Normal Subgroups and Quotient Groups

In this lesson you’ll learn what makes a subgroup normal, why normality is exactly what’s needed to multiply cosets, how to build the quotient group $G/N$, and what quotients are for. This is the traditional hard spot in a first course, so it’s worth going slowly.

We want to make the cosets of $H$ into a group. The obvious rule is

$$(aH)(bH) = abH$$

But there’s a catch. A coset has many names: if $a' \in aH$ then $a'H = aH$. For the rule to define anything at all, the answer has to come out the same whichever representative you pick. The operation must be well-defined, and for a general subgroup it isn’t.

Concretely, in $S_3$ with $H = \{e, s\}$: we have $rH = \{r, rs\}$, so $r$ and $rs$ name the same coset. Multiply on the left by $sH$:

• Using $r$: $(sH)(rH) = srH = r^2sH = \{r^2s, r^2\}$
• Using $rs$: $(sH)(rsH) = s(rs)H = (r^2s)sH = r^2 H = \{r^2, r^2s\}$

Those happen to agree. Try instead $(rH)(sH)$ two ways: with $r$, get $rsH = \{rs, r\}$; with $rs$, get $(rs)(s)H = rH = \{r, rs\}$. Also agree. But now compute $(rH)(rH)$: using $r$ twice gives $r^2H = \{r^2, r^2s\}$; using $r$ and $rs$ gives $r(rs)H = r^2sH = \{r^2s, r^2\}$. Still fine - but the general pattern breaks for other choices, and the theorem below says exactly when.

$N \le G$ is normal, written $N \trianglelefteq G$, if any of these equivalent conditions holds:

1. $gN = Ng$ for all $g \in G$ - left and right cosets coincide.
2. $gNg^{-1} = N$ for all $g \in G$ - conjugation preserves $N$.
3. $gng^{-1} \in N$ for all $g \in G$, $n \in N$ - the practical test.

Condition 3 is what you actually check. Note it does not say $gng^{-1} = n$; the conjugate may be a different element of $N$, it just has to stay inside.

Conjugation $n \mapsto gng^{-1}$ means “do $g$, then $n$, then undo $g$.” It’s the same operation viewed from a relabelled position. So normality says: the subgroup looks the same from everywhere in the group.

Some automatic cases:

• In an abelian group, every subgroup is normal, since $gng^{-1} = ngg^{-1} = n$.
• Any subgroup of index 2 is normal. With only two cosets, $N$ and everything else, there’s no room for left and right to differ.
• $\{e\}$ and $G$ are always normal.
• The center $Z(G)$ is always normal, since its elements commute with everything.

If $N \trianglelefteq G$, the set of cosets $G/N$ with $(aN)(bN) = abN$ is a group, of order $[G:N] = |G|/|N|$.

Normality is exactly what makes the operation well-defined. Here’s the check: suppose $aN = a'N$ and $bN = b'N$, so $a' = an_1$ and $b' = bn_2$. Then

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

Normality makes $b^{-1}n_1b \in N$, so $a'b' \in abN$, and the answer is the same coset. Without normality that middle term escapes and the rule collapses.

The quotient’s identity is $N$ itself, and $(aN)^{-1} = a^{-1}N$.

Forming $G/N$ deliberately forgets information: it declares everything in $N$ to be equivalent to the identity. What survives is whatever $N$ couldn’t see.

The clearest example is $\mathbb{Z}/n\mathbb{Z} = \mathbb{Z}_n$. Quotienting the integers by the multiples of $n$ throws away everything except the remainder. Clock arithmetic is a quotient group, and you’ve been using one since childhood.

That’s the standard use: a quotient keeps the feature you care about and discards the rest.

A group with no normal subgroups other than $\{e\}$ and itself is simple. Simple groups can’t be broken down by quotients, which makes them the atoms of finite group theory.

$\mathbb{Z}_p$ for prime $p$ is simple. $A_5$, of order 60, is the smallest non-abelian simple group. The classification of all finite simple groups took fifty years and ten thousand pages, and $A_5$‘s simplicity is precisely why the quintic has no solution formula.

∘	e	r	r²	s	rs	r²s
e	e	r	r²	s	rs	r²s
r	r	r²	e	rs	r²s	s
r²	r²	e	r	r²s	s	rs
s	s	r²s	rs	e	r²	r
rs	rs	s	r²s	r	e	r²
r²s	r²s	rs	s	r²	r	e

The shaded block is closed, and so is the unshaded complement viewed as a single coset. Collapsing each to a point gives S₃/A₃ ≅ ℤ₂: rotation or reflection, nothing else remembered.

Identity: e. Cells holding it mark inverse pairs. Shaded: e, r, r² (A₃, normal of index 2) - closed, every product stays inside .

$A_3$ has index 2 in $S_3$, so it is normal. The quotient $S_3/A_3$ has two elements: “even” and “odd.” All the detail about which rotation or reflection is discarded, and what remains is exactly parity.

Example 1: Show every subgroup of an abelian group is normal.

Solution. Let $G$ be abelian, $H \le G$, $g \in G$, $h \in H$.

$$ghg^{-1} = hgg^{-1} = he = h \in H$$

The first step uses commutativity. So condition 3 holds and $H \trianglelefteq G$. ∎

This is why normality never comes up in $\mathbb{Z}$ or $\mathbb{Z}_n$: it’s automatic.

Example 2: Show $\{e, s\}$ is not normal in $S_3$.

Solution. Test condition 3 with $g = r$, $n = s$, using $sr = r^2s$ so $sr^{-1} = sr^2 = rs$:

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

Is $r^2s \in \{e, s\}$? No.

Not normal. Geometrically: conjugating a reflection by a rotation gives a different reflection, because rotating the whole picture moves the mirror line.

Example 3: Show $A_n$ is normal in $S_n$.

Solution. $A_n$ has index 2 in $S_n$, so it is normal immediately.

For the direct argument, take $\sigma \in A_n$ (even) and any $\tau \in S_n$. Signs multiply, so

$$\operatorname{sign}(\tau\sigma\tau^{-1}) = \operatorname{sign}(\tau)\operatorname{sign}(\sigma)\operatorname{sign}(\tau)^{-1} = \operatorname{sign}(\sigma) = +1$$

The conjugate is even, hence in $A_n$. ∎

Example 4: Build $\mathbb{Z}_{12}/\langle 4 \rangle$.

Solution. $N = \langle 4 \rangle = \{0,4,8\}$, $|N| = 3$, so the quotient has order $12/3 = 4$.

Cosets: $\{0,4,8\}$, $\{1,5,9\}$, $\{2,6,10\}$, $\{3,7,11\}$, which we name $[0], [1], [2], [3]$.

Addition: $[1] + [1] = [2]$, $[2]+[2] = [4] = [0]$, $[1]+[3] = [4] = [0]$.

So $[1]$ has order 4 and generates:

$$\mathbb{Z}_{12}/\langle 4 \rangle \cong \mathbb{Z}_4$$

Pattern worth remembering: $\mathbb{Z}_n / \langle d \rangle \cong \mathbb{Z}_d$ when $d \mid n$.

Example 5: A quotient that loses non-abelianness.

Compute $D_4/\{e, r^2\}$.

Solution. $N = \{e, r^2\}$ is the center of $D_4$, so it’s normal. $|N| = 2$ and $|D_4| = 8$, so the quotient has order 4.

Cosets: $\{e,r^2\}$, $\{r, r^3\}$, $\{s, r^2s\}$, $\{rs, r^3s\}$.

In the quotient, $(rN)^2 = r^2N = N$, the identity. Likewise $(sN)^2 = N$ and $(rsN)^2 = N$. Every non-identity element squares to the identity, so

$$D_4/Z(D_4) \cong \mathbb{Z}_2 \times \mathbb{Z}_2$$

the Klein four-group. A non-abelian group has an abelian quotient. That’s typical: quotienting by the center forgets exactly the information that made things fail to commute.

Example 6: Why $A_4$ has no subgroup of order 6, revisited.

Solution. Suppose $H \le A_4$ with $|H| = 6$. Index 2 makes $H$ normal, so $A_4/H \cong \mathbb{Z}_2$.

In a group of order 2 every element satisfies $x^2 = e$, so for every $g \in A_4$ we get $(gH)^2 = H$, meaning $g^2 \in H$.

Now take any 3-cycle $g$. Then $g^2 = g^{-1}$ is also a 3-cycle, and as $g$ runs over the eight 3-cycles of $A_4$ so does $g^2$. So $H$ contains all eight 3-cycles plus $e$: at least 9 elements.

That contradicts $|H| = 6$. No such subgroup. ∎

The quotient did the work: assuming the subgroup existed forced a constraint on every element, and counting broke it.

Modular arithmetic. Every time you compute a day of the week, a clock time, or a checksum, you’re working in the quotient $\mathbb{Z}/n\mathbb{Z}$. The quotient construction is what makes “it’s the same as 3 o’clock” a legitimate equation rather than sloppiness.

Error-correcting codes. For a linear code $C \le \mathbb{Z}_2^n$, the quotient $\mathbb{Z}_2^n / C$ is the syndrome space. A decoder computes which coset a received word falls in, and the syndrome is the name of that coset. Syndrome decoding is the quotient group used as an algorithm.

Homology in topology. Homology groups are quotients: cycles modulo boundaries. The quotient is precisely what detects holes, because it declares “boundaries don’t count” and sees what’s left. This is the mechanism behind topological data analysis.

Physics and gauge theory. Physical states are often defined up to a symmetry, meaning the real state space is a quotient. Gauge invariance says two mathematically different field configurations describe the same physics, and the quotient by the gauge group is the physically meaningful object.

Cryptography. Elliptic curve cryptography works in a quotient of a group of points, and RSA works in $\mathbb{Z}/n\mathbb{Z}$. Quotients are what make these groups finite and computable.

Music theory. Pitch classes are a quotient: the group of all pitches modulo octave equivalence gives $\mathbb{Z}_{12}$. Saying “C is C regardless of octave” is forming a quotient group.