Normal Subgroups and Quotient Groups: The Condition for Multiplying Cosets

0. Key points

• If we try to equip the set $G/H$ of left cosets of a subgroup $H \le G$ with the law “multiply the representatives”, the answer may change when the representatives are changed. The pair $G = S_3$, $H = \{e, (1\,2)\}$ is an actual example.
• A necessary and sufficient condition for this law to be defined is that $gH = Hg$ for every $g \in G$, that is, that $H$ be a normal subgroup (Theorem 3.2). Normality is not some odd requirement that left and right cosets coincide; it is a restatement of the very wish to divide.
• For a normal subgroup $N$, the set $G/N$ becomes a group (the quotient group), and $\pi : G \to G/N,\ g \mapsto gN$ is a surjective homomorphism with kernel $N$. Conversely the kernel of a homomorphism is always a normal subgroup, so “normal subgroup” and “kernel of a homomorphism” are two faces of one concept.
• The subgroups of $\mathbb{Z}$ are exactly the $n\mathbb{Z}$, and the quotient group $\mathbb{Z}/n\mathbb{Z}$ is cyclic of order $n$. The order of $\bar{a}$ is $n/\gcd(a,n)$, and $\bar{a}$ is a generator precisely when $\gcd(a,n) = 1$.
• Passing to a quotient is an operation that discards information. The group $G/N$ need not be a subgroup of $G$, and all distinctions inside $N$ are forgotten completely.

1. Motivation: giving the set of cosets a law of composition

The hands of a clock come full circle at 12. Five hours after 9 o’clock is not 14 o’clock but 2 o’clock. What we are doing here is identifying integers under the relation “same remainder upon division by 12” and then adding in the identified world. That this computation is consistent is something we normally never question. Pause for a moment, however, and it is far from obvious. Behind it lies the fact that we may replace $9$ by $21$ and $5$ by $17$ without changing the answer, that is,

$$9 + 5 = 14 \equiv 2, \qquad 21 + 17 = 38 \equiv 2 \pmod{12}$$

so that the answer is determined independently of the choice of representatives. Only because of this does clock arithmetic mean anything. Gauss opened his Disquisitiones Arithmeticae (1801) by introducing the congruence symbol $a \equiv b \pmod{m}$ and setting out that congruences may be “computed with directly” as far as addition, subtraction and multiplication are concerned. This is the starting point of the idea of a quotient in modern algebra.

Group theory treats this situation in far greater generality. Given a group $G$ and a subgroup $H$, we saw in Subgroups and cosets (Lagrange’s theorem) that $G$ is partitioned into the left cosets $aH$ (Lemma 5.2［Subgroups and Cosets］). We write $G/H$ for the set of all cosets. This is the situation corresponding to $\mathbb{Z}$ and $12\mathbb{Z}$ in the case of the integers. It is then tempting to ask the naive question:

Can $G/H$ be given a law of composition by multiplying representatives? That is, may we set $(aH) \cdot (bH) := (ab)H$?

If this were possible, $G/H$ would itself be a group, and we would obtain a new group that views $G$ “coarsely, up to $H$”. The basic strategy of group theory — break a large group into smaller ones and study those — begins here.

The definition, however, fails in general. It can happen that $aH = a'H$ and $bH = b'H$ while $(ab)H \ne (a'b')H$ (Example 3.1). Isolating the condition under which it does not fail yields exactly the notion that Galois was the first to single out (as a subgroup whose left and right coset decompositions agree), that later came to be called an “invariant subgroup”, and that we call today a normal subgroup. The aim of this article is to derive that condition, to construct the quotient group, and to verify the most basic example $\mathbb{Z}/n\mathbb{Z}$ down to the last detail.

The idea of dividing by an equivalence relation to build a new object is of course not confined to group theory. The general theory at the level of sets is collected in Relations and equivalence relations: what does “the same” mean?. It clarifies matters to read this article as asking: when does a law of composition descend to the quotient?

2. Preliminaries: cosets and homomorphisms

Let us fix notation. Throughout, $G$ is a group with identity element $e$, and $H \le G$ means that $H$ is a subgroup of $G$.

Definition 2.1（Left coset, right coset, index）

For $H \le G$ and $a \in G$, the sets

$$aH := \{ah : h \in H\}, \qquad Ha := \{ha : h \in H\}$$

are called respectively the left coset and the right coset of $H$ containing $a$. The set of all left cosets is written $G/H$, the set of all right cosets $H \backslash G$. The number of elements of $G/H$ (a cardinal, not necessarily finite) is called the index of $H$ in $G$ and is written $[G : H]$.

We collect the basic properties of left cosets, since we shall use them repeatedly. The proof merely reconfirms the content of the previous chapter, but everything that follows rests on this proposition, so we write it out in full.

Proposition 2.2（Basic properties of left cosets）

Let $H \le G$ and $a, b \in G$.

1. $h \in H$ if and only if $hH = H$.
2. $aH = bH$ if and only if $a^{-1}b \in H$.
3. If $aH \cap bH \ne \varnothing$ then $aH = bH$. In particular the left cosets form a partition of $G$.
4. The map $H \to aH,\ h \mapsto ah$ is a bijection. In particular $|aH| = |H|$.

Proof（Proposition 2.2）

(1) Suppose $hH = H$. Since $H$ contains the identity $e$, we get $h = he \in hH = H$, so $h \in H$. Conversely let $h \in H$. As $H$ is closed under products, $hH \subseteq H$. For the reverse inclusion take $x \in H$. Since $H$ is closed under inverses we have $h^{-1} \in H$, hence $h^{-1}x \in H$, and $x = h(h^{-1}x) \in hH$. Thus $H \subseteq hH$, and $hH = H$ follows.

(2) Suppose $aH = bH$. Then $b = be \in bH = aH$, so there is $h \in H$ with $b = ah$. Multiplying on the left by $a^{-1}$ gives $a^{-1}b = h \in H$. Conversely suppose $a^{-1}b = h \in H$. Then $b = ah$ and

$$bH = (ah)H = a(hH) = aH$$

where the second equality is associativity of the product and the third uses (1).

(3) Take $x \in aH \cap bH$, so that $x = ah_1 = bh_2$ for some $h_1, h_2 \in H$. Then $a^{-1}b = h_1h_2^{-1} \in H$ (using that $H$ is a subgroup), so $aH = bH$ by (2). Moreover $e \in H$ gives $a = ae \in aH$, so every element lies in at least one left coset. Hence the left cosets partition $G$ into pairwise disjoint subsets.

(4) Surjectivity is the definition of $aH$. Injectivity follows because $ah = ah'$ gives $h = h'$ after multiplying on the left by $a^{-1}$ (cancellation in a group).

∎

Remark 2.3

Parts (2) and (3) of Proposition 2.2 say that the binary relation on $G$ defined by $a \sim b \overset{\text{def}}{\iff} a^{-1}b \in H$ is an equivalence relation（Definition 3.1）［関係と同値関係］ whose equivalence classes are exactly the left cosets. For $\mathbb{Z}$ with $H = n\mathbb{Z}$, the relation $a \sim b$ reads $b - a \in n\mathbb{Z}$, that is, $a \equiv b \pmod{n}$. Congruences are a special case of left cosets.

Discussing quotient groups requires the language of homomorphisms, so we set it up here. The subject is treated properly in the next chapter, The isomorphism theorems for groups.

Definition 2.4（Homomorphism and kernel）

Let $G, G'$ be groups and let $e'$ denote the identity of $G'$. A map $f : G \to G'$ satisfying

$$f(xy) = f(x)f(y) \qquad (\forall x, y \in G)$$

is called a group homomorphism. In that case

$$\ker f := \{x \in G : f(x) = e'\}, \qquad \operatorname{im} f := f(G)$$

are called respectively the kernel and the image of $f$. A bijective homomorphism is an isomorphism, and we write $G \cong G'$.

Remark 2.5

A homomorphism $f$ preserves the identity and inverses. Indeed, from $f(e) = f(ee) = f(e)f(e)$ we obtain $e' = f(e)$ upon multiplying both sides by $f(e)^{-1}$. Furthermore $e' = f(e) = f(xx^{-1}) = f(x)f(x^{-1})$ gives $f(x^{-1}) = f(x)^{-1}$. Also $\ker f$ is a subgroup of $G$: from $f(e) = e'$ we get $e \in \ker f$; if $x, y \in \ker f$ then $f(xy) = e'e' = e'$; and if $x \in \ker f$ then $f(x^{-1}) = f(x)^{-1} = (e')^{-1} = e'$.

3. When is the product of representatives defined?

Now to the heart of the matter. Given $H \le G$, we wish to put a law of composition on $G/H$. The rule we want to lay down is

$$(aH) \cdot (bH) := (ab)H$$

The left-hand side takes two sets as input, while the right-hand side produces its answer by way of the representatives $a$ and $b$. For a fixed set $aH$ there are $|H|$ possible choices of representative $a$, so unless we check that the answer does not depend on that choice, the formula defines no map at all. Let us first write down precisely what must be checked.

$$\text{(W)} \qquad aH = a'H \ \text{and}\ bH = b'H \ \Longrightarrow\ (ab)H = (a'b')H \qquad (\forall a, a', b, b' \in G)$$

When condition (W) holds, the rule above defines a map $G/H \times G/H \to G/H$, and we say that the operation is well defined. As the next example shows, (W) does not hold unconditionally.

Example 3.1（Failure of the product of representatives in the symmetric group on three letters）

Let $G = S_3$ (all permutations of $\{1,2,3\}$), with product the composition of maps, $(\sigma\tau)(x) = \sigma(\tau(x))$. Take the subgroup

$$H = \{e,\ (1\,2)\}$$

so that $|H| = 2$ and $[S_3 : H] = 3$.

First we compute the left cosets. For $(1\,3)(1\,2)$ we have $1 \mapsto 2 \mapsto 2$, $2 \mapsto 1 \mapsto 3$, $3 \mapsto 3 \mapsto 1$, giving $(1\,2\,3)$. Likewise $(2\,3)(1\,2)$ sends $1 \mapsto 2 \mapsto 3$, $2 \mapsto 1 \mapsto 1$, $3 \mapsto 3 \mapsto 2$, giving $(1\,3\,2)$. Hence

$$eH = \{e, (1\,2)\},\qquad (1\,3)H = \{(1\,3), (1\,2\,3)\},\qquad (2\,3)H = \{(2\,3), (1\,3\,2)\}.$$

On the other hand $(1\,2)(1\,3)$ sends $1 \mapsto 3 \mapsto 3$, $2 \mapsto 2 \mapsto 1$, $3 \mapsto 1 \mapsto 2$, giving $(1\,3\,2)$, so

$$H(1\,3) = \{(1\,3), (1\,3\,2)\} \ne (1\,3)H$$

and the left and right cosets do not agree.

Now we test (W). Since $(1\,3)H = (1\,2\,3)H$, the elements $a = b = (1\,3)$ and $a' = b' = (1\,2\,3)$ are representatives of the same coset. Yet

$$ab = (1\,3)(1\,3) = e \ \Longrightarrow\ (ab)H = H,$$$$a'b' = (1\,2\,3)(1\,2\,3) = (1\,3\,2) \ \Longrightarrow\ (a'b')H = (2\,3)H$$

(here $(1\,2\,3)(1\,2\,3)$ sends $1 \mapsto 2 \mapsto 3$, $2 \mapsto 3 \mapsto 1$, $3 \mapsto 1 \mapsto 2$, that is, it equals $(1\,3\,2)$), and $H \ne (2\,3)H$. Merely changing the representatives changed the resulting coset. For this $H$ there is no way to put a law of composition on $G/H$ by multiplying representatives.

Looking at the reason for the failure, one sees that the discrepancy between left and right cosets is what matters. Indeed, the following theorem holds. It is the most important statement in this article.

Theorem 3.2（Necessary and sufficient condition for the product of cosets to be defined）

Let $G$ be a group and $H \le G$. The following four conditions are equivalent.

1. Condition (W) holds; that is, the operation $(aH)(bH) := (ab)H$ on $G/H$ is well defined.
2. $aHa^{-1} \subseteq H$ for every $a \in G$, where $aHa^{-1} := \{aha^{-1} : h \in H\}$.
3. $aHa^{-1} = H$ for every $a \in G$.
4. $aH = Ha$ for every $a \in G$.

Proof（Theorem 3.2）

We prove $(1) \Rightarrow (2) \Rightarrow (3) \Rightarrow (4) \Rightarrow (1)$ in that order.

(1) $\Rightarrow$ (2). Take arbitrary $a \in G$ and $h \in H$. Since $h \in H$, part (1) of Proposition 2.2 gives $hH = H = eH$. Also $a^{-1}H = a^{-1}H$ holds trivially. Applying (W) with representatives $h$ and $e$ in the first slot and $a^{-1}$ and $a^{-1}$ in the second, we obtain

$$(h a^{-1})H = (e a^{-1})H = a^{-1}H$$

By part (2) of Proposition 2.2, this is equivalent to $(ha^{-1})^{-1} \cdot a^{-1} \in H$, that is, to

$$a h^{-1} a^{-1} \in H$$

As $h$ ranges over all of $H$ and $h \mapsto h^{-1}$ is a bijection of $H$ onto itself, $h^{-1}$ also ranges over all of $H$. Hence $aha^{-1} \in H$ for every $h \in H$, that is, $aHa^{-1} \subseteq H$.

(2) $\Rightarrow$ (3). Applying (2) to $a$ gives $aHa^{-1} \subseteq H$. Next, applying (2) to $a^{-1}$ gives $a^{-1}Ha \subseteq H$. Conjugating both sides by $a$ on the left and $a^{-1}$ on the right yields

$$H = a(a^{-1}Ha)a^{-1} \subseteq aHa^{-1}$$

(we used that $X \subseteq Y$ implies $aXa^{-1} \subseteq aYa^{-1}$). Combining the two inclusions gives $aHa^{-1} = H$. Note that the universally quantified hypothesis — that (2) holds for every element — is what does the essential work here.

(3) $\Rightarrow$ (4). Multiplying both sides of $aHa^{-1} = H$ on the right by $a$ gives $aH = Ha$. Since this is an equality of sets, let us be more careful: if $x \in aH$ then $x = ah$ and $x = (aha^{-1})a \in Ha$; conversely if $x \in Ha$ then $x = ha$, and applying (3) to $a^{-1}$ gives $a^{-1}ha \in H$, so $x = a(a^{-1}ha) \in aH$.

(4) $\Rightarrow$ (1). Suppose $aH = a'H$ and $bH = b'H$. By part (2) of Proposition 2.2 we may write $a' = ah_1$ and $b' = bh_2$ with $h_1, h_2 \in H$. Then

$$a'b' = a h_1 b h_2 .$$

Now $h_1 b \in Hb$, and applying hypothesis (4) to $b$ gives $Hb = bH$, so there is $h_3 \in H$ with $h_1 b = b h_3$. Therefore

$$a'b' = a (b h_3) h_2 = (ab)(h_3h_2) \in (ab)H$$

(here $h_3h_2 \in H$ because $H$ is closed under products). On the other hand $a'b' \in (a'b')H$, so $(ab)H \cap (a'b')H \ne \varnothing$, and part (3) of Proposition 2.2 gives $(ab)H = (a'b')H$.

∎

Equivalence of the four conditions on normality (proved in the direction of the arrows)

The value of this theorem lies in the fact that condition (2) is easy to check. Condition (1) is a universal statement about four elements, whereas (2) is nothing more than an inclusion between two subsets, expressing that conjugation does not carry elements of $H$ out of $H$. From now on we shall verify normality mainly in the form (2).

4. Normal subgroups

Definition 4.1（Normal subgroup）

A subgroup $N$ of a group $G$ satisfying

$$gNg^{-1} = N$$

for every $g \in G$ is called a normal subgroup of $G$, written $N \trianglelefteq G$. By Theorem 3.2, this is equivalent both to ”$gNg^{-1} \subseteq N$ for every $g \in G$” and to ”$gN = Ng$ for every $g \in G$”.

The element $g x g^{-1}$ is called the conjugate of $x$ by $g$. Setting $\varphi_g(x) := gxg^{-1}$, we have $\varphi_g(xy) = gxyg^{-1} = (gxg^{-1})(gyg^{-1}) = \varphi_g(x)\varphi_g(y)$, so $\varphi_g$ is a homomorphism, and since $\varphi_{g^{-1}}$ is its inverse it is an isomorphism of $G$ onto itself (an inner automorphism). A normal subgroup is therefore a subgroup fixed by every inner automorphism. This is the view behind the older name “invariant subgroup”. Think of conjugation as changing the coordinates from which one looks: in $S_n$ it amounts to relabelling the letters, in $\mathrm{GL}_n$ to changing the basis. Normal subgroups are those unaffected by such relabelling — subgroups intrinsic to the group.

Proposition 4.2（Tests for normality）

Let $G$ be a group.

1. If $G$ is abelian, then every subgroup of $G$ is normal.
2. If $H \le G$ satisfies $[G : H] = 2$, then $H \trianglelefteq G$.
3. The centre $Z(G) := \{z \in G : zx = xz \ (\forall x \in G)\}$ of $G$ is a normal subgroup of $G$.
4. If $f : G \to G'$ is a group homomorphism, then $\ker f \trianglelefteq G$.
5. If $\{N_\lambda\}_{\lambda \in \Lambda}$ is a family of normal subgroups of $G$ (with $\Lambda \ne \varnothing$), then $\bigcap_{\lambda \in \Lambda} N_\lambda \trianglelefteq G$.

Proof（Proposition 4.2）

(1) Let $H \le G$, $g \in G$ and $h \in H$. Since $G$ is abelian, $ghg^{-1} = hgg^{-1} = h \in H$. Hence $gHg^{-1} \subseteq H$ for every $g$, and part (2) of Theorem 3.2 gives $H \trianglelefteq G$.

(2) Take $g \in G$. If $g \in H$, then part (1) of Proposition 2.2 gives $gH = H$, and the same argument for right cosets gives $Hg = H$, so $gH = Hg$. Now suppose $g \notin H$. The left cosets partition $G$ (part (3) of Proposition 2.2) and there are $[G:H] = 2$ of them. One of them is $H = eH$, so the other is $G \setminus H$. Since $g \in gH$ and $g \notin H$ we have $gH \ne H$, whence $gH = G \setminus H$. Exactly the same argument for right cosets shows that the right cosets partition $G$ into two pieces, one being $H$ and the other $G \setminus H$, and since $Hg \ne H$ we get $Hg = G \setminus H$. Therefore $gH = Hg$, and part (4) of Theorem 3.2 gives $H \trianglelefteq G$.

(3) First we check that $Z(G) \le G$. The identity commutes with every element, so $e \in Z(G)$. For $z, w \in Z(G)$ and $x \in G$,

$$(zw)x = z(wx) = z(xw) = (zx)w = (xz)w = x(zw)$$

so $zw \in Z(G)$. If $z \in Z(G)$, multiplying $zx = xz$ by $z^{-1}$ on both sides gives $xz^{-1} = z^{-1}x$, so $z^{-1} \in Z(G)$. As for normality, for $g \in G$ and $z \in Z(G)$ we have $gzg^{-1} = zgg^{-1} = z \in Z(G)$, so $gZ(G)g^{-1} \subseteq Z(G)$, and Theorem 3.2 gives $Z(G) \trianglelefteq G$.

(4) By Remark 2.5 we have $\ker f \le G$. For $g \in G$ and $n \in \ker f$,

$$f(gng^{-1}) = f(g)f(n)f(g)^{-1} = f(g)\,e'\,f(g)^{-1} = e'$$

(the first equality uses the homomorphism property together with $f(g^{-1}) = f(g)^{-1}$ from Remark 2.5). Hence $gng^{-1} \in \ker f$, that is, $g(\ker f)g^{-1} \subseteq \ker f$, and Theorem 3.2 gives $\ker f \trianglelefteq G$.

(5) Put $N := \bigcap_\lambda N_\lambda$. Each $N_\lambda$ contains $e$, so $e \in N$. If $x, y \in N$ then $x, y \in N_\lambda$ for every $\lambda$, hence $xy \in N_\lambda$ and therefore $xy \in N$. Similarly $x^{-1} \in N$. Thus $N \le G$. For $g \in G$ and $x \in N$, we have $x \in N_\lambda$ and $N_\lambda \trianglelefteq G$ for every $\lambda$, so $gxg^{-1} \in N_\lambda$. As this holds for every $\lambda$, we get $gxg^{-1} \in N$. Hence $gNg^{-1} \subseteq N$ and $N \trianglelefteq G$.

∎

Example 4.3（Standard examples of normal subgroups）

(a) The special linear group. Let $K$ be a field and let $\det : \mathrm{GL}_n(K) \to K^{\times}$ be the determinant. Multiplicativity of the determinant, $\det(AB) = \det A \det B$ (see Theorem 6.1［Determinants and Their Properties］), says exactly that $\det$ is a group homomorphism, and its kernel is $\mathrm{SL}_n(K) = \{A : \det A = 1\}$. Part (4) of Proposition 4.2 gives $\mathrm{SL}_n(K) \trianglelefteq \mathrm{GL}_n(K)$.

(b) The alternating group. The sign map $\mathrm{sgn} : S_n \to \{\pm 1\}$ is a homomorphism whose kernel is the alternating group $A_n$. Hence $A_n \trianglelefteq S_n$. For $n \ge 2$ we have $[S_n : A_n] = 2$, so this also follows from part (2) of Proposition 4.2.

(c) Trivial examples. In any group $G$ we have $\{e\} \trianglelefteq G$ and $G \trianglelefteq G$, since $g\{e\}g^{-1} = \{geg^{-1}\} = \{e\}$ and $gGg^{-1} = G$. A nontrivial group with no normal subgroups other than these two is called a simple group.

(d) A non-example. The subgroup $H = \{e, (1\,2)\}$ of Example 3.1 is not normal in $S_3$. Indeed, computing $(1\,3)(1\,2)(1\,3)^{-1}$ and using $(1\,3)^{-1} = (1\,3)$, we get $1 \mapsto 3 \mapsto 3 \mapsto 1$, $2 \mapsto 2 \mapsto 1 \mapsto 3$, $3 \mapsto 1 \mapsto 2 \mapsto 2$, that is, $(2\,3) \notin H$. The index is 3, so part (2) of Proposition 4.2 does not apply, and in fact the subgroup is not normal.

Remark 4.4

Normality is not transitive. From $N \trianglelefteq H$ and $H \trianglelefteq G$ it does not follow that $N \trianglelefteq G$. Let us verify this in the dihedral group of order 8,

$$D_4 = \langle r, s \mid r^4 = s^2 = e,\ srs^{-1} = r^{-1} \rangle .$$

Put $H := \{e, r^2, s, r^2s\}$. From $sr^2s^{-1} = (srs^{-1})^2 = r^{-2} = r^2$ we get $sr^2 = r^2s$, so $H$ is closed under products; every element has order at most 2, so $H \le D_4$ with $|H| = 4$ and $[D_4 : H] = 2$. Part (2) of Proposition 4.2 therefore gives $H \trianglelefteq D_4$. Likewise $N := \{e, s\}$ is a subgroup of index 2 in $H$, so $N \trianglelefteq H$. But $srs^{-1} = r^{-1}$ yields $sr^{-1} = rs$, and

$$rsr^{-1} = r(sr^{-1}) = r(rs) = r^2s \notin N$$

so $N$ is not normal in $D_4$. The fact that a normal subgroup of a normal subgroup need not be normal in the whole group must always be kept in mind when decomposing a group in stages, that is, when working with composition series.

5. Construction of the quotient group G/N

Theorem 5.1（Quotient group）

Let $G$ be a group and $N \trianglelefteq G$. The operation

$$(aN)(bN) := (ab)N \qquad (a, b \in G)$$

on the set $G/N$ of left cosets is well defined, and $G/N$ is a group under it. Its identity element is $eN = N$, and the inverse of $aN$ is $a^{-1}N$. When $G$ is finite, $|G/N| = [G : N] = |G| / |N|$. This group $G/N$ is called the quotient group (or factor group) of $G$ by $N$.

Proof（Theorem 5.1）

Well-definedness. Since $N \trianglelefteq G$, condition (4) of Theorem 3.2 holds, and the implication (4) $\Rightarrow$ (1) of that theorem gives condition (W). Thus $(ab)N$ is determined independently of the choice of representatives.

Associativity. For $a, b, c \in G$, applying the definition of the operation twice on each side gives

$$\bigl((aN)(bN)\bigr)(cN) = \bigl((ab)N\bigr)(cN) = \bigl((ab)c\bigr)N,$$$$(aN)\bigl((bN)(cN)\bigr) = (aN)\bigl((bc)N\bigr) = \bigl(a(bc)\bigr)N .$$

These agree by associativity $(ab)c = a(bc)$ in $G$.

Identity. For every $a \in G$ we have $(eN)(aN) = (ea)N = aN$ and $(aN)(eN) = (ae)N = aN$. Hence $eN = N$ is the identity.

Inverses. $(aN)(a^{-1}N) = (aa^{-1})N = eN = N$ and $(a^{-1}N)(aN) = (a^{-1}a)N = eN = N$. Hence the inverse of $aN$ is $a^{-1}N$.

Therefore $G/N$ is a group. As for the order, $|G/N|$ is by definition the number of left cosets, that is, the index $[G:N]$. If $G$ is finite, part (3) of Proposition 2.2 shows that $G$ is partitioned into $[G:N]$ left cosets, and part (4) shows that each coset has $|N|$ elements, so $|G| = [G:N]\,|N|$, that is, $[G:N] = |G|/|N|$ (Lagrange's theorem（Theorem 6.1）［Subgroups and Cosets］).

∎

Note that each element of the quotient group is a subset of $G$. The coset $aN$ is a bag holding together all elements that differ from $a$ only by something in $N$, and the operation in the quotient group is the multiplication of such bags. Here is a picture.

The natural projection π collapses each coset to a single point. All distinctions inside N are forgotten completely

Proposition 5.2（The natural projection）

Let $N \trianglelefteq G$. The map

$$\pi : G \longrightarrow G/N, \qquad \pi(g) = gN$$

is a surjective homomorphism with $\ker \pi = N$. It is called the natural projection (or canonical surjection).

Proof（Proposition 5.2）

Homomorphism. For $a, b \in G$, the very definition of the operation in the quotient group gives

$$\pi(ab) = (ab)N = (aN)(bN) = \pi(a)\pi(b).$$

Surjectivity. By definition, every element of $G/N$ can be written as $gN$ for some $g \in G$, and this is $\pi(g)$.

Kernel. By Theorem 5.1 the identity of $G/N$ is $N = eN$. Hence

$$g \in \ker\pi \iff gN = eN \iff g^{-1}e \in N \iff g^{-1} \in N \iff g \in N$$

where the second equivalence is part (2) of Proposition 2.2 and the last holds because $N$ is closed under inverses. Therefore $\ker\pi = N$.

∎

Corollary 5.3

For a subgroup $N$ of a group $G$, the following are equivalent.

1. $N \trianglelefteq G$.
2. There exist a group $G'$ and a homomorphism $f : G \to G'$ with $N = \ker f$.

Proof（Corollary 5.3）

$(1) \Rightarrow (2)$: the map $\pi : G \to G/N$ of Proposition 5.2 satisfies $\ker \pi = N$, so we may take $G' = G/N$ and $f = \pi$. $(2) \Rightarrow (1)$: this is precisely part (4) of Proposition 4.2.

∎

This corollary matters conceptually. The definition of a normal subgroup, "$gNg^{-1} = N$", looks at first like a contrived condition, but it in fact means “the part that is crushed by a homomorphism”. Determining all homomorphisms out of $G$ and determining all normal subgroups of $G$ are one and the same task. The refinement of this correspondence is the isomorphism theorem of the next chapter, $G/\ker f \cong \operatorname{im} f$ (Theorem 5.1［群の準同型定理］), treated in The isomorphism theorems for groups.

Caution

$G/N$ is not a subgroup of $G$. Its elements are not elements of $G$ at all, but subsets of $G$. Moreover, $G$ need not even contain a subgroup isomorphic to $G/N$. For instance, take $G = \mathbb{Z}$ and $N = 2\mathbb{Z}$; then $G/N$ is a group of order 2, in which the non-identity element has order 2. But every non-identity element of $\mathbb{Z}$ has infinite order, so $\mathbb{Z}$ has no subgroup of order 2. Forming a quotient does not extract a part; it discards distinctions.

Remark 5.4

If $g \in G$ has finite order and $g^m = e$, then $(gN)^m = g^mN = eN = N$, so the order of $gN$ divides the order of $g$. It may, however, be strictly smaller. With $G = \mathbb{Z}$, $N = 12\mathbb{Z}$ and $g = 1$, the element $g$ has infinite order while $gN$ has order 12. Passing to a quotient can only lower orders, never raise them.

6. Example: the congruence group Z/nZ in full detail

Let us examine the most basic quotient group down to the last detail. The stage is the additive group $(\mathbb{Z}, +)$, so we write everything additively: cosets are $a + N$ and the operation is $(a+N) + (b+N) = (a+b)+N$.

Lemma 6.1

If $H$ is a subgroup of the additive group $\mathbb{Z}$, then there is exactly one integer $n \ge 0$ with $H = n\mathbb{Z} = \{nk : k \in \mathbb{Z}\}$.

Proof（Lemma 6.1）

Existence. If $H = \{0\}$, take $n = 0$, so that $H = 0\mathbb{Z}$. Assume from now on that $H \ne \{0\}$. Then there is $m \in H$ with $m \ne 0$, and since $H$ is closed under inverses (here $-m$), at least one of $m$ and $-m$ is positive. Hence $H$ contains positive integers. By the well-ordering of the natural numbers there is a least positive integer contained in $H$; call it $n$.

We show $n\mathbb{Z} \subseteq H$. We have $n \in H$, and $H$ is closed under addition and inverses, so by induction $nk \in H$ for every $k \ge 0$, and also $-(nk) = n(-k) \in H$ (formally one applies the principle of mathematical induction（Theorem 3.2）［Techniques of Proof］ directly). Hence $n\mathbb{Z} \subseteq H$.

We show the reverse inclusion. Take any $a \in H$ and write $a = qn + r$ with $q, r \in \mathbb{Z}$ and $0 \le r < n$ by the division algorithm. Since $qn \in n\mathbb{Z} \subseteq H$ and $a \in H$, we get $r = a - qn \in H$. If $r > 0$, then $r$ is a positive integer in $H$ with $r < n$, contradicting the minimality of $n$. Hence $r = 0$, that is, $a = qn \in n\mathbb{Z}$. Therefore $H = n\mathbb{Z}$.

Uniqueness. Suppose $n\mathbb{Z} = n'\mathbb{Z}$ with $n, n' \ge 0$. If $n = 0$ then $n\mathbb{Z} = \{0\}$, so $n'\mathbb{Z} = \{0\}$ and $n' = 0$. If $n > 0$, then $n$ is the least positive integer in $n\mathbb{Z}$, and the same holds for $n'$, so $n = n'$.

∎

Since $\mathbb{Z}$ is abelian, part (1) of Proposition 4.2 shows that all its subgroups are normal. Hence Theorem 5.1 yields, for every $n \ge 0$, the quotient group

$$\mathbb{Z}/n\mathbb{Z} = \{\,a + n\mathbb{Z} : a \in \mathbb{Z}\,\}$$

We abbreviate $a + n\mathbb{Z}$ by $\bar{a}$. Rewriting part (2) of Proposition 2.2 additively gives

$$\bar{a} = \bar{b} \iff -a + b \in n\mathbb{Z} \iff n \mid b - a \iff a \equiv b \pmod{n}$$

so the elements of $\mathbb{Z}/n\mathbb{Z}$ are exactly the congruence classes modulo $n$. And the assertion that the quotient operation $\bar{a} + \bar{b} = \overline{a+b}$ is well defined is nothing but the familiar statement of elementary number theory,

$$a \equiv a',\ b \equiv b' \pmod{n} \ \Longrightarrow\ a + b \equiv a' + b' \pmod{n}$$

(proved directly by deducing $n \mid (a+b) - (a'+b')$ from $n \mid a - a'$ and $n \mid b - b'$). What Theorem 3.2 tells us is that this naive fact held because $n\mathbb{Z}$ is a normal subgroup.

Example 6.2（The elements of Z/nZ and their orders）

Let $n \ge 1$.

Number of elements. For any $a \in \mathbb{Z}$ the division algorithm gives $a = qn + r$ with $0 \le r < n$, so $\bar{a} = \bar{r}$. Hence $\mathbb{Z}/n\mathbb{Z} = \{\bar{0}, \bar{1}, \ldots, \overline{n-1}\}$. Moreover if $0 \le r < r' < n$ then $0 < r' - r < n$, so $n \nmid r' - r$ and therefore $\bar{r} \ne \bar{r'}$. Thus these $n$ elements are distinct and $|\mathbb{Z}/n\mathbb{Z}| = n$, in agreement with the computation of $[G:N]$ in Theorem 5.1.

Cyclicity. Since $\bar{a} = a\bar{1}$ (add $\bar{1}$ to itself $a$ times; for $a < 0$, add $-\bar 1$ to itself $|a|$ times), we have $\mathbb{Z}/n\mathbb{Z} = \langle \bar{1} \rangle$, a cyclic group of order $n$.

Orders of elements. The order of $\bar{a}$ is $n/\gcd(a, n)$. Indeed, put $d := \gcd(a,n)$, $a = da'$, $n = dn'$ (so that $\gcd(a', n') = 1$). Then for $k \ge 1$,

$$k\bar{a} = \bar{0} \iff n \mid ka \iff dn' \mid k d a' \iff n' \mid ka' \iff n' \mid k$$

where the last equivalence uses $\gcd(a', n') = 1$ together with Euclid’s lemma (if $n' \mid ka'$ and $\gcd(n', a') = 1$ then $n' \mid k$). The least positive integer $k$ satisfying this is $k = n' = n/d$.

Generators. It follows that $\bar{a}$ generates $\mathbb{Z}/n\mathbb{Z}$ (that is, has order $n$) precisely when $\gcd(a,n) = 1$. Hence the number of generators equals Euler’s totient $\varphi(n)$.

Example 6.3（The clock group Z/12Z）

Let us compute explicitly for $n = 12$. Using the formula $\mathrm{ord}(\bar a) = 12/\gcd(a,12)$ of Example 6.2 we obtain the following table.

$a$	0	1	2	3	4	5	6	7	8	9	10	11
$\gcd(a,12)$	12	1	2	3	4	1	6	1	4	3	2	1
order of $\bar a$	1	12	6	4	3	12	2	12	3	4	6	12

The generators are $\bar1, \bar5, \bar7, \overline{11}$, four in number, matching $\varphi(12) = 12(1 - 1/2)(1-1/3) = 4$.

The subgroups are $\langle \bar{d}\rangle$ (of order $12/d$), one for each positive divisor $d$ of $12$, six in all. For instance

$$\langle \bar{3}\rangle = \{\bar{0}, \bar{3}, \bar{6}, \bar{9}\}, \qquad \langle \bar{4}\rangle = \{\bar{0}, \bar{4}, \bar{8}\} .$$

Since $\mathbb{Z}/12\mathbb{Z}$ is abelian these are all normal, so we may form quotients again. With $N := \langle\bar 3\rangle$ we get $|(\mathbb{Z}/12\mathbb{Z})/N| = 12/4 = 3$, and the cosets are the three sets

$$\bar0 + N = \{\bar0,\bar3,\bar6,\bar9\},\quad \bar1 + N = \{\bar1,\bar4,\bar7,\overline{10}\},\quad \bar2 + N = \{\bar2,\bar5,\bar8,\overline{11}\}$$

Since $\bar{a} + N$ depends only on the remainder of $a$ upon division by 3, this quotient group is isomorphic to $\mathbb{Z}/3\mathbb{Z}$; as an operation it corresponds to “take the remainder modulo 12, then take the remainder modulo 3”, which is the same as “take the remainder modulo 3 from the start”. This phenomenon, that a quotient of a quotient is again a quotient, is formulated in general as the third isomorphism theorem in the next chapter (Corollary 7.2［群の準同型定理］).

Tip

Some authors write $\mathbb{Z}_n$ for $\mathbb{Z}/n\mathbb{Z}$, but $\mathbb{Z}_p$ is widely used for the ring of $p$-adic integers, so to avoid confusion we write $\mathbb{Z}/n\mathbb{Z}$ throughout. Note also that $\mathbb{Z}/n\mathbb{Z}$ carries a well-defined multiplication $\bar{a}\,\bar{b} := \overline{ab}$, making it a ring. This is a consequence of $n\mathbb{Z}$ being an ideal of $\mathbb{Z}$ (Definition 3.1［イデアルと剰余環］), a topic treated in Ideals and quotient rings. What normal subgroups do for groups, ideals do for rings.

7. Further examples of quotient groups

Example 7.1（Quotients of the symmetric group, the general linear group and the real line）

(a) $S_3/A_3$. The subgroup $A_3 = \{e, (1\,2\,3), (1\,3\,2)\}$ has index 2, so part (2) of Proposition 4.2 gives $A_3 \trianglelefteq S_3$. The quotient group consists of the two elements $\{A_3,\ (1\,2)A_3\}$, where $A_3$ is the set of all even permutations and $(1\,2)A_3$ the set of all odd ones. The operation in the quotient is exactly the rule

$$(\text{even})(\text{even}) = \text{even},\quad (\text{even})(\text{odd}) = \text{odd},\quad (\text{odd})(\text{odd}) = \text{even}$$

and $S_3/A_3 \cong \mathbb{Z}/2\mathbb{Z}$. The “contents” of a permutation are forgotten entirely; only the parity survives.

(b) $\mathrm{GL}_n(\mathbb{R})/\mathrm{SL}_n(\mathbb{R})$. By Example 4.3 (a) we have $\mathrm{SL}_n(\mathbb{R}) \trianglelefteq \mathrm{GL}_n(\mathbb{R})$. Writing $S := \mathrm{SL}_n(\mathbb{R})$, part (2) of Proposition 2.2 gives

$$AS = BS \iff A^{-1}B \in S \iff \det(A^{-1}B) = 1 \iff \det A = \det B$$

so the cosets correspond bijectively to the values of the determinant. The operation is $(AS)(BS) = (AB)S$, and determinants multiply as $\det(AB) = \det A \det B$, so $\mathrm{GL}_n(\mathbb{R})/\mathrm{SL}_n(\mathbb{R}) \cong \mathbb{R}^{\times}$. The quotient group is the world in which a matrix is seen through its determinant alone.

(c) $\mathbb{R}/\mathbb{Z}$. Since $(\mathbb{R}, +)$ is abelian we have $\mathbb{Z} \trianglelefteq \mathbb{R}$ and may form $\mathbb{R}/\mathbb{Z}$. As $x + \mathbb{Z} = y + \mathbb{Z}$ is equivalent to $x - y \in \mathbb{Z}$, each coset has exactly one representative in the interval $[0,1)$, namely the fractional part of $x$. The operation is “add and discard the integer part”, that is, addition $\mathrm{mod}\ 1$. The map $x + \mathbb{Z} \mapsto e^{2\pi i x}$ is well defined (replacing $x$ by $x + m$ does not change the value, since $e^{2\pi i m} = 1$) and is a bijective homomorphism, so $\mathbb{R}/\mathbb{Z}$ is isomorphic to the multiplicative group of the unit circle in the complex plane. Winding an infinite line into a circle, one unit of length at a time, is in group-theoretic terms exactly the passage to a quotient.

What these examples have in common is that the quotient group represents what remains after the information we wish to discard has been discarded. In (a) the fine detail of a permutation is discarded and only the parity kept; in (b) the fine detail of a matrix is discarded and only the determinant kept; in (c) the integer part of a real number is discarded and only the fractional part kept. In each case the totality of the discarded information is precisely the normal subgroup $N$, and what remains is the quotient group $G/N$. What Corollary 5.3 said is that this splitting into “discarded” and “retained” is the same thing as a single homomorphism.

The general theory of groups starts from Introduction to group theory: the definition and examples, but the basic policy when facing the great problem of classifying groups is: find a normal subgroup $N$ and decompose into $N$ and $G/N$. The groups for which this policy finally stops working, the simple groups, are the “elementary particles” of the decomposition. Galois theory, which decides the solvability of an equation by decomposing a group (An invitation to Galois theory), is likewise an argument tracing a chain of normal subgroups (Theorem 5.4［ガロア理論への招待］).

8. Exercises

Exercise 8.1Easy

In $\mathbb{Z}/18\mathbb{Z}$, find the order of $\overline{12}$ and list all elements of $\langle \overline{12}\rangle$. Then find the order of the quotient group $(\mathbb{Z}/18\mathbb{Z})/\langle\overline{12}\rangle$.

Solution

By Example 6.2 the order of $\overline{12}$ is $18/\gcd(12,18) = 18/6 = 3$. Hence

$$\langle \overline{12}\rangle = \{\bar 0,\ \overline{12},\ \overline{24}\} = \{\bar 0,\ \overline{12},\ \bar 6\}$$

(since $24 = 18 + 6$, we have $\overline{24} = \bar 6$). As $\mathbb{Z}/18\mathbb{Z}$ is abelian, part (1) of Proposition 4.2 shows this subgroup is normal, and the order formula in Theorem 5.1 gives

$$\bigl|(\mathbb{Z}/18\mathbb{Z})/\langle\overline{12}\rangle\bigr| = \frac{18}{3} = 6 .$$

Note that $\langle\overline{12}\rangle = \langle\bar 6\rangle$, the subgroup corresponding to the divisor $6$ of $18$.

Exercise 8.2Standard

Let $N, M \trianglelefteq G$. Show that $N \cap M$ and $NM := \{nm : n \in N,\ m \in M\}$ are both normal subgroups of $G$.

Solution

The case of $N \cap M$. Apply part (5) of Proposition 4.2 with $\Lambda = \{1,2\}$, $N_1 = N$, $N_2 = M$; the claim follows at once.

The case of $NM$. First we show it is a subgroup. We have $e = ee \in NM$. To see closure under products, take $n_1m_1,\ n_2m_2 \in NM$. Since $M \trianglelefteq G$ we have $n_2^{-1}m_1n_2 \in M$; calling this element $m'$, we get $m_1 n_2 = n_2 m'$. Hence

$$(n_1m_1)(n_2m_2) = n_1(m_1n_2)m_2 = n_1n_2m'm_2 \in NM$$

(as $n_1n_2 \in N$ and $m'm_2 \in M$). As for inverses, $N \trianglelefteq G$ gives $m^{-1}n^{-1}m \in N$, so

$$(nm)^{-1} = m^{-1}n^{-1} = (m^{-1}n^{-1}m)\,m^{-1} \in NM .$$

Therefore $NM \le G$.

Next, normality. For $g \in G$, $n \in N$ and $m \in M$, inserting $g^{-1}g = e$ gives

$$g(nm)g^{-1} = (gng^{-1})(gmg^{-1}) \in NM$$

(here $gng^{-1} \in N$ because $N \trianglelefteq G$, and $gmg^{-1} \in M$ because $M \trianglelefteq G$). Hence $g(NM)g^{-1} \subseteq NM$ for every $g$, and part (2) of Theorem 3.2 gives $NM \trianglelefteq G$.

Note that if the normality of $M$ is dropped and $M$ is merely a subgroup, then $NM$ is still a subgroup (use $N \trianglelefteq G$ in place of $M \trianglelefteq G$ in the computation above) but need not be normal.

Exercise 8.3Standard

Let $G$ be a group. Show that if the quotient group $G/Z(G)$ is cyclic, then $G$ is abelian. Using this, show that if $|G| = p^2$ with $p$ prime, then $G$ is abelian. You may use Lagrange’s theorem and the fact that a group of prime order is cyclic. You may also assume $Z(G) \ne \{e\}$ (that the centre of a $p$-group is nontrivial).

Solution

Write $Z := Z(G)$. By part (3) of Proposition 4.2 we have $Z \trianglelefteq G$, so the quotient group $G/Z$ is defined.

First part. Suppose $G/Z = \langle gZ\rangle$. Take any $x \in G$. Then $xZ \in G/Z$, so $xZ = (gZ)^k = g^kZ$ for some integer $k$. By part (2) of Proposition 2.2 we have $(g^k)^{-1}x \in Z$, so putting $z := g^{-k}x \in Z$ we may write $x = g^kz$. Similarly any $y \in G$ can be written $y = g^lw$ with $w \in Z$. Since $z$ and $w$ commute with every element of $G$,

$$xy = g^kz\,g^lw = g^kg^lzw = g^{k+l}zw, \qquad yx = g^lw\,g^kz = g^lg^kwz = g^{k+l}wz .$$

From $z, w \in Z$ we get $zw = wz$ (apply the definition of the centre to $z \in Z$ and $w \in G$, say), so $xy = yx$. As $x$ and $y$ were arbitrary, $G$ is abelian.

Second part. Suppose $|G| = p^2$. Since $Z \le G$, Lagrange’s theorem gives $|Z| \in \{1, p, p^2\}$, and $|Z| \ne 1$ by hypothesis. If $|Z| = p^2$ then $Z = G$ and $G$ is abelian. We rule out the case $|Z| = p$. In that case Theorem 5.1 gives $|G/Z| = p^2/p = p$, so $G/Z$ is cyclic, a group of prime order being cyclic. By the first part $G$ is then abelian, that is, $Z = G$ and $|Z| = p^2$, contradicting $|Z| = p$. Hence $|Z| = p^2$ is the only possibility and $G$ is abelian.

Exercise 8.4Hard

Show that the alternating group $A_4$ (of order 12) has no subgroup of order 6. In particular, the converse of Lagrange’s theorem is false.

Solution

Suppose there were $H \le A_4$ with $|H| = 6$. The order formula of Theorem 5.1 gives $[A_4 : H] = 12/6 = 2$, so part (2) of Proposition 4.2 gives $H \trianglelefteq A_4$. Hence the quotient group $A_4/H$ is defined and has order 2.

In a group of order 2, $u^2$ is the identity for every element $u$ (clear if $u$ is the identity; otherwise $u$ has order 2 by Lagrange’s theorem). Let $\pi : A_4 \to A_4/H$ be the natural projection (Proposition 5.2). Then for every $x \in A_4$,

$$\pi(x^2) = \pi(x)^2 = H$$

so $x^2 \in \ker\pi = H$.

Now let $\sigma \in A_4$ be a 3-cycle. Since $\sigma^3 = e$, we have $(\sigma^2)^2 = \sigma^4 = \sigma$, and applying the fact above with $x = \sigma^2$ gives $\sigma = (\sigma^2)^2 \in H$. Thus every 3-cycle in $A_4$ lies in $H$.

The number of 3-cycles on $\{1,2,3,4\}$ is $4 \times 2 = 8$: there are $\binom{4}{3} = 4$ ways to choose the three letters moved, and $2$ ways to cycle the chosen three. A 3-cycle is a product of two transpositions ($(a\,b\,c) = (a\,b)(b\,c)$), hence even, so all of them lie in $A_4$. Therefore $|H| \ge 8 + 1 = 9$ (counting the identity as well), contradicting $|H| = 6$.

Hence no subgroup of order 6 exists. Since $6 \mid 12$ and yet there is no subgroup of order 6, the converse of Lagrange’s theorem — that for every divisor $d$ of $|G|$ there is a subgroup of order $d$ — is false.

References

• Matsuzaka Kazuo, Daisūkei Nyūmon (Introduction to Algebraic Systems), Iwanami Shoten, 1976 (in Japanese) — Chapter 3 (group theory) gives a detailed account of cosets, normal subgroups and quotient groups. One of the most careful introductions available in Japanese.
• Katsura Toshiyuki, Daisūgaku I: Gun to Kan (Algebra I: Groups and Rings), University of Tokyo Press, 2004 (in Japanese) — Chapter 1. A concise treatment of normal subgroups and the isomorphism theorems.
• Yukie Akihiko, Daisūgaku 1: Gunron Nyūmon (Algebra 1: Introduction to Group Theory), Nippon Hyoron Sha, 2010 (in Japanese) — Chapter 2. Rich in concrete examples, well suited to practising computations in $S_n$ and $D_n$.
• S. Lang, Algebra, 3rd revised ed., Springer GTM 211, 2002 — Chapter I (Groups). Organizes quotient groups and the isomorphism theorems in categorical language.
• D. S. Dummit and R. M. Foote, Abstract Algebra, 3rd ed., Wiley, 2004 — Chapter 3 (Quotient Groups and Homomorphisms). Contains the tests for normality together with a wealth of exercises.
• C. F. Gauss, Disquisitiones Arithmeticae, 1801 — Chapter 1. The original source for the notation of congruences and for their compatibility with addition, subtraction and multiplication.

Appendix: A normal subgroup is a union of conjugacy classes

The partition by conjugacy. The relation $x \approx y \iff \exists g \in G,\ y = gxg^{-1}$ on a group $G$ is an equivalence relation: it is reflexive because $x = exe^{-1}$; symmetric because $y = gxg^{-1}$ gives $x = g^{-1}y(g^{-1})^{-1}$; and transitive because $y = gxg^{-1}$ and $z = hyh^{-1}$ give $z = (hg)x(hg)^{-1}$. Its equivalence classes are called conjugacy classes. Directly from the definition, for a subgroup $N \le G$,

$$N \trianglelefteq G \iff N \text{ is a union of conjugacy classes}$$

Indeed, if $N \trianglelefteq G$ then all conjugates $gxg^{-1}$ of $x \in N$ lie in $N$, so $N$ contains the whole conjugacy class of $x$; conversely if $N$ is a union of conjugacy classes then $gNg^{-1} \subseteq N$ for every $g$, so $N$ is normal by Theorem 3.2. This reformulation is a practical tool when listing all normal subgroups of a finite group, because the constraint that the sizes of the classes must sum to $|N|$ and that $|N|$ must divide $|G|$ (Lagrange’s theorem) bites hard.

The case of $S_3$. The conjugacy classes of $S_3$ are three: $\{e\}$ (one element), the class of transpositions $\{(1\,2),(1\,3),(2\,3)\}$ (three elements), and the class of 3-cycles $\{(1\,2\,3),(1\,3\,2)\}$ (two elements). The unions containing $e$ whose size divides $6$ are $1$, $1+2 = 3$ and $1+3+2 = 6$, three possibilities in all (note that $1+3 = 4$ does not divide 6). The corresponding subsets are $\{e\}$, $A_3$ and $S_3$, all of which are subgroups. Hence the normal subgroups of $S_3$ are exactly $\{e\},\ A_3,\ S_3$. That the subgroup $\{e,(1\,2)\}$ of Example 3.1 was not normal is now immediate: it contains only part of the class of transpositions.

The case of $A_4$. Consider $V := \{e,\ (1\,2)(3\,4),\ (1\,3)(2\,4),\ (1\,4)(2\,3)\}$. Conjugation obeys the rule $\sigma (a\,b)(c\,d) \sigma^{-1} = (\sigma(a)\,\sigma(b))(\sigma(c)\,\sigma(d))$, so conjugating $(1\,2)(3\,4)$ by $\sigma = (1\,2\,3)$ yields $(2\,3)(1\,4)$, and by $\sigma = (1\,3\,2)$ yields $(3\,1)(2\,4)$. Thus the three double transpositions are conjugate to one another inside $A_4$, and $V$ is the union of the class $\{e\}$ with this three-element class. Once one checks that $V$ is a subgroup (that the product of two double transpositions is the third, and that every element has order 2), the criterion above gives $V \trianglelefteq A_4$. By Theorem 5.1 we have $|A_4/V| = 12/4 = 3$, and this quotient group is cyclic of order 3. Combined with Exercise 8.4, this reveals an asymmetric picture: $A_4$ has no subgroup of order 6, yet it has a normal subgroup of order 4.