Chapter 05

Hard

Normal Subgroups & Quotient Groups

00 · Symbol Glossary

$N \trianglelefteq G$N is a normal subgroup of G

Read " $N$ is normal in $G$ ." A special kind of subgroup — one where left and right cosets always coincide, making it possible to build a genuine quotient group.

$gNg^{-1}$Conjugate of N by g

The set $\{gng^{-1} : n\in N\}$ — take every element of $N$ , "sandwich" it between $g$ and $g^{-1}$ . Normality is exactly the statement that this sandwiching never moves $N$ at all.

$G/N$Quotient group

The set of cosets of $N$ in $G$ , equipped with the coset-multiplication operation $(aN)(bN)=abN$ — a bona fide group once $N$ is normal. Read " $G$ mod $N$ ."

$Z(G)$Center of G

$Z(G) = \{z\in G : zg=gz \text{ for all } g\in G\}$ — the elements that commute with everything. Always a normal subgroup, and a first measure of "how abelian" $G$ is: $Z(G)=G$ exactly when $G$ itself is abelian.

$gxg^{-1}$Conjugate of an element x by g

A single element's version of the sandwiching operation above. $x$ and $gxg^{-1}$ are called conjugate; this relation reappears as the engine behind Chapter 08's group actions and Chapter 09's Sylow theorems.

01 · The Problem With Coset Multiplication

Chapter 04 built $G/H$ , the set of cosets. The natural next question: can $G/H$ be given its own group structure, using the most obvious possible rule, $(aH)(bH) := abH$ ?

Common mistake — Assuming coset multiplication is automatically well-defined

Let $G=S_3$ and $H=\{e,(1\,2)\}$ (order 2, not normal — verified below). Take two ways of naming the same coset: $(1\,2)H = H$ and also, checking membership, $(1\,2) \cdot e = (1\,2) \in H$ , so indeed $(1\,2)H=H$ .

Now compute $(1\,3)H \cdot (1\,2)H$ two ways, using different representatives of the same cosets, and see if the rule " $aH\cdot bH := abH$ " gives a consistent answer. Using $a=(1\,3)$ , $b=(1\,2)$ : product coset is $(1\,3)(1\,2)H$ . But $(1\,3)H$ also equals $(1\,3)(1\,2)H$ if $(1\,2)\in H$ — pick a different representative of $b$ 's coset, say $b'=e$ (also representing $H$ since $eH=H$ ): product coset is $(1\,3)e\, H = (1\,3)H$ . Working out $(1\,3)(1\,2)$ in cycle notation gives a different coset than $(1\,3)$ alone represents. The "obvious" rule gives two different answers depending on which representative was used — it is not well-defined for this $H$ .

This is exactly the well-definedness question from Theorem 1.5 ( $\mathbb{Z}_n$ 's operations not depending on the chosen representative), reappearing in a harder setting: what property of $H$ guarantees the analogous rule works for cosets of a general group?

02 · Normal Subgroups

Definition — Normal Subgroup

$N\le G$ is normal, written $N\trianglelefteq G$ , if $gNg^{-1}=N$ for every $g\in G$ , where $gNg^{-1}=\{gng^{-1}:n\in N\}$ .

Theorem 5.1 — Equivalent Characterizations of Normality

For $N\le G$ , the following are equivalent: (a) $gNg^{-1}=N$ for all $g\in G$ ; (b) $gN=Ng$ for all $g\in G$ ; (c) every left coset of $N$ equals the corresponding right coset.

Proof

(a) $\Rightarrow$ (b): Multiply both sides of $gNg^{-1}=N$ on the right by $g$ : $gN = Ng$ .

(b) $\Rightarrow$ (a): Multiply both sides of $gN=Ng$ on the right by $g^{-1}$ : $gNg^{-1}=N$ .

(b) $\iff$ (c): identical statements, just naming " $gN$ " and " $Ng$ " as "the left coset by $g$ " and "the right coset by $g$ ."

A weaker-looking but equivalent condition

Definition 5.1's condition (a) can be checked one element at a time: $N$ is normal iff for every $g\in G$ and every $n\in N$ , the single conjugate $gng^{-1}$ lies back in $N$ (this gives $gNg^{-1}\subseteq N$ ; the reverse inclusion follows by applying the same fact to $g^{-1}$ ). This element-by-element version is usually the easiest to check in practice, and is the version used in most of the examples below.

03 · Recognizing Normal Subgroups

Example — Every subgroup of an abelian group is normal

If $G$ is abelian, $gng^{-1} = gg^{-1}n = n$ for every $n$ — conjugation does nothing at all. So $gNg^{-1}=N$ trivially, for any subgroup $N$ . In an abelian group, "subgroup" and "normal subgroup" are the same thing — this is precisely why normality never came up in Chapters 01–04, which worked mostly with $\mathbb{Z}_n$ and $\mathbb{Z}$ .

Example — $\{e\}$ and $G$ are always normal

$g\{e\}g^{-1} = \{geg^{-1}\} = \{e\}$ for any $g$ . And $gGg^{-1}=G$ since conjugation by $g$ is a bijection $G\to G$ (it has inverse "conjugate by $g^{-1}$ ") that happens to map $G$ onto itself. Both extremes of the subgroup lattice are always normal, in every group.

Theorem 5.2 — Index-2 Subgroups Are Always Normal

If $H\le G$ and $[G:H]=2$ , then $H\trianglelefteq G$ .

Proof

There are exactly two left cosets: $H$ itself and one other, which (since the cosets partition $G$ ) must be $G\setminus H$ (everything not in $H$ ). The same is true for right cosets: $H$ and $G\setminus H$ . So for $g\in H$ : $gH=H=Hg$ . For $g\notin H$ : $gH$ is the other left coset, necessarily $G\setminus H$ ; likewise $Hg=G\setminus H$ . Either way $gH=Hg$ , so by Theorem 5.1(b), $H\trianglelefteq G$ .

Example — The rotation subgroup of $D_n$ is normal

$R=\{e,r,\ldots,r^{n-1}\}\le D_n$ has $|R|=n$ and $|D_n|=2n$ , so $[D_n:R]=2$ . By Theorem 5.2, $R\trianglelefteq D_n$ — no case-by-case conjugation check needed.

Common mistake — A reflection subgroup of $D_3$ is not normal

Let $H=\{e,s\}\le D_3$ (a single reflection). Conjugate $s$ by the rotation $r$ : geometrically, "rotate, reflect across the original axis, rotate back" reflects across a different axis than $s$ did — so $rsr^{-1}$ is a different reflection, not in $H=\{e,s\}$ unless $n\le2$ . Algebraically, using $srs=r^{-1}$ (Chapter 02, Section 05) rearranged as $rs=sr^{-1}$ : $rsr^{-1} = sr^{-1}r^{-1}=sr^{-2}$ , which for $n=3$ is $sr$ , a different reflection, not equal to $s$ . $rHr^{-1}\neq H$ : not normal. This is exactly why $[D_n:H]=n$ here, not $2$ — Theorem 5.2 simply doesn't apply to this subgroup.

Fact — The Center Is Always Normal

$Z(G) = \{z\in G : zg=gz\ \forall g\in G\}$ satisfies $Z(G)\trianglelefteq G$ .

Proof

First, $Z(G)\le G$ : nonempty ( $e\in Z(G)$ , since $e$ commutes with everything); closed (if $z_1,z_2\in Z(G)$ , then $(z_1z_2)g = z_1(z_2g)=z_1(gz_2)=(z_1g)z_2=(gz_1)z_2=g(z_1z_2)$ for every $g$ , so $z_1z_2\in Z(G)$ ); closed under inverses (if $zg=gz$ , multiply both sides on the left and right by $z^{-1}$ to get $gz^{-1}=z^{-1}g$ , so $z^{-1}\in Z(G)$ ). By Theorem 3.1, $Z(G)\le G$ .

Normal: for any $g\in G$ and $z\in Z(G)$ , $gzg^{-1} = (gz)g^{-1} = (zg)g^{-1} = z(gg^{-1}) = z \in Z(G)$ , using $gz=zg$ since $z\in Z(G)$ . So $gZ(G)g^{-1}\subseteq Z(G)$ for every $g$ ; applying this with $g^{-1}$ in place of $g$ gives the reverse inclusion, so $gZ(G)g^{-1}=Z(G)$ .

04 · The Quotient Group

Normality is exactly the condition that repairs the well-definedness failure from Section 01.

Theorem 5.3 — G/N Is a Group When N Is Normal

If $N\trianglelefteq G$ , then coset multiplication $(aN)(bN) := abN$ is well-defined (independent of the choice of representatives $a,b$ ), and $G/N$ is a group under this operation, with $|G/N| = [G:N]$ (and $=|G|/|N|$ when $G$ is finite).

Proof

Well-defined. Suppose $aN=a'N$ and $bN=b'N$ ; we must show $abN=a'b'N$ . By Theorem 4.1, $a^{-1}a'\in N$ and $b^{-1}b'\in N$ , so $a'=an_1$ and $b'=bn_2$ for some $n_1,n_2\in N$ . Then:

a'b' = an_1bn_2 = ab(b^{-1}n_1b)n_2

Since $N$ is normal, $b^{-1}n_1b \in N$ (this is exactly condition (a) of Theorem 5.1, applied with $g=b^{-1}$ ), so $(b^{-1}n_1b)n_2 \in N$ by closure. Writing $n_3 := (b^{-1}n_1b)n_2 \in N$ , we get $a'b' = ab\,n_3$ , so $a'b' \in abN$ , and by Theorem 4.1, $a'b'N = abN$ . The product coset does not depend on the representatives chosen.

Group axioms. Closure: $(aN)(bN)=abN$ , and $abN$ is again a coset (a left coset by the element $ab$ ), so it's an element of $G/N$ . Associativity: $\big((aN)(bN)\big)(cN) = (ab)N\cdot cN = (ab)cN = a(bc)N = aN\cdot(bc)N = (aN)\big((bN)(cN)\big)$ , inherited directly from associativity in $G$ . Identity: $eN=N$ satisfies $(aN)(N) = aN = (N)(aN)$ . Inverses: $(aN)(a^{-1}N) = aa^{-1}N = eN = N$ , so $(aN)^{-1}=a^{-1}N$ .

Order. $G/N$ is, as a set, exactly the set of cosets whose count is $[G:N]$ by definition; this doesn't change by giving it a group operation.

Compare this proof to Theorem 1.5, side by side

The heart of this proof — writing $a'=an_1$ , $b'=bn_2$ , expanding the product, and showing the "extra" terms land back inside $N$ — is structurally identical to the well-definedness proof for $+$ and $\times$ on $\mathbb{Z}_n$ in Chapter 01. The one new ingredient, $b^{-1}n_1b\in N$ , is precisely where normality gets used; without it, the extra term $n_1$ would sit in the wrong place in the product and there would be no way to absorb it back into $N$ .

05 · Worked Quotient Groups

Example — $\mathbb{Z}/n\mathbb{Z}$, revisited

$G=(\mathbb{Z},+)$ , $N=n\mathbb{Z}$ . Since $G$ is abelian, $N\trianglelefteq G$ automatically. The quotient group $\mathbb{Z}/n\mathbb{Z}$ has elements $0+n\mathbb{Z}, 1+n\mathbb{Z}, \ldots, (n-1)+n\mathbb{Z}$ , with operation $(a+n\mathbb{Z})+(b+n\mathbb{Z}) = (a+b)+n\mathbb{Z}$ — exactly $\mathbb{Z}_n$ from Chapter 01, now revealed as a genuine quotient group rather than an ad hoc construction. This is the reason the notation " $\mathbb{Z}_n$ " and " $\mathbb{Z}/n\mathbb{Z}$ " are used interchangeably throughout mathematics.

Step-by-step — Building $D_4/R$ where $R$ is the rotation subgroup

Confirm normality:

[D_4:R]=8/4=2

, so by Theorem 5.2,

R\trianglelefteq D_4

List the cosets: exactly

[D_4:R]=2

of them:

R

itself, and

sR

(any reflection times

R

) — the set of all four reflections.

Multiply within the quotient:

(R)(R)=R

(identity element of the quotient).

(sR)(sR) = s^2R = eR = R

(since

s^2=e

(R)(sR)=sR=(sR)(R)

Recognize the structure: two elements,

R

acting as identity, and

sR

squaring to the identity — this is exactly the Cayley table of

\mathbb{Z}_2

Conclusion:

D_4/R

has order

2

and behaves exactly like

\mathbb{Z}_2

— the "shape" of the two-element group is unavoidable, since Corollary 4.6 already forces every group of prime order 2 to be cyclic.

Common mistake — Writing G/H when H isn't normal

$D_3/\{e,s\}$ is not valid notation — $\{e,s\}$ is not normal in $D_3$ (Section 03's FailBlock), so coset multiplication is not well-defined on this set of cosets, and " $D_3/\{e,s\}$ " does not denote a group. The set of cosets still exists (Chapter 04 built it for any subgroup), but only normal subgroups earn the right to the quotient-group notation and structure.

06 · Exercises

EXERCISE 5.1

Use the element-by-element version of normality: for $g\in D_4$ and $n\in\{e,r^2\}$ , is $gng^{-1}$ always back in $\{e,r^2\}$ ? Try $g=s$ (a reflection) and $n=r^2$ .

$N=\{e,r^2\}$ . For $n=e$ : $geg^{-1}=e\in N$ always. For $n=r^2$ : using the relation $srs=r^{-1}$ (equivalently $sr^2s = (srs)(srs)\cdot$ suitably, or directly $sr^2s^{-1} = (srs^{-1})^2 = (r^{-1})^2=r^{-2}=r^2$ since $r^4=e$ ), we get $sr^2s^{-1}=r^2 \in N$ . Since conjugating by a reflection also lands back in $N$ , and conjugating by any rotation power keeps $r^2$ fixed (rotations commute with each other), every conjugate of $r^2$ stays in $\{e,r^2\}$ .

$N=\{e,r^2\}\trianglelefteq D_4$ .

Show that $N=\{e,r^2\}$ is a normal subgroup of $D_4$ (where $r$ has order 4), by checking that conjugating $r^2$ by both a rotation and a reflection keeps it inside $N$ .

EXERCISE 5.2

Compute $|G/N|$ using Theorem 5.3, then recall Corollary 4.6 about groups of that specific order.

$|G/N| = |G|/|N| = 8/2 = 4$ .

Corollary 4.6 only forces cyclicity for prime order — $4$ is not prime, so $G/N$ need not be cyclic. (It could be $\mathbb{Z}_4$ or the noncyclic group $\mathbb{Z}_2\times\mathbb{Z}_2$ , a distinction that Chapter 07's direct products will make precise.)

If $|G|=8$ and $N\trianglelefteq G$ with $|N|=2$ , what is $|G/N|$ ? Does Corollary 4.6 guarantee $G/N$ is cyclic? Why or why not?

EXERCISE 5.3

Apply Theorem 5.2 directly — you only need to compute one number.

$A_n \le S_n$ with $|S_n|=n!$ and $|A_n|=n!/2$ , so $[S_n:A_n] = n!/(n!/2) = 2$ .

By Theorem 5.2, an index-2 subgroup is automatically normal: $A_n \trianglelefteq S_n$ , for every $n\geq2$ , with no case-by-case conjugation check required.

Using Theorem 5.2, explain why $A_n \trianglelefteq S_n$ for every $n\geq2$ , without checking any conjugations directly.

EXERCISE 5.4

Recall that $Z(G)=G$ exactly when every pair of elements commutes.

$Z(G) = \{z : zg=gz\ \forall g\in G\}$ . If $G$ is abelian, every element commutes with every other, so every $z\in G$ satisfies the defining condition: $Z(G)=G$ .

Conversely, if $Z(G)=G$ , then every element of $G$ commutes with every other element of $G$ (by definition of $Z(G)$ ) — which is exactly the statement that $G$ is abelian.

$Z(G)=G \iff G$ is abelian.

Prove that $Z(G)=G$ if and only if $G$ is abelian.

07 · Chapter Summary

Concept	Statement
Normal subgroup	$N\trianglelefteq G$ : $gNg^{-1}=N$ for all $g\in G$
Equivalent conditions	$gNg^{-1}=N \iff gN=Ng \iff$ left cosets = right cosets (Thm 5.1)
Abelian groups	Every subgroup is automatically normal
Index-2 shortcut	$[G:H]=2 \implies H\trianglelefteq G$ (Thm 5.2)
Center $Z(G)$	Always normal; $Z(G)=G \iff G$ abelian
Quotient group $G/N$	$(aN)(bN)=abN$ well-defined $\iff N\trianglelefteq G$ (Thm 5.3)
$	G/N
$\mathbb{Z}/n\mathbb{Z}$	The familiar $\mathbb{Z}_n$ , now seen as a genuine quotient group

Next: Chapter 06 — Group Homomorphisms & Isomorphism Theorems formalizes what it means for two groups to be "structurally the same," and shows that every quotient group $G/N$ arises naturally as the image of a structure-preserving map out of $G$ .