Abstract Algebra — 03 · Homomorphisms, Group Actions and the Sylow Theorems

Posts 01 and 02 studied groups one at a time: their elements, subgroups, cosets and quotients. This post studies the maps between groups and the ways a group can move a set. Homomorphisms explain what quotient groups really are (every quotient is the image of a homomorphism, and every image is a quotient). Group actions turn a group into a counting machine: the orbit–stabilizer theorem, the class equation and Burnside's lemma all come from one idea. The payoff is the Sylow theorems, which let you prove facts such as "every group of order 15 is cyclic" using nothing but divisibility. The chapter closes with the classification of finite abelian groups.

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

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Learning Outcome

• Define a group homomorphism, its kernel and its image, and prove that the kernel is a normal subgroup.
• Decide whether two groups are isomorphic by comparing invariants (order, element orders, abelian or not).
• Prove that $\operatorname{Aut}(\mathbb{Z}/n\mathbb{Z}) \cong (\mathbb{Z}/n\mathbb{Z})^\times$.
• State and prove the First Isomorphism Theorem, and state the Second, Third and Correspondence theorems.
• Prove Cayley's theorem: every group is a group of permutations.
• Define a group action, orbits and stabilizers, and prove the orbit–stabilizer theorem.
• Derive the class equation and use it to prove that a $p$-group has a nontrivial center.
• Count colourings up to symmetry with Burnside's lemma.
• State Cauchy's and Sylow's theorems and use them to analyse groups of order 12, 15 and 20.
• List all abelian groups of a given order, in both elementary-divisor and invariant-factor form.

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Chapter 1 — Homomorphisms

Terminology

• Homomorphism — a map $\varphi : G \to H$ between groups with $\varphi(ab) = \varphi(a)\varphi(b)$ for all $a, b \in G$. The product on the left is in $G$, the one on the right is in $H$.
• Kernel — $\ker \varphi = \{ g \in G : \varphi(g) = e_H \}$, the elements sent to the identity.
• Image — $\operatorname{im} \varphi = \varphi(G) = \{ \varphi(g) : g \in G \}$.
• Monomorphism / epimorphism — an injective / surjective homomorphism.
• Isomorphism — a bijective homomorphism. Write $G \cong H$ when one exists.
• Automorphism — an isomorphism $G \to G$. The automorphisms of $G$ form a group $\operatorname{Aut}(G)$ under composition.
• Inner automorphism — conjugation by a fixed element: $c_g(x) = gxg^{-1}$.

First properties

Let $\varphi : G \to H$ be a homomorphism. Then:

1. $\varphi(e_G) = e_H$. Proof: $\varphi(e) = \varphi(ee) = \varphi(e)\varphi(e)$; multiply both sides by $\varphi(e)^{-1}$.
2. $\varphi(g^{-1}) = \varphi(g)^{-1}$. Proof: $\varphi(g)\varphi(g^{-1}) = \varphi(gg^{-1}) = \varphi(e) = e_H$.
3. $\varphi(g^k) = \varphi(g)^k$ for every integer $k$ (induction, plus property 2 for negative $k$).
4. If $g$ has finite order $n$, then the order of $\varphi(g)$ divides $n$, because $\varphi(g)^n = \varphi(g^n) = e_H$.
5. $\ker \varphi$ is a subgroup of $G$ and $\operatorname{im}\varphi$ is a subgroup of $H$.

Property 4 is the workhorse for ruling maps out. There is no nontrivial homomorphism $\mathbb{Z}/3\mathbb{Z} \to \mathbb{Z}/4\mathbb{Z}$: the image of the generator would need an order dividing both 3 and 4, so it is the identity.

Examples worth knowing

Homomorphism	Kernel	Image
$\det : GL_n(\mathbb{R}) \to \mathbb{R}^\times$	$SL_n(\mathbb{R})$	all of $\mathbb{R}^\times$
$\operatorname{sgn} : S_n \to \{\pm 1\}$, $n \ge 2$	$A_n$	$\{\pm 1\}$
$\mathbb{Z} \to \mathbb{Z}/n\mathbb{Z}$, $k \mapsto k \bmod n$	$n\mathbb{Z}$	all of $\mathbb{Z}/n\mathbb{Z}$
$\exp : (\mathbb{R}, +) \to (\mathbb{R}_{>0}, \cdot)$	$\{0\}$	all of $\mathbb{R}_{>0}$ (isomorphism)
$(\mathbb{R}, +) \to S^1$, $t \mapsto e^{2\pi i t}$	$\mathbb{Z}$	the unit circle $S^1$
$c_g : G \to G$, $x \mapsto gxg^{-1}$	$\{e\}$	$G$ (automorphism)

Homomorphisms out of a cyclic group. A homomorphism $\varphi : \mathbb{Z}/n\mathbb{Z} \to \mathbb{Z}/m\mathbb{Z}$ is fixed by $a = \varphi(1)$, since $\varphi(k) = ka$. The only condition is that $n \cdot 1 = 0$ must go to $0$, that is $na \equiv 0 \pmod m$. The solutions are the multiples of $m/\gcd(n,m)$, so there are exactly $\gcd(n, m)$ homomorphisms. For $n = 12$, $m = 18$: $12a \equiv 0 \pmod{18}$ holds for $a \in \{0, 3, 6, 9, 12, 15\}$, which is $\gcd(12, 18) = 6$ maps.

Common pitfall: "$k \mapsto k \bmod m$" is not a well-defined map $\mathbb{Z}/n\mathbb{Z} \to \mathbb{Z}/m\mathbb{Z}$ unless $m \mid n$. In $\mathbb{Z}/4\mathbb{Z}$, $0 = 4$, but $0 \bmod 3 \ne 4 \bmod 3$. Always check that a map defined on representatives does not depend on the representative chosen.

Kernels are normal, and injectivity

Theorem. If $\varphi : G \to H$ is a homomorphism, then $\ker\varphi$ is a normal subgroup of $G$.

Proof. It is a subgroup: $\varphi(e) = e$; if $\varphi(a) = \varphi(b) = e$ then $\varphi(ab^{-1}) = \varphi(a)\varphi(b)^{-1} = e$. It is normal: for $k \in \ker\varphi$ and any $g \in G$,

$$\varphi(gkg^{-1}) = \varphi(g)\,\varphi(k)\,\varphi(g)^{-1} = \varphi(g)\,e\,\varphi(g)^{-1} = e,$$

so $gkg^{-1} \in \ker\varphi$. ∎

Theorem. $\varphi$ is injective if and only if $\ker\varphi = \{e\}$.

Proof. If $\varphi$ is injective, only $e$ maps to $e$. Conversely, if $\ker\varphi = \{e\}$ and $\varphi(a) = \varphi(b)$, then $\varphi(ab^{-1}) = e$, so $ab^{-1} = e$ and $a = b$. ∎

More precisely, the fibres of $\varphi$ are exactly the cosets of the kernel: $\varphi(a) = \varphi(b)$ if and only if $a\ker\varphi = b\ker\varphi$. This is the whole content of the First Isomorphism Theorem in Chapter 2.

Isomorphism invariants

Isomorphic groups are "the same group with renamed elements", so any property defined only from the group operation is shared. To prove two groups are not isomorphic, find a property one has and the other lacks:

• the order $|G|$;
• being abelian or cyclic;
• the number of elements of each order;
• the size of the center, the number of subgroups of a given order, and so on.

$\mathbb{Z}/4\mathbb{Z}$ has an element of order 4; in $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$ every non-identity element has order 2. So they are not isomorphic, although both have order 4 and are abelian. On the other hand $\mathbb{Z}/6\mathbb{Z} \cong \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z}$ via $k \mapsto (k \bmod 2, k \bmod 3)$; this is the Chinese Remainder Theorem from post 00. In general $\mathbb{Z}/m\mathbb{Z} \times \mathbb{Z}/n\mathbb{Z}$ is cyclic if and only if $\gcd(m, n) = 1$.

Watch out: to prove two groups are isomorphic you must produce a map and check it. Matching every invariant you can think of is evidence, not proof.

Automorphisms of cyclic groups

Theorem. $\operatorname{Aut}(\mathbb{Z}/n\mathbb{Z}) \cong (\mathbb{Z}/n\mathbb{Z})^\times$, the group of units modulo $n$. In particular $\lvert\operatorname{Aut}(\mathbb{Z}/n\mathbb{Z})\rvert = \phi(n)$, Euler's totient function from post 00. (This post writes Euler's function as $\phi$ to keep $\varphi$ free for homomorphisms.)

Proof. Every endomorphism of $\mathbb{Z}/n\mathbb{Z}$ is $\mu_a(k) = ak$ for $a = \mu_a(1)$ (see "Homomorphisms out of a cyclic group" with $m = n$). The map $\mu_a$ is a bijection exactly when its image contains $1$, that is when $ab \equiv 1 \pmod n$ for some $b$, that is when $a$ is a unit. Define $\Psi : (\mathbb{Z}/n\mathbb{Z})^\times \to \operatorname{Aut}(\mathbb{Z}/n\mathbb{Z})$, $a \mapsto \mu_a$. It is surjective by what was just said and injective since $\mu_a(1) = a$ recovers $a$. It is a homomorphism because $\mu_a \circ \mu_b (k) = abk = \mu_{ab}(k)$. ∎

Example: $(\mathbb{Z}/8\mathbb{Z})^\times = \{1, 3, 5, 7\}$ and each element squares to $1$ ($3^2 = 9$, $5^2 = 25$, $7^2 = 49$ are all $\equiv 1$). So $\operatorname{Aut}(\mathbb{Z}/8\mathbb{Z}) \cong \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$, not cyclic. In the other direction, $\operatorname{Aut}(\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}) \cong S_3$: an automorphism may permute the three non-identity elements in any of the $3! = 6$ ways.

Inner automorphisms. The map $G \to \operatorname{Aut}(G)$, $g \mapsto c_g$, is a homomorphism ($c_g c_h = c_{gh}$). Its kernel is the set of $g$ with $gxg^{-1} = x$ for all $x$, which is the center $Z(G)$ from post 02. Its image is $\operatorname{Inn}(G)$. By the First Isomorphism Theorem (next chapter), $\operatorname{Inn}(G) \cong G/Z(G)$. For $D_4$ (order 8) the center is $\{e, r^2\}$, so $\operatorname{Inn}(D_4)$ has order $8/2 = 4$.

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Chapter 2 — The isomorphism theorems

The First Isomorphism Theorem

Theorem (First Isomorphism Theorem). Let $\varphi : G \to H$ be a homomorphism with kernel $K$. Then $K$ is normal in $G$ and

$$G/K \;\cong\; \operatorname{im}\varphi, \qquad gK \mapsto \varphi(g).$$

Proof. Normality of $K$ was proved in Chapter 1. Define $\bar\varphi : G/K \to \operatorname{im}\varphi$ by $\bar\varphi(gK) = \varphi(g)$.

• Well defined. If $gK = g'K$ then $g' = gk$ for some $k \in K$, so $\varphi(g') = \varphi(g)\varphi(k) = \varphi(g)$.
• Homomorphism. $\bar\varphi(gK \cdot g'K) = \bar\varphi(gg'K) = \varphi(gg') = \varphi(g)\varphi(g') = \bar\varphi(gK)\,\bar\varphi(g'K)$.
• Injective. If $\bar\varphi(gK) = e$ then $\varphi(g) = e$, so $g \in K$ and $gK = K$, the identity of $G/K$. A homomorphism with trivial kernel is injective.
• Surjective. Every element of $\operatorname{im}\varphi$ is $\varphi(g) = \bar\varphi(gK)$. ∎

Read it as a factorisation: every homomorphism is "quotient map, then isomorphism, then inclusion", $G \to G/K \xrightarrow{\;\cong\;} \operatorname{im}\varphi \hookrightarrow H$. For finite $G$ it also gives the counting rule $|G| = |\ker\varphi| \cdot |\operatorname{im}\varphi|$.

Worked applications of the table in Chapter 1:

• $GL_n(\mathbb{R}) / SL_n(\mathbb{R}) \cong \mathbb{R}^\times$ (via $\det$).
• $S_n / A_n \cong \{\pm 1\} \cong \mathbb{Z}/2\mathbb{Z}$ (via $\operatorname{sgn}$), so $[S_n : A_n] = 2$ and $|A_n| = n!/2$.
• $\mathbb{Z}/n\mathbb{Z}$ really is $\mathbb{Z}$ modulo the kernel $n\mathbb{Z}$ of reduction.
• $\mathbb{R}/\mathbb{Z} \cong S^1$ (via $t \mapsto e^{2\pi i t}$): "the real line wrapped around a circle".

Trick: to identify a quotient $G/N$, do not work with cosets directly. Find a surjective homomorphism out of $G$ whose kernel is exactly $N$; its target is the answer.

The Second and Third Isomorphism Theorems

Theorem (Second, or "diamond"). Let $H \le G$ and $N \trianglelefteq G$. Then $HN$ is a subgroup of $G$, $H \cap N$ is normal in $H$, and

$$HN/N \;\cong\; H/(H \cap N).$$

(Proof idea: $H \to HN/N$, $h \mapsto hN$, is surjective with kernel $H \cap N$; apply the First theorem.) For finite groups it gives $|HN| = |H|\,|N| / |H \cap N|$.

Theorem (Third). Let $N \subseteq K$ be normal subgroups of $G$. Then $K/N$ is normal in $G/N$ and

$$(G/N)\big/(K/N) \;\cong\; G/K.$$

(Proof idea: $G/N \to G/K$, $gN \mapsto gK$, is well defined and surjective with kernel $K/N$.) Example: $(\mathbb{Z}/12\mathbb{Z}) / (4\mathbb{Z}/12\mathbb{Z}) \cong \mathbb{Z}/4\mathbb{Z}$. The subgroup $4\mathbb{Z}/12\mathbb{Z} = \{0, 4, 8\}$ has order 3, and $12/3 = 4$.

Theorem (Correspondence). Let $N \trianglelefteq G$ and $\pi : G \to G/N$ the quotient map. Then $K \mapsto K/N = \pi(K)$ is a bijection between subgroups $K$ of $G$ that contain $N$ and subgroups of $G/N$. It preserves inclusion, indices ($[G:K] = [G/N : K/N]$) and normality.

So the subgroup lattice of $G/N$ is exactly the part of the lattice of $G$ that sits above $N$.

Cayley's theorem

Theorem (Cayley). Every group $G$ is isomorphic to a subgroup of the symmetric group $\operatorname{Sym}(G)$. If $|G| = n$, then $G$ is isomorphic to a subgroup of $S_n$.

Proof. For $g \in G$ let $\lambda_g : G \to G$, $\lambda_g(x) = gx$. It is a bijection with inverse $\lambda_{g^{-1}}$, so $\lambda_g \in \operatorname{Sym}(G)$. The map $\Lambda : G \to \operatorname{Sym}(G)$, $g \mapsto \lambda_g$, is a homomorphism because $\lambda_g\lambda_h(x) = g(hx) = (gh)x = \lambda_{gh}(x)$. Its kernel is trivial: if $\lambda_g$ is the identity then $g = ge = \lambda_g(e) = e$. By the First Isomorphism Theorem $G \cong \operatorname{im}\Lambda \le \operatorname{Sym}(G)$. ∎

Cayley's theorem is historically important (it says the abstract definition of a group loses nothing compared with "groups of permutations") but it is wasteful: it embeds a group of order 8 in $S_8$, of order 40320. Group actions on smaller sets give much better information, which is the next chapter.

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Chapter 3 — Group actions

Terminology

• Action — a group $G$ acts on a set $X$ if there is a map $G \times X \to X$, $(g, x) \mapsto g \cdot x$, with $e \cdot x = x$ and $g \cdot (h \cdot x) = (gh) \cdot x$.
• Permutation representation — equivalently, a homomorphism $\rho : G \to \operatorname{Sym}(X)$, with $\rho(g)(x) = g\cdot x$. The two descriptions are the same thing.
• Faithful action — one whose homomorphism $\rho$ is injective (only $e$ acts as the identity).
• Orbit — $G\cdot x = \{ g\cdot x : g \in G \}$, everywhere $x$ can be moved to.
• Stabilizer — $G_x = \{ g \in G : g\cdot x = x \}$, the elements that fix $x$. It is a subgroup of $G$.
• Transitive action — one with a single orbit.
• Fixed set — $X^g = \{ x \in X : g\cdot x = x \}$, the points a given $g$ fixes.

Examples

1. $S_n$ acts on $\{1, \dots, n\}$ by $\sigma\cdot i = \sigma(i)$. Transitive and faithful.
2. $D_4$ (order 8, the symmetries of a square, notation from post 01) acts on the four vertices. Label them $1, 2, 3, 4$ counter-clockwise; then $r = (1\,2\,3\,4)$ and, taking $s$ to be the reflection in the diagonal through vertices 1 and 3, $s = (2\,4)$.
3. The rotation group of a cube (24 elements) acts on the 6 faces, the 8 vertices and the 12 edges.
4. Any group acts on itself by left multiplication, $g\cdot x = gx$ (this is Cayley's action), and on the left cosets $G/H$ by $g\cdot xH = gxH$.
5. Any group acts on itself by conjugation, $g\cdot x = gxg^{-1}$. Orbits are conjugacy classes $\operatorname{cl}(x)$, and the stabilizer of $x$ is the centralizer $C_G(x) = \{ g : gx = xg \}$.

The orbits of an action partition $X$: the relation "$y = g\cdot x$ for some $g$" is an equivalence relation (reflexive via $e$, symmetric via $g^{-1}$, transitive via products), and orbits are its classes (equivalence relations are from post 00).

The orbit–stabilizer theorem

Theorem (orbit–stabilizer). Let a finite group $G$ act on $X$ and let $x \in X$. Then

$$|G\cdot x| \;=\; [G : G_x] \;=\; \frac{|G|}{|G_x|}.$$

Proof. Define $f$ from the set of left cosets of $G_x$ to the orbit by $f(gG_x) = g\cdot x$.

• Well defined and injective. For $g, h \in G$: $g\cdot x = h\cdot x \iff h^{-1}g\cdot x = x \iff h^{-1}g \in G_x \iff gG_x = hG_x$. Reading the chain left to right gives injectivity; right to left gives well-definedness.
• Surjective. Every orbit element is $g\cdot x = f(gG_x)$.

So the orbit is in bijection with the cosets of $G_x$, and there are $[G:G_x] = |G|/|G_x|$ of them by Lagrange's theorem (post 02). ∎

Worked example: $D_4$ on the square. The orbit of vertex 1 is all four vertices (rotations reach each one). The stabilizer of vertex 1 is $\{e, s\}$: the only non-identity symmetry fixing vertex 1 is the reflection in the diagonal through it. So $|G\cdot 1| \cdot |G_1| = 4 \cdot 2 = 8 = |D_4|$. The proof's bijection is concrete here. The four cosets of $\{e, s\}$ are $\{e, s\}$, $\{r, rs\}$, $\{r^2, r^2s\}$, $\{r^3, r^3s\}$, and both elements of $r^k\{e, s\}$ send vertex 1 to vertex $k+1$.

Worked example: the cube. The rotation group acts transitively on faces, vertices and edges. A face is fixed by the 4 rotations about the axis through its centre; a vertex by the 3 rotations about the long diagonal through it; an edge by the identity and the half-turn about the axis through its midpoint. All three counts agree:

$$6 \cdot 4 \;=\; 8 \cdot 3 \;=\; 12 \cdot 2 \;=\; 24.$$

This is also the standard way to find the order of a symmetry group: pick an object, count where it can go, and count what fixes it.

Trick: points in the same orbit have conjugate stabilizers: $G_{g\cdot x} = g\,G_x\,g^{-1}$. So all stabilizers along one orbit have the same size, as the formula requires.

The class equation

Apply orbit–stabilizer to conjugation. The orbit of $x$ is $\operatorname{cl}(x)$ and its stabilizer is $C_G(x)$, so $|\operatorname{cl}(x)| = [G : C_G(x)]$, which divides $|G|$. An element has a class of size 1 exactly when it commutes with everything, that is when $x \in Z(G)$. Splitting $G$ into classes and collecting the size-1 classes gives the class equation:

$$|G| \;=\; |Z(G)| \;+\; \sum_{i=1}^{k} [G : C_G(x_i)],$$

where $x_1, \dots, x_k$ are representatives of the classes with more than one element. Each term in the sum is a divisor of $|G|$ bigger than 1.

Group	Order	Class sizes	Center
$S_3$	6	1, 2, 3	trivial
$D_4$	8	1, 1, 2, 2, 2	$\{e, r^2\}$
$A_4$	12	1, 3, 4, 4	trivial
$S_4$	24	1, 3, 6, 6, 8	trivial
$A_5$	60	1, 12, 12, 15, 20	trivial

For $D_4$ the classes are $\{e\}$, $\{r^2\}$, $\{r, r^3\}$, $\{s, r^2s\}$, $\{rs, r^3s\}$. In $S_n$ two permutations are conjugate exactly when they have the same cycle type; in $S_4$ the classes are the identity (1), double transpositions (3), transpositions (6), 4-cycles (6), 3-cycles (8).

p-groups have a nontrivial center

A $p$-group is a group of order $p^n$ for a prime $p$ and some $n \ge 1$.

Theorem. If $|G| = p^n$ with $n \ge 1$, then $Z(G) \ne \{e\}$.

Proof. In the class equation, each $[G : C_G(x_i)]$ is a divisor of $p^n$ bigger than 1, so it is divisible by $p$. The left side $p^n$ is divisible by $p$. Hence $|Z(G)| = p^n - \sum [G : C_G(x_i)]$ is divisible by $p$. Since $e \in Z(G)$, $|Z(G)| \ge 1$, so $|Z(G)| \ge p$. ∎

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

Proof. Let $gZ(G)$ generate $G/Z(G)$. Every element of $G$ has the form $g^i z$ with $z \in Z(G)$. Two such elements commute: $g^i z \cdot g^j w = g^{i+j} z w = g^j w \cdot g^i z$, since $z, w$ commute with everything. ∎

Corollary. Every group of order $p^2$ is abelian, hence isomorphic to $\mathbb{Z}/p^2\mathbb{Z}$ or $\mathbb{Z}/p\mathbb{Z} \times \mathbb{Z}/p\mathbb{Z}$.

Proof. $|Z(G)|$ is $p$ or $p^2$. If it were $p$, then $G/Z(G)$ would have order $p$, hence be cyclic, hence $G$ abelian by the lemma, so $Z(G) = G$ has order $p^2$: a contradiction. So $Z(G) = G$. The two possibilities come from Chapter 5. ∎

$D_4$ shows that the corollary stops at $p^2$: it has order $2^3$, is not abelian, and its center has order $2$.

Burnside's lemma

Theorem (Burnside's lemma, also called the Cauchy–Frobenius lemma). If a finite group $G$ acts on a finite set $X$, the number of orbits is

$$\#\text{orbits} \;=\; \frac{1}{|G|} \sum_{g \in G} |X^g|,$$

the average number of points fixed by an element.

Proof. Count the pairs $(g, x)$ with $g\cdot x = x$ in two ways. By $g$ first: $\sum_g |X^g|$. By $x$ first: $\sum_x |G_x| = \sum_x |G|/|G\cdot x|$. Group the $x$ by orbits. An orbit $O$ contributes $|O| \cdot |G|/|O| = |G|$. So the total is $|G|$ times the number of orbits. ∎

The typical use is counting colourings "up to symmetry". If $g$ permutes the positions with $c(g)$ cycles, a colouring with $k$ colours is fixed by $g$ exactly when it is constant on each cycle, so $|X^g| = k^{c(g)}$.

Worked example: necklaces. Colour 6 beads on a circle black or white; two necklaces are the same if one is a rotation of the other. The group is the cyclic group of rotations by $0, 1, \dots, 5$ steps. A rotation by $j$ steps has $\gcd(j, 6)$ cycles on the beads: $6, 1, 2, 3, 2, 1$ cycles for $j = 0, \dots, 5$. So

$$\#\text{necklaces} = \frac{2^6 + 2^1 + 2^2 + 2^3 + 2^2 + 2^1}{6} = \frac{64 + 2 + 4 + 8 + 4 + 2}{6} = \frac{84}{6} = 14.$$

If flips are also allowed (the dihedral group of order 12, so "bracelets"), the count drops to 13. With 3 colours and rotations only, the same formula gives $(729 + 3 + 9 + 27 + 9 + 3)/6 = 780/6 = 130$.

Worked example: colouring a cube's faces. The 24 rotations of the cube, sorted by type, with the number of cycles each makes on the 6 faces:

Rotation type	How many	Cycles on faces
identity	1	6
90° or 270° about a face axis	6	3 (two fixed faces, one 4-cycle)
180° about a face axis	3	4 (two fixed faces, two 2-cycles)
120° or 240° about a vertex diagonal	8	2 (two 3-cycles)
180° about an edge axis	6	3 (three 2-cycles)

With 3 colours:

$$\frac{3^6 + 6\cdot 3^3 + 3\cdot 3^4 + 8\cdot 3^2 + 6\cdot 3^3}{24} = \frac{729 + 162 + 243 + 72 + 162}{24} = \frac{1368}{24} = 57.$$

With 2 colours the same table gives $(64 + 48 + 48 + 32 + 48)/24 = 240/24 = 10$.

Common pitfall: dividing the number of colourings by $|G|$ ($3^6/24$ is not even an integer) is wrong because symmetric colourings have smaller orbits. Burnside's average of fixed points corrects for exactly this.

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Chapter 4 — Cauchy and Sylow

Lagrange's theorem says the order of a subgroup divides $|G|$. The converse is false: $A_4$ has order 12 but no subgroup of order 6. The theorems in this chapter recover as much of the converse as is true, for prime powers.

Terminology

• Sylow $p$-subgroup — if $|G| = p^k m$ with $p$ prime and $p \nmid m$, a subgroup of order exactly $p^k$ (the largest power of $p$ allowed by Lagrange).
• $n_p$ — the number of Sylow $p$-subgroups of $G$.
• Normalizer — $N_G(H) = \{ g : gHg^{-1} = H \}$, the stabilizer of $H$ when $G$ acts on its subgroups by conjugation.
• Simple group — a nontrivial group whose only normal subgroups are $\{e\}$ and itself (post 02).

The theorems

Theorem (Cauchy). If a prime $p$ divides $|G|$, then $G$ has an element of order $p$.

(A short proof, due to McKay, lets $\mathbb{Z}/p\mathbb{Z}$ rotate the $p$-tuples $(g_1, \dots, g_p)$ with $g_1 g_2 \cdots g_p = e$ and counts fixed points modulo $p$.)

Theorem (Sylow). Let $|G| = p^k m$ with $p$ prime, $k \ge 1$ and $p \nmid m$.

1. Existence. $G$ has a subgroup of order $p^k$. More generally, every $p$-subgroup of $G$ is contained in some Sylow $p$-subgroup.
2. Conjugacy. Any two Sylow $p$-subgroups are conjugate: if $P, Q$ are Sylow, then $Q = gPg^{-1}$ for some $g \in G$.
3. Counting. $n_p \equiv 1 \pmod p$, and $n_p$ divides $m$. Moreover $n_p = [G : N_G(P)]$ for any Sylow $p$-subgroup $P$.

Two consequences are used constantly:

• $n_p = 1$ if and only if the Sylow $p$-subgroup is normal. (Conjugates of a Sylow subgroup are Sylow, and by part 2 all Sylow subgroups are conjugates of one of them.)
• When $p$ divides $|G|$ only once, the Sylow $p$-subgroups have prime order $p$. Two distinct subgroups of prime order meet only in $\{e\}$, so the $n_p$ of them contain $n_p(p-1)$ elements of order $p$.

Sylow counts in familiar groups (checked by listing subgroups):

Group	Order	$n_2$	$n_3$	$n_5$
$S_3$	$2 \cdot 3$	3	1	—
$A_4$	$2^2 \cdot 3$	1	4	—
$S_4$	$2^3 \cdot 3$	3	4	—
$A_5$	$2^2 \cdot 3 \cdot 5$	5	10	6

Every entry obeys part 3: for $A_5$, $n_5 = 6 \equiv 1 \pmod 5$ and $6 \mid 12$; $n_3 = 10 \equiv 1 \pmod 3$ and $10 \mid 20$.

Application 1 — every group of order 15 is cyclic

Let $|G| = 15 = 3 \cdot 5$.

• $n_3$ divides 5 and $n_3 \equiv 1 \pmod 3$. The divisors of 5 are 1 and 5, and $5 \equiv 2$. So $n_3 = 1$.
• $n_5$ divides 3 and $n_5 \equiv 1 \pmod 5$. The divisors of 3 are 1 and 3, and $3 \not\equiv 1$. So $n_5 = 1$.

So $G$ has a normal subgroup $P$ of order 3 and a normal subgroup $Q$ of order 5. Their intersection is a subgroup of both, so its order divides $\gcd(3, 5) = 1$: $P \cap Q = \{e\}$. For $a \in P$, $b \in Q$, the commutator $aba^{-1}b^{-1}$ lies in $Q$ (as $aba^{-1} \in Q$) and in $P$ (as $ba^{-1}b^{-1} \in P$), so it is $e$ and $ab = ba$. Then $ab$ has order $\operatorname{lcm}(3, 5) = 15$, and $G = \langle ab \rangle \cong \mathbb{Z}/15\mathbb{Z}$.

The same argument proves: if $p < q$ are primes and $p \nmid q - 1$, every group of order $pq$ is cyclic. (For $p = 2$, $q = 3$ it fails, correctly: $S_3$ has order 6 and $2 \mid 3 - 1$.)

Application 2 — no group of order 12 is simple

Let $|G| = 12 = 2^2 \cdot 3$. Then $n_3 \mid 4$ with $n_3 \equiv 1 \pmod 3$, so $n_3 \in \{1, 4\}$; and $n_2 \mid 3$ with $n_2$ odd, so $n_2 \in \{1, 3\}$.

• If $n_3 = 1$, the Sylow 3-subgroup is normal and $G$ is not simple.
• If $n_3 = 4$, the four subgroups of order 3 contain $4 \cdot 2 = 8$ elements of order 3. That leaves $12 - 8 = 4$ elements. A Sylow 2-subgroup has 4 elements, none of order 3, so it must be exactly these 4 elements. There is room for only one, so $n_2 = 1$ and it is normal.

Either way $G$ has a proper nontrivial normal subgroup. $A_4$ is the second case: $n_3 = 4$ and $n_2 = 1$, the normal subgroup being $V_4 = \{e, (1\,2)(3\,4), (1\,3)(2\,4), (1\,4)(2\,3)\}$.

Application 3 — order 20

If $|G| = 20 = 2^2 \cdot 5$, then $n_5 \mid 4$ and $n_5 \equiv 1 \pmod 5$ force $n_5 = 1$. So every group of order 20 has a normal subgroup of order 5 and is not simple.

Trick: the recipe for "no simple group of order $n$" is: list the allowed $n_p$ for each prime; if some prime forces $n_p = 1$, you are done; if not, count elements of prime order and look for a contradiction, as for order 12.

Watch out: Sylow part 3 gives only candidates. $n_2 \in \{1, 3\}$ for order 12 does not mean both values occur in every group; $A_4$ has $n_2 = 1$, while the dihedral group of order 12 has $n_2 = 3$.

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Chapter 5 — Finite abelian groups

For abelian groups the story is complete: they are classified by lists of numbers.

Theorem (fundamental theorem of finite abelian groups). Every finite abelian group $A$ is isomorphic to a direct product of cyclic groups of prime-power order,

$$A \;\cong\; \mathbb{Z}/p_1^{a_1}\mathbb{Z} \times \cdots \times \mathbb{Z}/p_r^{a_r}\mathbb{Z},$$

and the list of prime powers $p_i^{a_i}$ (the elementary divisors) is unique up to order. Equivalently, $A \cong \mathbb{Z}/d_1\mathbb{Z} \times \cdots \times \mathbb{Z}/d_s\mathbb{Z}$ with $d_1 \mid d_2 \mid \cdots \mid d_s$ and $d_1 > 1$, and the invariant factors $d_i$ are unique.

To go from elementary divisors to invariant factors, write the prime powers of each prime in decreasing order, then multiply across: the largest power of every prime gives $d_s$, the next largest gives $d_{s-1}$, and so on. This uses $\mathbb{Z}/m\mathbb{Z} \times \mathbb{Z}/n\mathbb{Z} \cong \mathbb{Z}/mn\mathbb{Z}$ for coprime $m, n$ (Chapter 1).

Counting. The number of abelian groups of order $p_1^{e_1} \cdots p_r^{e_r}$ is the product of the partition numbers $p(e_1) \cdots p(e_r)$, since each prime's part is chosen independently as a partition of its exponent.

Worked example: order 72. $72 = 2^3 \cdot 3^2$. The exponent 3 has three partitions ($3$, $2+1$, $1+1+1$) and the exponent 2 has two ($2$, $1+1$), so there are $3 \cdot 2 = 6$ abelian groups of order 72:

Elementary divisors	Invariant factors
$8, 9$	$\mathbb{Z}/72$
$8, 3, 3$	$\mathbb{Z}/3 \times \mathbb{Z}/24$
$4, 2, 9$	$\mathbb{Z}/2 \times \mathbb{Z}/36$
$4, 2, 3, 3$	$\mathbb{Z}/6 \times \mathbb{Z}/12$
$2, 2, 2, 9$	$\mathbb{Z}/2 \times \mathbb{Z}/2 \times \mathbb{Z}/18$
$2, 2, 2, 3, 3$	$\mathbb{Z}/2 \times \mathbb{Z}/6 \times \mathbb{Z}/6$

(Here $\mathbb{Z}/n$ abbreviates $\mathbb{Z}/n\mathbb{Z}$.) Check one row: for $4, 2, 3, 3$ the powers of 2 in decreasing order are $4, 2$ and of 3 are $3, 3$; multiplying across gives $4\cdot 3 = 12$ and $2 \cdot 3 = 6$, so $\mathbb{Z}/6 \times \mathbb{Z}/12$, and indeed $6 \mid 12$.

Common pitfall: $\mathbb{Z}/2 \times \mathbb{Z}/36$ and $\mathbb{Z}/4 \times \mathbb{Z}/18$ are the same group (both have elementary divisors $4, 2, 9$). Always reduce to one normal form before comparing.

The same theorem, stated for finitely generated modules over a principal ideal domain, is the structure theory behind the Jordan normal form; it is planned as a later post.

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Practice Exercises

1. How many homomorphisms are there from $\mathbb{Z}/20\mathbb{Z}$ to $\mathbb{Z}/8\mathbb{Z}$? List the possible images of $1$.

Solution

A homomorphism is fixed by $a = \varphi(1)$, subject to $20a \equiv 0 \pmod 8$, i.e. $4a \equiv 0 \pmod 8$, i.e. $a$ even. So $a \in \{0, 2, 4, 6\}$: four homomorphisms, which is $\gcd(20, 8) = 4$.

2. Show that $\mathbb{Z}/8\mathbb{Z}$, $\mathbb{Z}/4\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$ and $(\mathbb{Z}/2\mathbb{Z})^3$ are pairwise non-isomorphic.

Solution

Compare the largest element order. $\mathbb{Z}/8\mathbb{Z}$ has an element of order 8. In $\mathbb{Z}/4\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$ the order of $(a, b)$ is $\operatorname{lcm}$ of the orders, at most $\operatorname{lcm}(4, 2) = 4$, and $(1, 0)$ has order 4. In $(\mathbb{Z}/2\mathbb{Z})^3$ every non-identity element has order 2. The maximal orders 8, 4, 2 differ, so no two are isomorphic. (These are the three abelian groups of order 8, from the partitions $3$, $2+1$, $1+1+1$.)

3. $S_4$ acts on the set of its three "pairings" $\{\{1,2\},\{3,4\}\}$, $\{\{1,3\},\{2,4\}\}$, $\{\{1,4\},\{2,3\}\}$. Use orbit–stabilizer to find the order of the stabilizer of one pairing, and find the kernel of the action.

Solution

The action is transitive (for instance $(2\,3)$ swaps the first two pairings, $(2\,4)$ swaps the first and third), so the orbit has size 3 and the stabilizer has order $24/3 = 8$. It is a Sylow 2-subgroup (a copy of $D_4$). The kernel is the set of permutations fixing all three pairings: $V_4 = \{e, (1\,2)(3\,4), (1\,3)(2\,4), (1\,4)(2\,3)\}$. By the First Isomorphism Theorem, $S_4/V_4$ is isomorphic to the image in $S_3$, which has order $24/4 = 6$, so $S_4/V_4 \cong S_3$.

4. How many ways are there to colour the 4 vertices of a square with 3 colours, up to rotation? Up to all 8 symmetries of $D_4$?

Solution

Rotations by $0^\circ, 90^\circ, 180^\circ, 270^\circ$ have $4, 1, 2, 1$ cycles on the vertices: $(3^4 + 3 + 3^2 + 3)/4 = (81 + 3 + 9 + 3)/4 = 96/4 = 24$.

For $D_4$ add the reflections. The two diagonal reflections fix 2 vertices and swap 2: 3 cycles each. The two reflections through edge midpoints swap two pairs: 2 cycles each. Total $(96 + 2\cdot 27 + 2 \cdot 9)/8 = (96 + 54 + 18)/8 = 168/8 = 21$.

5. Prove that no group of order 35 is simple, and in fact that every group of order 35 is cyclic.

Solution

$35 = 5 \cdot 7$. $n_5 \mid 7$ and $n_5 \equiv 1 \pmod 5$: the divisors are 1, 7 and $7 \equiv 2$, so $n_5 = 1$. $n_7 \mid 5$ and $n_7 \equiv 1 \pmod 7$: the divisors are 1, 5 and $5 \not\equiv 1$, so $n_7 = 1$. Both Sylow subgroups are normal, meet trivially, and commute elementwise, exactly as for order 15, so $G \cong \mathbb{Z}/5 \times \mathbb{Z}/7 \cong \mathbb{Z}/35$. (This is the "$p \nmid q - 1$" criterion with $5 \nmid 6$.)

6. List the abelian groups of order 36 in both normal forms.

Solution

$36 = 2^2 \cdot 3^2$, and $p(2) \cdot p(2) = 2 \cdot 2 = 4$ groups:

• $4, 9$ → $\mathbb{Z}/36$
• $4, 3, 3$ → $\mathbb{Z}/3 \times \mathbb{Z}/12$
• $2, 2, 9$ → $\mathbb{Z}/2 \times \mathbb{Z}/18$
• $2, 2, 3, 3$ → $\mathbb{Z}/6 \times \mathbb{Z}/6$

7. Let $G$ be a finite group and $H$ a subgroup of index $n$. Using the action of $G$ on the cosets $G/H$, show that $G$ has a normal subgroup $K \subseteq H$ with $[G:K]$ dividing $n!$. Deduce that a simple group of order 60 has no subgroup of index 2, 3 or 4.

Solution

The action gives a homomorphism $\rho : G \to \operatorname{Sym}(G/H) \cong S_n$. Its kernel $K$ is normal; anything in $K$ fixes the coset $H$, so $K \subseteq H$. By the First Isomorphism Theorem $G/K$ embeds in $S_n$, so $[G:K]$ divides $n!$ by Lagrange.

If $G$ is simple of order 60 and $[G:H] = n \in \{2, 3, 4\}$, then $K$ is $\{e\}$ or $G$. $K = G$ is impossible since $K \subseteq H \ne G$. So $K = \{e\}$ and $60$ divides $n! \le 24$, a contradiction.

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Cheat-sheet

Idea	Statement
Homomorphism	$\varphi(ab) = \varphi(a)\varphi(b)$; then $\varphi(e) = e$, $\varphi(g^{-1}) = \varphi(g)^{-1}$
Kernel	always a normal subgroup; trivial exactly when $\varphi$ is injective
Maps between cyclic groups	$\gcd(n, m)$ homomorphisms $\mathbb{Z}/n \to \mathbb{Z}/m$, fixed by the image of 1
Automorphisms	$\operatorname{Aut}(\mathbb{Z}/n) \cong (\mathbb{Z}/n)^\times$, order $\phi(n)$; $\operatorname{Inn}(G) \cong G/Z(G)$
First Isomorphism Theorem	$G/\ker\varphi \cong \operatorname{im}\varphi$
Second	$HN/N \cong H/(H \cap N)$ when $N$ is normal
Third	$(G/N)/(K/N) \cong G/K$
Cayley	$G$ embeds in $\operatorname{Sym}(G)$; order $n$ embeds in $S_n$
Orbit–stabilizer	orbit size times stabilizer size $= \lvert G\rvert$
Class equation	$\lvert G\rvert = \lvert Z(G)\rvert + \sum [G : C_G(x_i)]$
$p$-groups	nontrivial center; order $p^2$ implies abelian
Burnside	orbits = average number of fixed points; colourings fixed by $g$: $k^{c(g)}$
Cauchy	prime $p$ divides the order, so an element of order $p$ exists
Sylow	exist; all conjugate; $n_p \equiv 1 \bmod p$ and $n_p \mid m$
Normal Sylow	$n_p = 1$ exactly when the Sylow $p$-subgroup is normal
Order $pq$, $p < q$	cyclic if $p \nmid q - 1$
Finite abelian groups	product of cyclic prime-power groups; count = product of partition numbers

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What's next

Post 04 moves from one operation to two. A ring has addition and multiplication tied by the distributive law; the integers, the integers modulo $n$, polynomials and matrices are the models. Every idea of this post has a ring version: ring homomorphisms and their kernels (now called ideals), quotient rings, and a First Isomorphism Theorem with the same proof. Prime and maximal ideals then explain why $\mathbb{Z}/p\mathbb{Z}$ is a field, and the same construction applied to polynomials builds $\mathbb{C}$ and the finite fields that post 05 needs.

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References

• D. S. Dummit and R. M. Foote, Abstract Algebra, 3rd edition, Wiley, 2004 — Chapters 3.3 (isomorphism theorems), 4 (group actions, class equation, Sylow) and 5.2 (finite abelian groups).
• M. Artin, Algebra, 2nd edition, Pearson, 2011 — Chapters 6 (symmetry, group actions, counting) and 7 (class equation, Sylow theorems).
• J. A. Gallian, Contemporary Abstract Algebra, 10th edition, CRC Press, 2021 — homomorphisms, Burnside's lemma worked examples, Sylow theorems.
• J. J. Rotman, An Introduction to the Theory of Groups, 4th edition, Springer GTM 148, 1995 — full proofs of Sylow and the basis theorem.
• MIT OpenCourseWare, 18.701 Algebra I (Fall 2010) and 18.702 Algebra II (Spring 2011) — lecture notes and problem sets.
• K. Conrad, Expository papers ("blurbs") — short notes on group actions, Sylow theorems and finite abelian groups.
• J. H. McKay, "Another proof of Cauchy's group theorem", American Mathematical Monthly 66 (1959), p. 119.

Note: this post is in progress. Planned additions: full proofs of the three Sylow theorems (via the action on subsets of size $p^k$ and on Sylow subgroups by conjugation), semidirect products and the classification of groups of order 12 and of order $pq$, and the Pólya enumeration theorem as a refinement of Burnside's lemma.

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