A subgroup $H$ of a group $G$ cuts $G$ into translated copies of itself, the cosets of $H$ . Because the cosets all have the same size and never overlap, the size of $H$ must divide the size of $G$ . That is Lagrange's theorem, the first real structure theorem of group theory, and it re-proves Fermat's and Euler's theorems from post 00 in one line. This post then asks when the cosets can themselves be multiplied to form a new, smaller group. The answer is exactly when $H$ is a normal subgroup, and the new group is the quotient group $G/H$ . Clock arithmetic $\mathbb{Z}/n\mathbb{Z}$ turns out to be the quotient of $\mathbb{Z}$ by $n\mathbb{Z}$ . The post closes with the center, the commutator subgroup, and simple groups, the groups that have no quotients to take. It uses the notation of post 01: permutations compose right to left, and $D_n$ has order $2n$ .

Series map: 00 Integers, Relations & Modular Arithmetic → 01 Groups & Subgroups → 02 Cosets, Lagrange & Quotient Groups (this post) → 03 Homomorphisms, Group Actions & Sylow → 04 Rings, Ideals & Fields → 05 Field Extensions & Galois Theory.

Learning Outcome

Compute left and right cosets and prove that the left cosets of $H$ partition $G$ into sets of size $\lvert H \rvert$ .
State and prove Lagrange's theorem and its corollaries: element orders divide $\lvert G \rvert$ , groups of prime order are cyclic, Fermat and Euler.
Use the index $[G : H]$ and the tower law.
Explain why the converse of Lagrange's theorem fails, using $A_4$ .
Prove the equivalent conditions for a subgroup to be normal, and test normality with the conjugation rule for cycles.
Construct the quotient group $G/N$ , prove that its operation is well defined, and see exactly how it breaks for a non-normal subgroup.
Compute the center and the commutator subgroup of small groups, and use " $G/Z(G)$ cyclic implies $G$ abelian".
Define simple groups and explain what " $A_5$ is simple" means.

Chapter 1 — Cosets

Terminology

For a subgroup $H \leq G$ and $g \in G$ , the left coset is $gH = \{gh : h \in H\}$ and the right coset is $Hg = \{hg : h \in H\}$ . In additive notation: $g + H$ .
$g$ is a representative of the coset $gH$ . Any element of the coset can serve.
The index $[G : H]$ is the number of left cosets of $H$ in $G$ (possibly infinite).

First examples

In $\mathbb{Z}$ with $H = 3\mathbb{Z}$ , the cosets are $0 + 3\mathbb{Z}$ , $1 + 3\mathbb{Z}$ , $2 + 3\mathbb{Z}$ : exactly the congruence classes mod 3 from post 00. So $[\mathbb{Z} : 3\mathbb{Z}] = 3$ .

In $\mathbb{R}^2$ under addition, let $H$ be a line through the origin. The coset $v + H$ is the parallel line through the point $v$ . The cosets are all the lines parallel to $H$ : they fill the plane and never cross. This picture — parallel copies of the subgroup, filling the group without overlap — is the right mental model in every example.

Cosets in $S_3$

Use $r = (1 \ 2 \ 3)$ and $s = (2 \ 3)$ as in post 01, so $S_3 = \{e, r, r^2, s, rs, r^2 s\}$ with $r^2 = (1 \ 3 \ 2)$ , $rs = (1 \ 2)$ , $r^2 s = (1 \ 3)$ , and $sr = r^2 s$ .

Take $H = \langle s \rangle = \{e, (2 \ 3)\}$ . The left cosets are

$H = \{e, (2 \ 3)\}, \qquad rH = \{r, rs\} = \{(1 \ 2 \ 3), (1 \ 2)\}, \qquad r^2 H = \{r^2, r^2 s\} = \{(1 \ 3 \ 2), (1 \ 3)\}.$

The right cosets are $H$ , $Hr = \{r, sr\} = \{r, r^2 s\} = \{(1 \ 2 \ 3), (1 \ 3)\}$ and $Hr^2 = \{r^2, sr^2\} = \{r^2, rs\} = \{(1 \ 3 \ 2), (1 \ 2)\}$ . Both families split $S_3$ into three pairs, but into different pairs: $rH \neq Hr$ .

Now take the rotation subgroup $A_3 = \langle r \rangle = \{e, r, r^2\}$ . Its cosets are $A_3$ itself and $sA_3 = \{s, sr, sr^2\} = \{s, rs, r^2 s\}$ , the three reflections (transpositions). Here $sA_3 = A_3 s$ : left and right cosets agree.

The six elements of S3 split three ways: into the left cosets of H = {e, (2 3)}, which pair (1 2 3) with (1 2) and (1 3 2) with (1 3); into the right cosets of H, which pair (1 2 3) with (1 3) and (1 3 2) with (1 2); and into the two cosets of A3, the rotations and the reflections, which are the same on both sides

The coset lemma

Lemma 1.1. Let $H \leq G$ and $a, b \in G$ .

$a \in aH$ .
$aH = bH$ if and only if $a^{-1}b \in H$ . In particular $aH = H$ if and only if $a \in H$ .
Two left cosets are either equal or disjoint.
$\lvert aH \rvert = \lvert H \rvert$ .

Proof. Define $a \sim b$ if $a^{-1}b \in H$ . This is an equivalence relation: $a^{-1}a = e \in H$ ; if $a^{-1}b \in H$ then its inverse $b^{-1}a \in H$ ; if $a^{-1}b, b^{-1}c \in H$ then their product $a^{-1}c \in H$ . Its class of $a$ is $\{b : a^{-1}b \in H\} = \{b : b = ah \text{ for some } h \in H\} = aH$ . So the left cosets are exactly the equivalence classes, and parts 1–3 are Theorem 2.3 of post 00 (classes partition the set; one common element makes two classes equal). For part 4, the map $H \to aH$ , $h \mapsto ah$ , is onto by definition and one-to-one by cancellation. $\blacksquare$

The same holds for right cosets with $ba^{-1} \in H$ in part 2. The map $gH \mapsto Hg^{-1}$ is a well-defined bijection from left cosets to right cosets (because $(gh)^{-1} = h^{-1}g^{-1}$ ), so the number of left cosets equals the number of right cosets, even when the cosets themselves differ.

Common pitfall: $aH = bH$ does not mean $a = b$ , and it is tested by $a^{-1}b \in H$ , not by $ab^{-1} \in H$ (that is the test for right cosets). In $S_3$ above, $rH = (rs)H$ because $r^{-1}(rs) = s \in H$ .

Chapter 2 — Lagrange's theorem

Theorem 2.1 (Lagrange). If $G$ is a finite group and $H \leq G$ , then $\lvert H \rvert$ divides $\lvert G \rvert$ , and

$[G : H] = \frac{\lvert G \rvert}{\lvert H \rvert}.$

Proof. By Lemma 1.1 the left cosets of $H$ partition $G$ , and each has exactly $\lvert H \rvert$ elements. If there are $k = [G : H]$ of them, counting the elements of $G$ coset by coset gives $\lvert G \rvert = k \lvert H \rvert$ . $\blacksquare$

The proof is three lines because the work was done in the coset lemma. Lagrange's theorem is a restriction: it says which subgroup orders are impossible. A group of order 12 can only have subgroups of orders 1, 2, 3, 4, 6, 12; a group of order 8 has no subgroup (and so no element) of order 3.

Corollaries

Corollary 2.2. In a finite group, the order of every element divides $\lvert G \rvert$ .

Proof. $\lvert g \rvert = \lvert \langle g \rangle \rvert$ (post 01, Lemma 3.1), and $\langle g \rangle$ is a subgroup. $\blacksquare$

Corollary 2.3. $g^{\lvert G \rvert} = e$ for every $g$ in a finite group $G$ .

Proof. $\lvert G \rvert = m \lvert g \rvert$ for some integer $m$ , so $g^{\lvert G \rvert} = (g^{\lvert g \rvert})^m = e$ . $\blacksquare$

Corollary 2.4. A group of prime order $p$ is cyclic, and every non-identity element generates it. In particular it is abelian and, up to isomorphism, it is $\mathbb{Z}/p\mathbb{Z}$ .

Proof. Take $g \neq e$ . Then $\lvert g \rvert$ divides $p$ and is not 1, so $\lvert g \rvert = p$ and $\langle g \rangle = G$ . $\blacksquare$

Corollary 2.5 (Euler and Fermat, again). If $\gcd(a, n) = 1$ then $a^{\varphi(n)} \equiv 1 \pmod n$ ; if $p$ is prime and $p \nmid a$ then $a^{p-1} \equiv 1 \pmod p$ .

Proof. $(\mathbb{Z}/n\mathbb{Z})^\times$ is a group of order $\varphi(n)$ containing $[a]$ ; apply Corollary 2.3. Fermat is $n = p$ , $\varphi(p) = p - 1$ . $\blacksquare$

Compare the proof in post 00, which rearranged a product of all the units: that argument is Lagrange's theorem in disguise, specialised to one group. Lagrange also explains the remark there that $\varphi(n)$ need not be the smallest exponent that works: the order of each element divides $\varphi(n)$ but can be smaller. In $(\mathbb{Z}/20\mathbb{Z})^\times$ , of order $\varphi(20) = 8$ , the elements $3, 7, 13, 17$ have order 4 and $9, 11, 19$ have order 2, so in fact $a^4 \equiv 1 \pmod{20}$ for every $a$ coprime to 20.

Application: the groups of order 4

Proposition 2.6. Every group of order 4 is cyclic or is the Klein four group $V$ .

Proof. Element orders divide 4. If some element has order 4, the group is cyclic. Otherwise every non-identity element has order 2; write $G = \{e, a, b, c\}$ . The product $ab$ is not $e$ (since $a^{-1} = a \neq b$ ), not $a$ (else $b = e$ ), and not $b$ (else $a = e$ ). So $ab = c$ , and the same argument fills in the rest of the table: it is the table of $V$ . $\blacksquare$

Index and the tower law

Proposition 2.7 (tower law). If $K \leq H \leq G$ with $G$ finite, then $[G : K] = [G : H][H : K]$ .

Proof. $\dfrac{\lvert G \rvert}{\lvert K \rvert} = \dfrac{\lvert G \rvert}{\lvert H \rvert} \cdot \dfrac{\lvert H \rvert}{\lvert K \rvert}$ . $\blacksquare$

(The law also holds for infinite groups when the indices are finite, by a direct coset argument.) Example: in $\mathbb{Z}/12\mathbb{Z} \geq \langle 2 \rangle \geq \langle 4 \rangle$ , of orders $12, 6, 3$ : $[\mathbb{Z}/12\mathbb{Z} : \langle 4 \rangle] = 4 = 2 \cdot 2$ .

Trick: if $H$ and $K$ are finite subgroups with $\gcd(\lvert H \rvert, \lvert K \rvert) = 1$ , then $H \cap K = \{e\}$ , because $\lvert H \cap K \rvert$ divides both orders. This quick consequence of Lagrange is used constantly in post 03's Sylow arguments.

Chapter 3 — The converse of Lagrange fails

Lagrange says a subgroup's order divides $\lvert G \rvert$ . The converse — "for every divisor $d$ of $\lvert G \rvert$ there is a subgroup of order $d$ " — is true for cyclic groups (post 01, Theorem 5.2) but false in general. The smallest counterexample is the alternating group $A_4$ .

$A_4$ is the group of even permutations of $\{1, 2, 3, 4\}$ , of order $4!/2 = 12$ . Sorting its elements by cycle type:

type	elements	count
identity	$e$	1
double transpositions	$(1 \ 2)(3 \ 4)$ , $(1 \ 3)(2 \ 4)$ , $(1 \ 4)(2 \ 3)$	3
3-cycles	$(1 \ 2 \ 3)$ , $(1 \ 3 \ 2)$ , $(1 \ 2 \ 4)$ , $(1 \ 4 \ 2)$ , $(1 \ 3 \ 4)$ , $(1 \ 4 \ 3)$ , $(2 \ 3 \ 4)$ , $(2 \ 4 \ 3)$	8

Proposition 3.1. $A_4$ has no subgroup of order 6, although 6 divides 12.

Proof. Suppose $H \leq A_4$ with $\lvert H \rvert = 6$ . Then $[A_4 : H] = 2$ , so there are only two left cosets, $H$ and the complement. Let $\sigma$ be any 3-cycle. The three cosets $H$ , $\sigma H$ , $\sigma^2 H$ cannot all be different, so two coincide: $\sigma^i H = \sigma^j H$ with $0 \leq i < j \leq 2$ , which by Lemma 1.1 means $\sigma^{j-i} \in H$ , with $j - i \in \{1, 2\}$ . If $\sigma^2 \in H$ then also $\sigma = \sigma^4 = (\sigma^2)^2 \in H$ . Either way $\sigma \in H$ . So $H$ contains all eight 3-cycles, too many for a set of 6 elements. $\blacksquare$

In fact $A_4$ has subgroups of orders 1, 2, 3, 4 and 12 only: three of order 2 (one per double transposition), four of order 3 (one per pair of inverse 3-cycles), and one of order 4, the Klein four group $V = \{e, (1 \ 2)(3 \ 4), (1 \ 3)(2 \ 4), (1 \ 4)(2 \ 3)\}$ .

Partial converses do hold, and they are the subject of post 03: Cauchy's theorem (if a prime $p$ divides $\lvert G \rvert$ , there is an element of order $p$ ) and the Sylow theorems (subgroups of every prime-power order dividing $\lvert G \rvert$ exist).

Chapter 4 — Normal subgroups

Terminology

The conjugate of $x$ by $g$ is $gxg^{-1}$ . For a subset, $gHg^{-1} = \{ghg^{-1} : h \in H\}$ . If $H \leq G$ then $gHg^{-1}$ is again a subgroup, of the same order.
A subgroup $N \leq G$ is normal, written $N \trianglelefteq G$ , if $gNg^{-1} \subseteq N$ for every $g \in G$ .

Equivalent conditions

Theorem 4.1. For a subgroup $N \leq G$ the following are equivalent:

$gNg^{-1} \subseteq N$ for all $g \in G$ ;
$gNg^{-1} = N$ for all $g \in G$ ;
$gN = Ng$ for all $g \in G$ ;
every left coset of $N$ is a right coset of $N$ .

Proof. (1 ⇒ 2) Apply (1) to $g^{-1}$ : $g^{-1}Ng \subseteq N$ . Conjugating by $g$ gives $N \subseteq gNg^{-1}$ . (2 ⇒ 3) $gN = gNg^{-1}g = Ng$ . (3 ⇒ 4) Clear. (4 ⇒ 1) Suppose $gN = Nx$ for some $x$ . Since $g \in gN$ , $g \in Nx$ , so $Nx$ and $Ng$ are right cosets sharing $g$ , hence equal. So $gN = Ng$ . Then for $n \in N$ , $gn \in Ng$ , so $gn = n'g$ for some $n' \in N$ and $gng^{-1} = n' \in N$ . $\blacksquare$

Watch out: $gN = Ng$ is an equality of sets. It does not say $gn = ng$ for each $n$ ; it says $gn = n'g$ for some possibly different $n' \in N$ . Normal subgroups need not commute elementwise with $G$ .

Examples

Every subgroup of an abelian group is normal: $gng^{-1} = n$ .
Index 2 subgroups are normal. If $[G : N] = 2$ , the left cosets are $N$ and $G \setminus N$ , and so are the right cosets. So for $g \notin N$ , $gN = G \setminus N = Ng$ . Hence $A_n \trianglelefteq S_n$ , the rotations $\langle r \rangle \trianglelefteq D_n$ , and $A_3 \trianglelefteq S_3$ (as the coset computation in Chapter 1 showed directly).
$SL_n(\mathbb{R}) \trianglelefteq GL_n(\mathbb{R})$ : $\det(gAg^{-1}) = \det g \cdot \det A \cdot (\det g)^{-1} = \det A = 1$ .
In $Q_8$ every subgroup is normal. The subgroups are $\{1\}$ , $\{\pm 1\}$ , $\langle i \rangle$ , $\langle j \rangle$ , $\langle k \rangle$ and $Q_8$ ; the three cyclic subgroups of order 4 have index 2, and $\{\pm 1\}$ consists of elements that commute with everything. So $Q_8$ is a non-abelian group all of whose subgroups are normal.

The conjugation rule for cycles

In $S_n$ there is a fast way to conjugate: for any $\sigma \in S_n$ ,

$\sigma \, (a_1 \ a_2 \ \dots \ a_k) \, \sigma^{-1} = (\sigma(a_1) \ \sigma(a_2) \ \dots \ \sigma(a_k)).$

Proof. The right side sends $\sigma(a_i) \mapsto \sigma(a_{i+1})$ . The left side does too: $\sigma^{-1}$ takes $\sigma(a_i)$ to $a_i$ , the cycle takes it to $a_{i+1}$ , and $\sigma$ takes that to $\sigma(a_{i+1})$ . On a point $\sigma(b)$ with $b$ outside the cycle, both sides give $\sigma(b)$ . $\blacksquare$

So conjugation relabels the entries and keeps the cycle type. Consequences:

Non-example. $H = \{e, (2 \ 3)\}$ is not normal in $S_3$ : with $\sigma = (1 \ 2 \ 3)$ , $\sigma (2 \ 3) \sigma^{-1} = (\sigma(2) \ \sigma(3)) = (3 \ 1) = (1 \ 3) \notin H$ . This matches the different left and right cosets found in Chapter 1.
$V \trianglelefteq S_4$ : $V$ consists of $e$ and all three permutations of type "two disjoint transpositions", and conjugation preserves type.
Normality is not transitive. $K = \{e, (1 \ 2)(3 \ 4)\}$ is normal in $V$ (index 2, or: $V$ is abelian), and $V \trianglelefteq A_4$ , but $K$ is not normal in $A_4$ : $(1 \ 2 \ 3)(1 \ 2)(3 \ 4)(1 \ 2 \ 3)^{-1} = (2 \ 3)(1 \ 4) = (1 \ 4)(2 \ 3) \notin K$ .

Worked example: the normal subgroups of $D_4$

$D_4$ has ten subgroups (post 01). Sort them by order and test each:

order	subgroups	normal?	reason
1	$\{e\}$	yes	trivial
2	$\{e, r^2\}$	yes	$r^2$ is in the center: $r^2$ commutes with $r$ , and $s r^2 s = r^{-2} = r^2$
2	$\{e, s\}$ , $\{e, rs\}$ , $\{e, r^2 s\}$ , $\{e, r^3 s\}$	no	$r s r^{-1} = r^2 s$ , and in general $r (r^k s) r^{-1} = r^{k+2} s$
4	$\langle r \rangle$ , $\{e, r^2, s, r^2 s\}$ , $\{e, r^2, rs, r^3 s\}$	yes	index 2
8	$D_4$	yes	whole group

So $D_4$ has six normal subgroups out of ten. The computation $r (r^k s) r^{-1} = r^{k+1} (s r^{-1}) = r^{k+1} r s = r^{k+2} s$ uses $s r^{-1} = r s$ , which is $srs = r^{-1}$ read with $r$ replaced by $r^{-1}$ .

Products of subgroups

For subgroups $H, K \leq G$ let $HK = \{hk : h \in H, k \in K\}$ .

Proposition 4.2. If $H$ and $K$ are finite, $\lvert HK \rvert = \dfrac{\lvert H \rvert \lvert K \rvert}{\lvert H \cap K \rvert}$ . If moreover one of $H, K$ is normal in $G$ , then $HK$ is a subgroup.

Proof. $HK$ is the union of the left cosets $hK$ for $h \in H$ , each of size $\lvert K \rvert$ . Two of them coincide, $h_1 K = h_2 K$ , exactly when $h_1^{-1} h_2 \in K$ , that is (both being in $H$ ) when $h_1^{-1} h_2 \in H \cap K$ , i.e. when $h_1$ and $h_2$ lie in the same left coset of $H \cap K$ in $H$ . So the number of distinct cosets $hK$ is $[H : H \cap K] = \lvert H \rvert / \lvert H \cap K \rvert$ , giving the formula. If $K \trianglelefteq G$ , then $hK = Kh$ for each $h$ , so $HK = KH$ , and $(h_1 k_1)(h_2 k_2)^{-1} = h_1 k_1 k_2^{-1} h_2^{-1} \in h_1 K h_2^{-1} = h_1 h_2^{-1} K \subseteq HK$ ; the one-step test applies. The case $H \trianglelefteq G$ is symmetric. $\blacksquare$

Without normality $HK$ can fail to be a subgroup, and Lagrange detects it. In $S_3$ with $H = \{e, (1 \ 2)\}$ and $K = \{e, (1 \ 3)\}$ : $\lvert HK \rvert = 2 \cdot 2 / 1 = 4$ , namely $\{e, (1 \ 2), (1 \ 3), (1 \ 2)(1 \ 3) = (1 \ 3 \ 2)\}$ . Since 4 does not divide 6, $HK$ cannot be a subgroup.

Chapter 5 — Quotient groups

Let $N \trianglelefteq G$ . The quotient group (factor group) $G/N$ is the set of cosets of $N$ with the operation

$(aN)(bN) = (ab)N.$

The formula uses representatives $a$ and $b$ , so the first job is to check that the answer does not depend on which representatives are chosen — the same worry as for $\mathbb{Z}/n\mathbb{Z}$ in post 00.

Theorem 5.1. If $N \trianglelefteq G$ , the operation $(aN)(bN) = abN$ is well defined, and it makes $G/N$ a group with identity $eN = N$ and inverses $(aN)^{-1} = a^{-1}N$ . If $G$ is finite, $\lvert G/N \rvert = [G : N] = \lvert G \rvert / \lvert N \rvert$ .

Proof. Suppose $aN = a'N$ and $bN = b'N$ , so $a' = an_1$ and $b' = bn_2$ with $n_1, n_2 \in N$ . Then

$a'b' = a n_1 b n_2 = ab \, (b^{-1} n_1 b) \, n_2.$

By normality $b^{-1} n_1 b \in N$ , so $a'b' \in abN$ and $a'b'N = abN$ . The group axioms are inherited from $G$ : $(aN \cdot bN) \cdot cN = (ab)c N = a(bc) N = aN \cdot (bN \cdot cN)$ , $N \cdot aN = aN$ , and $aN \cdot a^{-1}N = N$ . $\blacksquare$

Where it breaks without normality

Take the non-normal $H = \{e, s\}$ in $S_3$ . Two names for the same coset: $eH = sH = H$ . Multiply each by $rH$ :

with representative $e$ : $(eH)(rH) = rH = \{(1 \ 2 \ 3), (1 \ 2)\}$ ;
with representative $s$ : $(sH)(rH) = srH = r^2 s H = r^2 H = \{(1 \ 3 \ 2), (1 \ 3)\}$ .

Same inputs, different outputs: the "operation" is not a function. Normality is exactly what is needed. Indeed, if the coset product is well defined for $H$ , then for $h \in H$ and $g \in G$ , $(hg)H = (hH)(gH) = (eH)(gH) = gH$ , so $g^{-1}hg \in H$ , and $H$ is normal.

Examples of quotients

$\mathbb{Z}/n\mathbb{Z}$ is the quotient of $\mathbb{Z}$ by the subgroup $n\mathbb{Z}$ : the cosets $a + n\mathbb{Z}$ are the congruence classes, and the coset sum is the addition of classes from post 00. This is where the notation comes from.
$S_3/A_3$ has two elements, $A_3$ (rotations) and $sA_3$ (reflections). "Rotation times reflection is a reflection; reflection times reflection is a rotation": the quotient is $\mathbb{Z}/2\mathbb{Z}$ . Likewise $S_n / A_n \cong \mathbb{Z}/2\mathbb{Z}$ records only the sign.
$(\mathbb{Z}/12\mathbb{Z})/\langle 4 \rangle$ : the cosets of $\{0, 4, 8\}$ are $\{0, 4, 8\}$ , $\{1, 5, 9\}$ , $\{2, 6, 10\}$ , $\{3, 7, 11\}$ . The coset $1 + \langle 4 \rangle$ has order 4 (the first multiple of 1 landing in $\{0, 4, 8\}$ is $4$ ), so the quotient is cyclic of order 4.
$Q_8 / \{\pm 1\}$ has order 4. Every $q \in Q_8$ has $q^2 = \pm 1$ , so every coset squares to the identity coset, and by Proposition 2.6 the quotient is $V$ .
$\mathbb{R}/\mathbb{Z}$ : cosets $x + \mathbb{Z}$ are "real numbers up to whole turns", added mod 1. It behaves like the circle group $S^1$ via $x + \mathbb{Z} \mapsto e^{2\pi i x}$ (post 03 makes this an isomorphism with the first isomorphism theorem).

A quotient table: $D_4 / Z(D_4)$

The center of $D_4$ is $Z = \{e, r^2\}$ (Chapter 6 below; it is the normal subgroup of order 2 found in Chapter 4). Its four cosets are

$Z = \{e, r^2\}, \quad rZ = \{r, r^3\}, \quad sZ = \{s, r^2 s\}, \quad rsZ = \{rs, r^3 s\}.$

Multiply cosets through representatives, then look up which coset the product lies in:

$\cdot$	$Z$	$rZ$	$sZ$	$rsZ$
$Z$	$Z$	$rZ$	$sZ$	$rsZ$
$rZ$	$rZ$	$Z$	$rsZ$	$sZ$
$sZ$	$sZ$	$rsZ$	$Z$	$rZ$
$rsZ$	$rsZ$	$sZ$	$rZ$	$Z$

For instance $(sZ)(rZ) = srZ = r^3 s Z$ , and $r^3 s \in rsZ$ ; while $(rZ)(sZ) = rsZ$ . In $D_4$ , $sr \neq rs$ , but in the quotient the two products land in the same coset: the quotient is abelian, a copy of $V$ . Passing to $G/N$ forgets the information inside $N$ , here the difference between $r^3 s$ and $rs$ .

The correspondence theorem (statement)

Theorem 5.2 (correspondence). Let $N \trianglelefteq G$ . The subgroups of $G/N$ correspond one-to-one to the subgroups $H$ of $G$ that contain $N$ , via $H \mapsto H/N$ . The correspondence preserves inclusion, and $H/N$ is normal in $G/N$ exactly when $H$ is normal in $G$ .

Example: the subgroups of $\mathbb{Z}/12\mathbb{Z}$ containing $\langle 4 \rangle = \{0, 4, 8\}$ are $\langle 4 \rangle$ , $\langle 2 \rangle$ and the whole group, of orders 3, 6 and 12. They match the three subgroups of the quotient $(\mathbb{Z}/12\mathbb{Z})/\langle 4 \rangle \cong \mathbb{Z}/4\mathbb{Z}$ , of orders $3/3 = 1$ , $6/3 = 2$ and $12/3 = 4$ . The proof is cleanest with homomorphisms and is part of post 03.

Order of an element in $G/N$ . $gN$ has order the smallest $k \geq 1$ with $g^k \in N$ , since $(gN)^k = g^k N$ equals $N$ exactly when $g^k \in N$ . The order of $gN$ therefore divides the order of $g$ .

Common pitfall: $G/N$ is not a subgroup of $G$ . Its elements are subsets of $G$ (cosets), and its identity is the whole of $N$ . Sometimes $G$ happens to contain a subgroup isomorphic to $G/N$ (for $S_3/A_3$ , the subgroup $\{e, s\}$ ), but often it does not: $Q_8/\{\pm 1\} \cong V$ , while $Q_8$ has only one element of order 2 and so contains no copy of $V$ .

Chapter 6 — Center, commutators and simple groups

The center

The center of $G$ is $Z(G) = \{z \in G : zg = gz \text{ for all } g \in G\}$ , the elements that commute with everything.

Proposition 6.1. $Z(G)$ is a normal subgroup of $G$ . $G$ is abelian if and only if $Z(G) = G$ .

Proof. $e \in Z(G)$ . If $y, z \in Z(G)$ , then $yz^{-1}g = y g z^{-1} = g y z^{-1}$ (from $zg = gz$ we get $gz^{-1} = z^{-1}g$ ). So the one-step test passes. Normality: $gzg^{-1} = zgg^{-1} = z$ . $\blacksquare$

$G$	$Z(G)$
abelian $G$	$G$
$S_3$ (and $S_n$ , $n \geq 3$ )	$\{e\}$
$D_4$	$\{e, r^2\}$
$Q_8$	$\{\pm 1\}$
$GL_n(\mathbb{R})$	non-zero scalar matrices $\lambda I$

Theorem 6.2. If $G/Z(G)$ is cyclic, then $G$ is abelian.

Proof. Write $Z = Z(G)$ and let $gZ$ generate $G/Z$ . Every element of $G$ lies in some coset $g^k Z$ , so has the form $g^k z$ with $z \in Z$ . Two such elements commute: $(g^a z_1)(g^b z_2) = g^{a+b} z_1 z_2 = (g^b z_2)(g^a z_1)$ , because the $z_i$ commute with everything and powers of $g$ commute with each other. $\blacksquare$

So for a non-abelian group, $G/Z(G)$ is never cyclic; in particular $\lvert G/Z(G) \rvert$ is never a prime. For $D_4$ : $\lvert D_4/Z(D_4) \rvert = 8/2 = 4$ , and since every $g \in D_4$ has $g^2 \in \{e, r^2\}$ , the quotient is $V$ , not cyclic, as the theorem requires.

The commutator subgroup

The commutator of $a$ and $b$ is $[a, b] = aba^{-1}b^{-1}$ . It equals $e$ exactly when $ab = ba$ , so it measures failure to commute. The commutator subgroup (derived subgroup) $G'$ is the subgroup generated by all commutators.

Theorem 6.3. $G' \trianglelefteq G$ , the quotient $G/G'$ is abelian, and for any normal subgroup $N$ : $G/N$ is abelian if and only if $G' \leq N$ .

Proof. Conjugation respects products and $g[a, b]g^{-1} = [gag^{-1}, gbg^{-1}]$ , so a conjugate of a product of commutators is again one; hence $G'$ is normal. For $N \trianglelefteq G$ : $G/N$ is abelian iff $abN = baN$ for all $a, b$ , iff $(ba)^{-1}(ab) = a^{-1}b^{-1}ab = [a^{-1}, b^{-1}] \in N$ for all $a, b$ . As $a, b$ range over $G$ so do $a^{-1}, b^{-1}$ , so this says every commutator lies in $N$ , that is $G' \leq N$ . Taking $N = G'$ shows $G/G'$ is abelian. $\blacksquare$

$G/G'$ is the largest abelian quotient of $G$ . Examples: $G' = \{e\}$ exactly when $G$ is abelian; $S_3' = A_3$ ; $D_4' = \{e, r^2\}$ (for instance $[r, s] = rsr^{-1}s^{-1} = r \cdot r = r^2$ ); $Q_8' = \{\pm 1\}$ ; $A_4' = V$ ; $S_4' = A_4$ .

Simple groups

A group $G \neq \{e\}$ is simple if its only normal subgroups are $\{e\}$ and $G$ . A simple group has no non-trivial quotients: it cannot be broken into a normal subgroup and a smaller quotient group.

Proposition 6.4. An abelian group is simple if and only if it is cyclic of prime order.

Proof. Groups of prime order have no subgroups except $\{e\}$ and $G$ (Lagrange), so they are simple. Conversely let $G$ be abelian and simple; all its subgroups are normal, so it has none except $\{e\}$ and $G$ . For $g \neq e$ , $\langle g \rangle = G$ , so $G$ is cyclic. It is not infinite ( $\langle g^2 \rangle$ would be a proper non-trivial subgroup of $\mathbb{Z}$ ), so $G \cong \mathbb{Z}/n\mathbb{Z}$ , and $n$ must be prime since there is a subgroup for every divisor of $n$ (post 01, Theorem 5.2). $\blacksquare$

The first non-abelian simple group is $A_5$ , of order 60.

Theorem 6.5 (stated). $A_5$ is simple. More generally, $A_n$ is simple for every $n \geq 5$ .

A sketch for $A_5$ , whose missing step comes in post 03: a normal subgroup is a union of conjugacy classes (sets $\{gxg^{-1} : g \in G\}$ ), and it contains the identity. The conjugacy classes of $A_5$ have sizes $1, 12, 12, 15, 20$ . The sums that include the class $\{e\}$ of size 1 are $1, 13, 16, 21, 25, 28, 33, 36, 40, 45, 48, 60$ , and by Lagrange a subgroup's order must divide 60. Only $1$ and $60$ do. So the only normal subgroups are $\{e\}$ and $A_5$ .

Simple groups are the "atoms" of finite group theory: the Jordan–Hölder theorem says every finite group is built from a uniquely determined list of simple groups, much as an integer is built from primes. The classification of finite simple groups, completed through the work of many mathematicians, lists them all: the cyclic groups of prime order, the alternating groups $A_n$ for $n \geq 5$ , the groups of Lie type (built from matrix groups over finite fields), and 26 sporadic groups. In post 05 the simplicity of $A_5$ is the group-theoretic fact behind the theorem that the general quintic equation cannot be solved by radicals.

Practice Exercises

1. In $D_4$ (order 8, $r^4 = s^2 = e$ , $srs = r^{-1}$ ), list the left and the right cosets of $H = \{e, s\}$ . Is $H$ normal?

Solution

Left: $\{e, s\}$ , $\{r, rs\}$ , $\{r^2, r^2 s\}$ , $\{r^3, r^3 s\}$ . Right: $\{e, s\}$ , $\{r, sr\} = \{r, r^3 s\}$ , $\{r^2, sr^2\} = \{r^2, r^2 s\}$ , $\{r^3, sr^3\} = \{r^3, rs\}$ (using $sr^k = r^{-k}s$ ). Since $rH = \{r, rs\} \neq \{r, r^3 s\} = Hr$ , $H$ is not normal. Note that $[D_4 : H] = 4 = 8/2$ , as Lagrange requires.

2. Let $H$ and $K$ be subgroups of a finite group with $\gcd(\lvert H \rvert, \lvert K \rvert) = 1$ . Prove $H \cap K = \{e\}$ . Deduce that in a group of order 15, a subgroup of order 3 and a subgroup of order 5 meet only in $e$ .

Solution

$H \cap K$ is a subgroup of both $H$ and $K$ , so by Lagrange its order divides $\lvert H \rvert$ and $\lvert K \rvert$ , hence divides their gcd, 1. For the second part, 3 and 5 are coprime.

3. Without computing any powers, explain why $a^{8} \equiv 1 \pmod{20}$ for every $a$ with $\gcd(a, 20) = 1$ . Then show that the exponent 8 can be replaced by 4.

Solution

$(\mathbb{Z}/20\mathbb{Z})^\times$ has order $\varphi(20) = 8$ ; apply Corollary 2.3. Its elements are $1, 3, 7, 9, 11, 13, 17, 19$ . By Lagrange each order divides 8. None has order 8: $3^2 = 9$ , $3^4 = 81 \equiv 1$ ; $7^2 = 49 \equiv 9$ , so $7^4 \equiv 81 \equiv 1$ ; $9^2 = 81 \equiv 1$ ; $11^2 = 121 \equiv 1$ ; $13 \equiv -7$ and $17 \equiv -3$ have fourth powers equal to those of $7$ and $3$ ; $19 \equiv -1$ . So every order divides 4.

4. Use the conjugation rule to decide whether $H = \{e, (1 \ 2)(3 \ 4)\}$ is normal in $S_4$ , and whether $V$ is.

Solution

$H$ is not normal: conjugating by $(2 \ 3)$ gives $(1 \ 3)(2 \ 4) \notin H$ . $V$ is normal: it consists of $e$ and all permutations with cycle type two disjoint transpositions, and conjugation preserves cycle type.

5. In $(\mathbb{Z}/18\mathbb{Z})/\langle 6 \rangle$ , find the order of the quotient, the order of $4 + \langle 6 \rangle$ , and whether the quotient is cyclic.

Solution

$\langle 6 \rangle = \{0, 6, 12\}$ has order 3, so the quotient has order $18/3 = 6$ . The order of $4 + \langle 6 \rangle$ is the least $k$ with $4k \in \{0, 6, 12\}$ modulo 18: $4, 8, 12$ , so $k = 3$ . The coset $1 + \langle 6 \rangle$ has order 6 (the least $k$ with $k \in \{0, 6, 12\}$ is 6), so the quotient is cyclic, isomorphic to $\mathbb{Z}/6\mathbb{Z}$ .

6. Prove that $\lvert G/Z(G) \rvert$ is never a prime number.

Solution

If $\lvert G/Z(G) \rvert = p$ is prime, $G/Z(G)$ is cyclic (Corollary 2.4), so $G$ is abelian (Theorem 6.2). Then $Z(G) = G$ and $\lvert G/Z(G) \rvert = 1$ , not $p$ . Contradiction.

7. Find the commutator subgroup of $D_4$ and identify $D_4/D_4'$ .

Solution

$[r, s] = rsr^{-1}s^{-1} = r(sr^{-1}s) = r \cdot r = r^2$ , so $\{e, r^2\} \leq D_4'$ . The quotient $D_4/\{e, r^2\}$ has order 4, and every group of order 4 is abelian (Proposition 2.6), so by Theorem 6.3 $D_4' \leq \{e, r^2\}$ . Hence $D_4' = \{e, r^2\} = Z(D_4)$ , and $D_4/D_4' \cong V$ (every element squares into $\{e, r^2\}$ ).

Cheat-sheet

Idea	Statement
Left coset	$gH = \{gh : h \in H\}$
Same coset	$aH = bH$ iff $a^{-1}b \in H$
Coset facts	cosets partition $G$ ; each has $\lvert H \rvert$ elements
Lagrange	$\lvert H \rvert$ divides $\lvert G \rvert$ ; $[G:H] = \lvert G \rvert / \lvert H \rvert$
Element orders	$\lvert g \rvert$ divides $\lvert G \rvert$ ; $g^{\lvert G \rvert} = e$
Prime order	cyclic, $\cong \mathbb{Z}/p\mathbb{Z}$
Tower law	$[G:K] = [G:H][H:K]$
Converse fails	$A_4$ (order 12) has no subgroup of order 6
Normal	$gNg^{-1} = N$ iff $gN = Ng$ for all $g$
Index 2	always normal
Conjugating a cycle	$\sigma (a_1 \dots a_k) \sigma^{-1} = (\sigma(a_1) \dots \sigma(a_k))$
Quotient	$G/N$ , $(aN)(bN) = abN$ , needs $N$ normal
Order in $G/N$	least $k$ with $g^k \in N$
Center	$Z(G)$ normal; $G/Z(G)$ cyclic implies abelian
Commutators	$G/N$ abelian iff $G' \leq N$
Simple	only $\{e\}$ and $G$ normal; $A_5$ simple

What's next

Post 03 introduces homomorphisms, maps between groups that respect the operation. Their kernels are exactly the normal subgroups, and the first isomorphism theorem says $G/\ker\phi \cong \operatorname{im}\phi$ , which turns statements such as $GL_n(\mathbb{R})/SL_n(\mathbb{R}) \cong \mathbb{R}^\times$ and $\mathbb{R}/\mathbb{Z} \cong S^1$ into one-line proofs. Group actions give the orbit–stabiliser theorem and the class equation (which supplies the conjugacy class sizes used for $A_5$ above), and the Sylow theorems give the partial converse of Lagrange's theorem that Chapter 3 promised.

References

David S. Dummit and Richard M. Foote, Abstract Algebra, 3rd edition, Wiley, 2004 — Sections 3.1–3.2 (quotient groups, Lagrange's theorem) and 3.4 (composition series, Jordan–Hölder).
Joseph A. Gallian, Contemporary Abstract Algebra, 10th edition, CRC Press, 2021 — Chapters 7 (cosets and Lagrange's theorem) and 9 (normal subgroups and factor groups).
Michael Artin, Algebra, 2nd edition, Pearson, 2011 — Chapter 2, sections on cosets, normal subgroups and quotient groups.
John B. Fraleigh, A First Course in Abstract Algebra, 7th edition, Addison-Wesley, 2003 — Sections 10 (cosets and Lagrange), 14–15 (factor groups, simple groups).
Thomas W. Judson, Abstract Algebra: Theory and Applications (open textbook), judsonbooks.org — chapters on cosets and Lagrange's theorem and on normal subgroups and factor groups.
MIT OpenCourseWare, 18.701 Algebra I (Fall 2010) — lectures on cosets, Lagrange, and quotient groups.
Keith Conrad, Expository papers on group theory (University of Connecticut) — including notes on the simplicity of $A_n$ and on conjugation in $S_n$ .

Note: this post is in progress. Planned additions: a full proof that $A_n$ is simple for $n \geq 5$ , a proof of the correspondence theorem, and a drawing of the cosets of a line in $\mathbb{R}^2$ .