Factor groups

Again let $N \lhd G$ and consider the set

$$\{aN, bN, ...\}$$ (6.2)

of all cosets. This set will be denoted by $G/N$. We claim that a binary operation can be introduced such that $G/N$ with respect to this operation is a group called the factor group (or quotient group) of $G$ with respect to $N$. Thus we define

$$aN \cdot bN = (aN) (bN) = abN,$$ (6.3)

i.e., we define the operation to be the ordinary product of complexes. The operation is well-defined for if $a' N= aN$ and $b' N = bN$, then, as we know $b' = bn_1$ and $a' = an_2$ where $n_1, n_2 \in N$. Hence

$$a' b' N = an_2bn_1N = an_2bN$$ (6.4)

since $n_1 \in N$. But $b^{-1}n_2b = n_3 \in N$, since $N$ is normal, so the right hand side of (6.4) can be written as

$$an_2bN =(ab)(b^{-1}n_2b)N=abn_3N=abN.$$ (6.5)

Thus (6.4) and (6.5) show that if $N \lhd G$ then $a' b' N = abN$, i.e., the operation of coset multiplication defined in (6.3) is, indeed, well-defined.

The associative law is, of course, true because coset multiplication as defined in (6.3) uses the ordinary group operation which is by definition associative.

We claim $N$ serves as the identity element of $G/N$. Indeed,

$$aN \cdot N = aN^2 = aN,$$

and

$$N \cdot aN = aN^2 = aN.$$

The inverse of $aN$ is $a^{-1}N$ since

$$aN a^{-1}N = aa^{-1}N2 = N.$$

Similarly $a^{-1}N aN = N$.

To emphasize: the elements of $G/N$ are complexes (subsets) of $G$. If $\vert G\vert < \infty$, then $\vert G/N\vert=[G:N]$, i.e., the member of cosets of $N$ in $G$. It is also to be emphasized that in order for $G/N$ to be a group $N$ must be a normal subgroup of $G$. Again, if $G$ is finite from Lagrange's Theorem $[G : N] = G / N$ , (see equation 4.9) thus

$$G/N = G / N$$ (6.6)

As some of our examples will show, it is possible to have infinite $G$, infinite $N$, but finite $G/N$.

We now consider some examples.

Example 6.2.1   Let $6\mathbb {Z}= \{..., 12, 6, 0, 6, 12, ...\}$, i.e., $6\mathbb {Z}$ is the subgroup $\langle 6\rangle$ of the group $\mathbb {Z}$ of integers under addition. Since $\mathbb {Z}$ is abelian, $6\mathbb {Z}\lhd \mathbb {Z}$. To construct $\mathbb {Z}/6\mathbb {Z}$, we first find all the (left) cosets of $6\mathbb {Z}$ in $\mathbb {Z}$. Consider the following 6 cosets:

$$\begin{array}{c} 0 + 6\mathbb {Z}= \{ ...,-12,-6,0,6,12 ,...... ... \\ 5 + 6\mathbb {Z}= \{ ..., -7,-1,5,11,17,...\}. \end{array}$$

From the above, it is evident that $\mathbb {Z}= 6\mathbb {Z}\cup (1 + 6\mathbb {Z}) \cup ... \cup (5 + 6\mathbb {Z})$ (disjoint) which shows that these are all the cosets of $6\mathbb {Z}$ in $\mathbb {Z}$. (This is also clear from the Division Algorithm, for if $n \in\mathbb {Z}$, then $n = 6q + r$ where $0 \leq r < 6$. Thus $n + 6\mathbb {Z}= 6q + r + 6\mathbb {Z}= r + 6\mathbb {Z}$. We also note that this shows that if $[n]$ is the equivalence class of n under the equivalence relation of congruence modulo 6 (see Section 1.2), then $[n] = [r] = r + 6\mathbb {Z}$.) Now that we know the elements of the factor group, we write its Cayley table

$\mathbb {Z}/6\mathbb {Z}$	$0 + 6\mathbb {Z}$	$1 + 6\mathbb {Z}$	$2 + 6\mathbb {Z}$	$3 + 6\mathbb {Z}$	$4 + 6\mathbb {Z}$	$5 + 6\mathbb {Z}$
$0 + 6\mathbb {Z}$	$0 + 6\mathbb {Z}$	$1 + 6\mathbb {Z}$	$2 + 6\mathbb {Z}$	$3 + 6\mathbb {Z}$	$4 + 6\mathbb {Z}$	$5 + 6\mathbb {Z}$
$1 + 6\mathbb {Z}$	$1 + 6\mathbb {Z}$	$2 + 6\mathbb {Z}$	$3 + 6\mathbb {Z}$	$4 + 6\mathbb {Z}$	$5 + 6\mathbb {Z}$	$0 + 6\mathbb {Z}$
$2 + 6\mathbb {Z}$	$2 + 6\mathbb {Z}$	$3 + 6\mathbb {Z}$	$4 + 6\mathbb {Z}$	$5 + 6\mathbb {Z}$	$0 + 6\mathbb {Z}$	$1 + 6\mathbb {Z}$
$3 + 6\mathbb {Z}$	$3 + 6\mathbb {Z}$	$4 + 6\mathbb {Z}$	$5 + 6\mathbb {Z}$	$0 + 6\mathbb {Z}$	$1 + 6\mathbb {Z}$	$2 + 6\mathbb {Z}$
$4 + 6\mathbb {Z}$	$4 + 6\mathbb {Z}$	$5 + 6\mathbb {Z}$	$0 + 6\mathbb {Z}$	$1 + 6\mathbb {Z}$	$2 + 6\mathbb {Z}$	$3 + 6\mathbb {Z}$
$5 + 6\mathbb {Z}$	$5 + 6\mathbb {Z}$	$0 + 6\mathbb {Z}$	$1 + 6\mathbb {Z}$	$2 + 6\mathbb {Z}$	$3 + 6\mathbb {Z}$	$4 + 6\mathbb {Z}$

Here we note that since the group operation is +, (6.3) becomes $(a + 6\mathbb {Z}) + (b + 6\mathbb {Z}) = (a + b) + 6\mathbb {Z}$. It is easy to see that $\vert\mathbb {Z}/6\mathbb {Z}\cong \mathbb {Z}_ 6$. (See problem 4 for Section 5.2. As a matter of fact $\mathbb {Z}/6\mathbb {Z}= \mathbb {Z}_6$ from the above parenthesized remark.) More generally, if $n \in \mathbb {N}$ and we let $n\mathbb {Z}= \langle n\rangle = \{0, \pm n, \pm 2n, ...\}$, then $\mathbb {Z}/n\mathbb {Z}= \mathbb {Z}_n$. $\Box$

Example 6.2.2   Consider $G = S_3$ and $N = \langle (123)\rangle$. As already remarked $\langle (1,2,3)\rangle = A_3 =\{(1), (1,2,3), (1,3,2)\} \lhd S_3$, and so $\vert G/N\vert = 6/3 = 2$. The elements of $G/N$ are $N = \langle (1,2,3)\rangle = \{(1), (1,2,3), (1,3,2)\}$ and $(1,2)N = \{(1,2),(2,3), (1,3)\}$. The group $G/N = \{N, (1,2)N\}$ is a group of order 2 where the element (coset) $N$ is the identity and $(1,2)N\cdot (12)N = (12)^2 N = N$. $\Box$

Example 6.2.3   Let $G = GL(n,\mathbb {R})$ and $N = SL(n,\mathbb {R})$ (see Example 2.2.5). If $A \in G$, $B \in N$, $\det (ABA^{-1}) = \det A \det B (\det A)^{-1} = 1$. This implies $N \lhd G$. Then $G/N = \{AN \ \vert\ A \in G\}$. We claim that $X \in AN$ if and only if $X = AB$, where $B \in N$ is such that the element $AN$ of $G/N$ consists of all $n\times n$ matrices in $G$ with the same determinant as $A$. Indeed, if $X = AB$ then $\det (X) = \det (A) \det (B) = det (A)$. It still remains to show that if $C \in GL(n,\mathbb {R})$ and $\det (C)= \det (A)$, then $C \in AN$, i.e., the other inclusion. (This is left as exercise 1 for this section). This proves the claim. Thus $G$ is the disjoint union $G = \coprod_A AN$ (disjoint), where the union is taken over matrices $A$ with different determinants. If we choose for each nonzero real number $\alpha$ ( $\alpha \in \mathbb {R}-\{0\}$) an $A_\alpha\in G$ such that $\det (A_\alpha) = \alpha$ and let $\Lambda =\mathbb {R}-\{0\}$, then $G = \coprod_{\alpha \in \Lambda} A_\alpha N$. Moreover, if $\alpha\in\Lambda$ and $\beta \in \Lambda$ are distinct then in $G/N$, $(A_\alpha N) (A_\beta N) = A_{\alpha\beta}N$. As a matter of fact, if we just think of $A_\alpha$ and $A_\beta$ as representatives of their respective cosets (equivalence class representatives), then we can suppress the $N$, and think of this multiplication in $G/N$ as being given by $A_\alpha A_\beta = A_{\alpha\beta}$. $\Box$

When we create the factor group $G/N$, it is important to understand that we are really defining every element of $N$ to be the identity. This is apparent from the previous example where we just suppressed the $N$. In Example 6.2.1, we are saying that any multiple of $6$ is $0$ in the factor group $\mathbb {Z}/6\mathbb {Z}$. That is why $8 + 6\mathbb {Z}= 2 + 6 + 6\mathbb {Z}= 2 + 6\mathbb {Z}$, etc. In Example 6.2.2, we have $(1,2)N = (2,3)N$, since $(2,3) = (1,2) (1,2,3)$ in $S_3$ and going to the factor group makes $(1,2,3)$ the identity. Group theorists often refer to the process of creating the factor group $G/N$ as ``killing'' $N$.

Example 6.2.4   Let $G = S_4$ and $N = V_4$. We first note that for the same reason as in Example 6.1.3, $V_4 \lhd S_4$ (i.e., $gV_4g^{- 1} = V_4$ for all $g \in S_4$, Why?). To construct $S_4/V_4$, we first find all the (left) cosets of $V_4$ in $S_4$. Consider the following $6 = \vert S_4/V_4\vert = 4!/4$ cosets:

$$\begin{array}{c} V_4= \\ (1,2,3)V_4=\{(1,2,3),(1,3,4),(2,4,... ...\\ (2,3)V_4=\{(1,4),(2,4),(1,3,4,2),(1,2,4,3) \}. \end{array}$$

We can therefore write the Cayley table for $S_4/V_4$.

$S_4/V_4$	$V_4$	$(1,2,3)V_4$	$(1,3,2)V_4$	$(1,2)V_4$	$(1,3)V_4$	$(2,3)V_4$
$V_4$	$V_4$	$(1,2,3)V_4$	$(1,3,2)V_4$	$(1,2)V_4$	$(1,3)V_4$	$(2,3)V_4$
$(1,2,3)V_4$	$(1,2,3)V_4$	$(1,3,2)V_4$	$V_4$	$(1,3)V_4$	$(2,3)V_4$	$(1,2)V_4$
$(1,3,2)V_4$	$(1,3,2)V_4$	$V_4$	$(1,2,3)V_4$	$(2,3)V_4$	$(1,2)V_4$	$(1,3)V_4$
$(1,2)V_4$	$(1,2)V_4$	$(2,3)V_4$	$(1,3)V_4$	$V_4$	$(1,3,2)V_4$	$(1,2,3)V_4$
$(1,3)V_4$	$(1,3)V_4$	$(1,2)V_4$	$(2,3)V_4$	$(1,2,3)V_4$	$V_4$	$(1,3,2)V_4$
$(2,3)V_4$	$(2,3)V_4$	$(1,3)V_4$	$(1,2)V_4$	$(1,3,2)V_4$	$(1,2,3)V_4$	$V_4$

The reader should note that this table gives a non-abelian group of order 6. As a matter of fact, $S_4/V_4 \cong S_3$, which can be seen immediately from the above if we think of killing off $V_4$. $\Box$