f)))= \phi(gg'\cdot ker(f))=f(gg')=f(g)f(g')= \phi(g\cdot ker(f))\phi(g'\cdot ker(f))$ , for all $g,g'\in G$ . This implies $\phi$ is a homomorphism.

It is clear that $\phi$ is surjective. To show that $\phi$ is a bijection, it suffices to prove $\phi$ is an injection. Suppose that $\phi(g\cdot ker(f))=\phi(g'\cdot ker(f))$ , for some $g,g'\in G$ . Then $f(g)=f(g')$ , so $f(g^{-1}g')=1$ . By definition of the kernel, this implies $g^{-1}g'\in ker(f)$ , so $g\cdot ker(f)=g'\cdot ker(f)$ . This implies $\phi$ is injective.

$\Box$

Example 5.18.7   If $f:k\mathbb{Z}/lcm(m,k)\mathbb{Z} \rightarrow \mathbb{Z}/{m\over{gcd(k,m)}}\mathbb{Z}$ is the ``mod ${m\over{gcd(k,m)}}$ map'' then $f$ is an isomorphism.

In particular, if $m>1$ and $k>1$ are relatively prime integers then $k\mathbb{Z}/mk\mathbb{Z} \cong \mathbb{Z}/m\mathbb{Z}$ .

Indeed, by replacing $m$ by $m/gcd(k,m)$ if necessary, we may assume that $m>1$ and $k>1$ are relatively prime integers. It suffices to show that the kernel is trivial since both $k\mathbb{Z}/mk\mathbb{Z}$ and $\mathbb{Z}/m\mathbb{Z}$ have the same cardinality. If $f$ sends $\overline{ka}$ to $\overline{0}$ (where $ka\in k\mathbb{Z}$ ) then $m\vert ka$ , so $mk\vert ka$ since $m,k$ are relatively prime.

proof: $\Box$

Example 5.18.8   Let $f:\mathbb{Z}\rightarrow \mathbb{Z}/k\mathbb{Z}$ denote the map which sends an integer $a$ to its residue $\overline{a}=a ({\rm mod} k)$ . Let $g:\mathbb{Z}\rightarrow \mathbb{Z}/m\mathbb{Z}$ denote the map which sends an integer $a$ to its residue $\overline{a}=a ({\rm mod} m)$ , where $k>1$ , $m>1$ are integers. Let $H=ker(f)$ , $N=ker(g)$ , so that $N\cap H=m\mathbb{Z}\cap k\mathbb{Z} =lcm(k,m)\mathbb{Z}$ .

Example 5.18.9   Let $f_1:\mathbb{Z}\rightarrow \mathbb{Z}/km\mathbb{Z}$ denote the map which sends an integer $a$ to its residue $\overline{a}=a ({\rm mod} km)$ . Let $f_2:\mathbb{Z}\rightarrow \mathbb{Z}/m\mathbb{Z}$ denote the map which sends an integer $a$ to its residue $\overline{a}=a ({\rm mod} m)$ , where $k>1$ , $m>1$ are integers. Let $N_1=ker(f_1)$ , $N_2=ker(f_2)$ , so that $N_1=km\mathbb{Z}\subset N_2=m\mathbb{Z}$ .

Then $N_2/N_1=m\mathbb{Z}/km\mathbb{Z}$ is a subgroup of $\mathbb{Z}/km\mathbb{Z}$ and the map $f:\mathbb{Z}/km\mathbb{Z} \rightarrow \mathbb{Z}/m\mathbb{Z}$ defined by sending $a ({\rm mod} km)\in \mathbb{Z}/km\mathbb{Z}$ to $a ({\rm mod} m)\in \mathbb{Z}/m\mathbb{Z}$ has kernel $ker(f)=m\mathbb{Z}/km\mathbb{Z}$ . By the first isomorphism theorem, we therefore have

$$(\mathbb{Z}/km\mathbb{Z})/(m\mathbb{Z}/km\mathbb{Z}) \cong \mathbb{Z}/m\mathbb{Z}.$$

Exercise 5.18.10   Let $S_n$ act on the set $\mathbb{Z}_n = \{1,2, ..., n\}$ by permutations in the usual way. Let $x\in \mathbb{Z}_n$ be arbitary.

(a) Describe the elements in the stabilizer $stab_{S_n}(x)$ as permutations in $2\times n$ array form (recall (5.3) was the definition). Is this a normal subgroup of $S_n$ ?

(b) Compute the orbit $S_n*x$ .

(c) Show that there is a 1-1 correspondence between the elements in the orbit $S_n*x$ and elements in $stab_{S_n}(x)$ . (Hint: See the proof of Proposition 5.13.2.)

Exercise 5.18.11   Let $S_n$ act on the set $\mathbb{Z}_n = \{1,2, ..., n\}$ by permutations in the usual way. Let $X=\mathbb{Z}_n^k$ denote the set of all $k$ -tuples of $\mathbb{Z}_n$ . The group $S_n$ acts on $X$ as well: if $x=(a_1,...,a_k)\in X$ and $g\in S_n$ then we define $g*x=(g*a_1,...,g*a_k)$ . Show that the stabilizer $stab_{S_n}(x)$ is a normal subgroup of $S_n$ if and only if $x$ has the form $x=(i,i,...,i)\in X$ for some $i\in \mathbb{Z}_n$ . In this case, compute the quotient group $S_n/stab_{S_n}(x)$ .

Exercise 5.18.12   Find all the normal subgroups $N$ of $A_4$ . For each one, compute $A_4/N$ .