Finite Groups and Symmetry

Introduction

A finite group is a group of transformations: this is Cayley's theorem for the abstract statement, and it is the concrete content of every classification below. This article treats the finite groups that arise as groups of transformations of a small set, and the way they are classified. It is the companion of Group Actions and Structure, which supplies the class equation, the Sylow theorems and the composition series used throughout, and of Transformation Groups, which supplies the action-theoretic frame and the principle that a transformation group is determined by what it preserves.

The plan is concrete. The cyclic and dihedral groups are treated first, with their subgroups and conjugacy classes; then the symmetric and alternating groups, with cycles, parity, conjugacy by cycle type and the simplicity of $A_n$ for $n \geq 5$; then the quaternion group $Q_8$ and the dicyclic groups; and finally the classification of the groups of small order.

Everything is a statement about groups of transformations of a set. The symmetry groups of the regular polygons and of the regular solids, their realisation inside the orthogonal group and the crystallographic refinement are groups of isometries and require a distance, so they belong to Part II, where a form and a distance are available, and they are not used here. No physical object is invoked anywhere. Throughout, $R$ is a commutative ring with identity $1 \neq 0$ and $F$, $K$ are fields. Finite groups are written multiplicatively with identity $e$; $C_n$ is the cyclic group of order $n$, $D_n$ the dihedral group of order $2n$, $S_n$ and $A_n$ the symmetric and alternating groups, $V_4$ the Klein four group and $Q_8$ the quaternion group, as in the earlier articles of this category.

Cyclic and Dihedral Groups

Cyclic Groups

Definition. A group is cyclic if it is generated by one element, $G = \langle g \rangle$; if $G$ is finite of order $n$ then $G \cong \mathbb{Z}/n\mathbb{Z}$ by $g^k \mapsto k$.

Proposition. Let $C_n = \langle g \rangle$ be cyclic of order $n$.

1. For each divisor $d$ of $n$ there is exactly one subgroup of order $d$, namely $\langle g^{n/d} \rangle$, and these are all the subgroups.
2. The generators of $C_n$ are the elements $g^k$ with $\gcd(k, n) = 1$, so there are $\varphi(n)$ of them, where $\varphi$ is Euler's function.
3. $\operatorname{Aut}(C_n) \cong (\mathbb{Z}/n\mathbb{Z})^\times$, of order $\varphi(n)$, acting by $g \mapsto g^k$ for $k$ coprime to $n$.
4. $C_n$ is a lattice of subgroups ordered by divisibility.

Proof. (1) If $H \leq C_n$ is nontrivial, let $k$ be the least positive integer with $g^k \in H$; division with remainder by $k$ shows every element of $H$ is a power of $g^k$, so $H = \langle g^k \rangle$; and $\langle g^k \rangle$ has order $n/\gcd(n,k)$, so the subgroups correspond to the divisors of $n$. (2) $g^k$ generates exactly when $\gcd(k,n) = 1$, since $\langle g^k \rangle = \langle g^{\gcd(n,k)} \rangle$. (3) An automorphism sends $g$ to a generator and is determined by that image; conversely $g \mapsto g^k$ with $\gcd(k,n)=1$ is an automorphism. (4) is the definition of the subgroup lattice.

Example. $C_4$ has three subgroups, of orders $1, 2, 4$; $C_6$ has four, of orders $1, 2, 3, 6$; the subgroups of $C_n$ form a lattice isomorphic to the divisibility lattice of $n$, and the number of subgroups of $C_n$ is the number of divisors of $n$. Every group of order $p$ with $p$ prime is cyclic, and a group of order $p^2$ is $C_{p^2}$ or $C_p \times C_p$, as used in the classification below.

Dihedral Groups

Definition. For $n \geq 3$ the dihedral group $D_n$ is the group of order $2n$ with presentation

$$ D_n = \langle r, s \mid r^n = s^2 = e, \ s r s = r^{-1} \rangle . $$

Every element is uniquely $r^i$ or $r^i s$ with $0 \leq i < n$, the cyclic subgroup $C_n = \langle r \rangle$ is normal of index $2$, and $D_n \cong C_n \rtimes C_2$ with the nontrivial element of $C_2$ acting by inversion. For $n = 3$, $D_3 \cong S_3$. The group is the symmetry group of the regular $n$-gon; that realisation, which needs a distance, belongs to Part II.

Theorem (conjugacy classes). In $D_n$ the conjugacy classes are: $\{e\}$; the pairs $\{r^k, r^{-k}\}$ for $1 \leq k < n/2$, together with $\{r^{n/2}\}$ when $n$ is even; and the elements outside $\langle r \rangle$, which form one class if $n$ is odd and two classes if $n$ is even. Hence the number of conjugacy classes is $(n+3)/2$ for odd $n$ and $(n+6)/2$ for even $n$.

Proof. Conjugation is computed from the relations: $r^j \, r^k \, r^{-j} = r^k$ and $r^j s \, r^k \, s^{-1} r^{-j} = r^{-k}$, so the elements of $\langle r \rangle$ fall into the pairs $\{r^k, r^{-k}\}$ and the class of $r^{n/2}$ is a singleton when $n$ is even; the number of pairs is $(n-1)/2$ for odd $n$ and $(n-2)/2 + 1$ for even $n$ counting the central $r^{n/2}$. For the elements outside $\langle r \rangle$, $r^j (r^k s) r^{-j} = r^{k+2j} s$, so the $r^k s$ with $k$ of fixed parity are conjugate; if $n$ is odd every $k$ is reached by $2j$ as $j$ varies, giving one class of $n$ elements, and if $n$ is even the even and odd $k$ give two classes of $n/2$ each. Summing the class sizes gives $|D_n| = 2n$ in both cases.

Example. $D_4$ has $5$ conjugacy classes, of sizes $1, 1, 2, 2, 2$, and $6$ normal subgroups, of orders $1, 2, 4, 4, 4, 8$; $D_6 \cong D_3 \times C_2$ has $6$ classes of sizes $1, 1, 2, 2, 3, 3$ and $7$ normal subgroups, of orders $1, 2, 3, 6, 6, 6, 12$. The normal subgroups of $D_n$ are: the subgroups of $\langle r \rangle$ (all normal, since $\langle r \rangle$ is abelian and normal and inversion preserves each cyclic subgroup $\langle r^d \rangle$), the subgroups of index $2$, and $D_n$ itself.

Remark (subgroups of $D_n$). Every subgroup of $D_n$ is either a subgroup of the cyclic subgroup $\langle r \rangle$, hence some $C_d$ with $d \mid n$, or is generated by an element of $\langle r \rangle$ together with one outside it, $\langle r^d, r^j s \rangle$, of order $2n/d$; in particular all subgroups of odd order are cyclic, and the number of subgroups of order $2$ is $n$ when $n$ is odd and $n + 1$ when $n$ is even (the $n$ elements $r^j s$ together with $\langle r^{n/2} \rangle$). For $D_4$ this gives one subgroup of order $1$, five of order $2$, three of order $4$ (namely $\langle r \rangle$ and the two subgroups $\langle r^2, s\rangle$, $\langle r^2, rs \rangle$), and one of order $8$.

Symmetric and Alternating Groups

Cycles, Parity and Generation

Definition. A cycle $(a_1 a_2 \cdots a_k)$ in $S_n$ is the permutation sending $a_1 \mapsto a_2 \mapsto \cdots \mapsto a_k \mapsto a_1$ and fixing the other points; a transposition is a cycle of length $2$. Every permutation is a product of disjoint cycles, and this cycle type is unique up to the order of the factors; it is recorded as the partition of $n$ given by the cycle lengths, where fixed points contribute parts equal to $1$. The sign is $\operatorname{sgn}(a) = (-1)^{n - c(a)}$, where $c(a)$ is the number of cycles of $a$ counted with fixed points, and $A_n = \ker \operatorname{sgn}$ is the alternating group, of order $n!/2$ for $n \geq 2$.

Proposition. The order of a permutation is the least common multiple of the lengths of its cycles; $\operatorname{sgn}$ is a homomorphism onto $\{\pm 1\}$; a $k$-cycle has sign $(-1)^{k-1}$; and $S_n$ is generated by the transpositions, by the transpositions $(1\,i)$ alone, by the two elements $(1\,2)$ and $(1\,2\,\cdots\,n)$, and $A_n$ is generated by the $3$-cycles.

Proof. Disjoint cycles commute, so a power of a permutation is the identity exactly when each cycle length divides the exponent, giving the order formula. The sign is multiplicative because the number of cycles changes by exactly $1$ when a permutation is multiplied by a transposition, which also proves surjectivity. A transposition is a $k$-cycle with $k = 2$. For generation, $(1\,i)$ conjugates and $(1\,2\,\cdots\,n)$ cycles the indices $2, \ldots, n$, and together they produce all transpositions $(i\,j)$; the same computation with three factors gives the $3$-cycles, which are the even products of two transpositions.

Conjugacy Classes of $S_n$ and $A_n$

Theorem. Two permutations of $S_n$ are conjugate if and only if they have the same cycle type. Hence the conjugacy classes of $S_n$ are indexed by the partitions of $n$, and if the cycle type has $m_k$ cycles of length $k$ then the class has size

$$ \frac{n!}{\prod_{k} k^{m_k} m_k!} . $$

Proof. Conjugation relabels the points, so it preserves the partition into cycles; conversely, given two permutations with the same cycle structure, the bijection matching the cycles in order conjugates one to the other. For the size, the centraliser of the permutation consists of the permutations preserving the cycle decomposition, and its order is $\prod_k k^{m_k} m_k!$: each of the $m_k$ cycles of length $k$ contributes its $k$ powers and the $m_k$ cycles may be permuted. The class size is the index of the centraliser.

Example. In $S_4$ the partitions of $4$ are $1+1+1+1$, $2+1+1$, $2+2$, $3+1$, $4$, giving class sizes $1, 6, 3, 8, 6$, which sum to $24$; in $S_5$ there are $p(5) = 7$ partitions and class sizes $1, 10, 15, 20, 20, 24, 30$, which sum to $120$. The number of conjugacy classes of $S_n$ is the partition number $p(n)$.

Theorem (classes of $A_n$). Let $a \in A_n$. The $S_n$-class of $a$ either is contained in $A_n$ and is a single $A_n$-class, or splits into exactly two $A_n$-classes of equal size. It splits if and only if all parts of the cycle type of $a$ are odd and distinct.

Proof. The $A_n$-class of $a$ has size $|A_n| / |C_{A_n}(a)|$, where $C_{A_n}(a) = C_{S_n}(a) \cap A_n$ is the centraliser in $A_n$, while the $S_n$-class has size $|S_n|/|C_{S_n}(a)|$. Since $|S_n| = 2|A_n|$, the $A_n$-class has either the same size as the $S_n$-class, when $C_{S_n}(a)$ contains an odd permutation, or half that size in the contrary case, and in the second case the $S_n$-class is the union of two $A_n$-classes of equal size. The centraliser contains an odd permutation exactly when the cycle type fails to consist of distinct odd parts: two equal parts are exchanged by an odd transposition of the two cycles, and a part of even length is an odd permutation commuting with $a$, while if all parts are odd and distinct every permutation commuting with $a$ permutes the cycles of equal length — of which there are none — and acts within cycles by powers of the cycles of odd order, all even permutations.

Example. In $A_4$ the class sizes are $1, 3, 4, 4$. The $S_4$-class of double transpositions has size $3$ and does not split, giving the normal Klein four group $V_4 = \{e\} \cup \{$double transpositions$\}$; the eight $3$-cycles form one $S_4$-class of size $8$ which splits into two $A_4$-classes of size $4$. In $A_5$ the class sizes are $1, 12, 12, 15, 20$: the class of $5$-cycles has size $24$ in $S_5$ and splits, since a $5$-cycle has distinct odd parts, and the class of double transpositions has size $15$ and stays. In both cases the class sizes sum to the group order.

Simplicity of $A_n$

Theorem. $A_n$ is simple for $n \geq 5$. Consequently $A_n = [A_n, A_n]$, the derived series of $S_n$ is $S_n \triangleright A_n \triangleright A_n \triangleright \cdots$ and stabilises at $A_n$, and $S_n$ is solvable if and only if $n \leq 4$.

Proof sketch. A normal subgroup $N \neq \{e\}$ of $A_n$ contains a $3$-cycle: take $a \neq e$ in $N$ moving as few points as possible, decompose $a$ into disjoint cycles and, if a cycle has length $\geq 4$ or if $a$ has two cycles of length $\geq 2$, produce a commutator of $a$ with a suitable $3$-cycle that is a nontrivial element of $N$ moving fewer points, a contradiction; otherwise $a$ is a $3$-cycle, or a double transposition which is converted into a $3$-cycle by one more commutator. Once $N$ contains a $3$-cycle, the $3$-cycles are all conjugate in $A_n$ for $n \geq 5$, so $N$ contains all of them, and the $3$-cycles generate $A_n$; hence $N = A_n$. For $n \leq 4$ the groups $S_1, S_2, S_3, S_4$ and $A_4$ are solvable, exhibited by the composition series $\{e\} \trianglelefteq C_2 \trianglelefteq V_4 \trianglelefteq A_4 \trianglelefteq S_4$ with cyclic factors $C_2, C_2, C_3, C_2$.

Corollary. $A_5$ has no proper nontrivial normal subgroup: its only normal subgroups are $\{e\}$ and $A_5$, and it has $5$ conjugacy classes of sizes $1, 12, 12, 15, 20$. The group $A_5$ is the smallest nonabelian simple group, of order $60$; every group of order less than $60$ is solvable.

Remark. The simplicity of $A_n$ makes the composition series of $S_n$ transparent: $\{e\} \trianglelefteq A_n \trianglelefteq S_n$ for $n \geq 5$, with factors $A_n$ and $C_2$, whereas for $n = 4$ the series passes through $V_4$, as in Group Actions and Structure. The symmetric groups therefore supply one infinite family of finite simple groups, the alternating groups, alongside the cyclic groups of prime order.

The Quaternion Group and the Dicyclic Groups

The Quaternion Group

Definition. The quaternion group is

$$ Q_8 = \{\pm 1, \pm e_1, \pm e_2, \pm e_3\} , $$

with $e_1^2 = e_2^2 = e_3^2 = -1$ and $e_1 e_2 = e_3$, $e_2 e_3 = e_1$, $e_3 e_1 = e_2$. Equivalently

$$ Q_8 = \langle e_1, e_2 \mid e_1^4 = e, \ e_1^2 = e_2^2, \ e_2 e_1 = e_1^{-1} e_2 \rangle . $$

Theorem. $|Q_8| = 8$; the centre is $Z(Q_8) = \{\pm 1\}$; the conjugacy classes are $\{1\}$, $\{-1\}$, $\{\pm e_1\}$, $\{\pm e_2\}$, $\{\pm e_3\}$, of sizes $1, 1, 2, 2, 2$; the class equation is $8 = 1 + 1 + 2 + 2 + 2$; every subgroup is normal, namely $\{1\}$, the three subgroups $\langle e_1 \rangle, \langle e_2 \rangle, \langle e_3 \rangle$ of order $4$, the subgroup $\{\pm 1\}$ of order $2$, and $Q_8$; and $Q_8$ is not a semidirect product, not isomorphic to $D_4$, and not abelian.

Proof. The multiplication table of the eight listed units is closed by the quaternion relations, and $-1$ is the unique element of order $2$ and is central, so $\{\pm 1\} \subseteq Z(Q_8)$ with equality because $e_1$ does not commute with $e_2$. Conjugation by an element of $Q_8$ fixes $\pm 1$ and sends each $e_k$ to $\pm e_k$: for a unit $u$ of order $4$, $u e_k u^{-1} = u e_k (-u) = -u e_k u = \pm e_k$, the sign being determined by whether $u e_k = \pm e_k u$. Thus conjugation by $e_1$ fixes $e_1$ and inverts $e_2$ and $e_3$, and the other cases are analogous, so direct computation in the multiplication table gives the pairs $\{\pm e_k\}$ as classes. Since every subgroup of order $4$ contains $-1$ and hence the unique involution, and every subgroup containing $-1$ and one of $\pm e_k$ contains the pair, the list of subgroups is complete; the three order-$4$ subgroups are normal because they contain the unique element of order $2$ and are of index $2$. Finally $Q_8$ is not a semidirect product $C_4 \rtimes C_2$ because the unique involution is central, so no complement to a normal $C_4$ exists; and for the same reason $Q_8 \not\cong D_4$, whose involutions are $r^2$ together with the four elements of $D_4$ outside $\langle r \rangle$ that are involutions, of which only $r^2$ is central, whereas $Q_8$ has the single involution $-1$.

Remark. A nonabelian group in which every subgroup is normal is called Hamiltonian; $Q_8$ is the smallest, and the Hamiltonian groups are exactly the products $Q_8 \times E \times A$ with $E$ elementary abelian of exponent $2$ and $A$ abelian of odd order, a classical theorem of Dedekind and Baer.

Dicyclic Groups

Definition. For $n \geq 2$ the dicyclic group of order $4n$ is

$$ \operatorname{Dic}_n = \langle a, b \mid a^{2n} = e, \ b^2 = a^n, \ b a b^{-1} = a^{-1} \rangle , $$

so that $Q_8 = \operatorname{Dic}_2$ and the group of order $12$ exhibited in Group Actions and Structure as $C_3 \rtimes C_4$ with the generator acting by inversion is $\operatorname{Dic}_3$.

Proposition. $\operatorname{Dic}_n$ has order $4n$; its centre is $\{e, a^n\}$ of order $2$; the element $a^n$ is the unique involution; and $\operatorname{Dic}_n$ contains a normal cyclic subgroup $\langle a \rangle$ of index $2$ with the generator of the quotient acting by inversion.

Proof. Every element is $a^k$ or $a^k b$ with $0 \leq k < 2n$, and $b a^k b^{-1} = a^{-k}$, so there are at most $4n$ elements and the presentation supplies exactly that many, since $a$ has order $2n$ and $b \notin \langle a \rangle$. The relation $b^2 = a^n$ shows $b^2$ is central and of order $2$; it is the only involution because $a^k$ has order $2$ only for $k = n$, and $(a^k b)^2 = a^k b a^k b = a^k a^{-k} b^2 = a^n$.

The dicyclic groups are exactly the finite groups with a unique involution that are not cyclic: the classical theorem of Burnside states that a finite group with exactly one involution is either cyclic of even order or one of the generalized quaternion groups, which are the groups $\operatorname{Dic}_n$ under a different indexing.

The Classification of the Small Orders

The group-theoretic classification is explicit for small orders. The count is as follows; the entries are the number of groups of order $n$ up to isomorphism.

Theorem. The entries for $n \leq 15$ are as displayed.

Proof. For $n = p$ prime the group is cyclic. For $n = p^2$ the group is abelian, hence $C_{p^2}$ or $C_p \times C_p$, by the theorem of Group Actions and Structure on the centre of a $p$-group together with the lemma that $G/Z(G)$ is never cyclic unless trivial: for $n = 4$ this gives $2$ groups, for $n = 9$ also $2$. For $n = 2p$ with $p$ an odd prime, a group of order $2p$ is $C_{2p}$ or the dihedral group $D_p$, giving $2$ groups for $n = 6, 10, 14$; the argument is that a normal subgroup of order $p$ has cyclic automorphism group $C_{p-1}$ even, so the complement acts either trivially, giving $C_{2p}$, or by the unique element of order $2$, giving $D_p$. For $n = pq$ with $p < q$ primes and $p \nmid q - 1$ the group is cyclic: the Sylow numbers force $n_q = n_p = 1$, giving $C_{pq}$; this covers $n = 15$. For $n = 8$ the abelian groups are $C_8, C_4 \times C_2, C_2^3$ by the structure theorem, and the nonabelian groups are exactly two: a nonabelian group of order $8$ has $\{e\} \neq Z(G)$ and $G/Z(G)$ is not cyclic, so $|Z(G)| = 2$ and $G/Z(G) \cong V_4$; choose $u$ of order $4$ (which must exist, else all elements have order dividing $2$ and $G$ is abelian), then $\langle u \rangle$ has index $2$ and is normal, and an element $v$ outside it satisfies $v u v^{-1} = u^{-1}$ (otherwise $G$ is abelian) and $v^2 = u^a$ with $v u^a v^{-1} = (v u v^{-1})^a = u^{-a} = v^2 = u^a$, so $2a \equiv 0 \pmod 4$ and $a \in \{0, 2\}$; the case $a = 0$ is $D_4$ and the case $a = 2$ is $Q_8$. For $n = 12$ the five groups are $C_{12}, C_6 \times C_2, A_4, D_6, \operatorname{Dic}_3$, as derived in Group Actions and Structure. The count $14$ for order $16$ is the classical enumeration and is cited as standard.

Remark. The table shows the irregularities of the classification: for prime order there is one group, for order $p^2$ there are two, for order $p^3$ there are five, and order $16$ is the first with as many as $14$, more than any smaller order. The numbers $g(n)$ for $n \leq 16$ are $1, 1, 1, 2, 1, 2, 1, 5, 2, 2, 1, 5, 1, 2, 1, 14$, and the pattern is not monotone in the number of prime factors: $g(8) = 5$ while $g(9) = 2$, and $g(12) = 5$ while $g(16) = 14$. For odd $p$ the five groups of order $p^3$ are the three abelian ones $C_{p^3}$, $C_{p^2} \times C_p$, $C_p^3$ and two nonabelian ones, distinguished by whether the exponent is $p$ or $p^2$; for $p = 2$ the count is again $5$, but the two nonabelian groups $D_4$ and $Q_8$ both have exponent $4$, so the criterion degenerates in the smallest case.

Example. The groups of order $8$ are the three abelian ones and $D_4$, $Q_8$; those of order $12$ are the two abelian ones, $A_4$, $D_6$ and $\operatorname{Dic}_3$; and among the groups of order at most $15$ the nonabelian simple groups do not occur, the smallest nonabelian simple group being $A_5$ of order $60$. Every group of order $< 60$ is solvable, and every group of order $p^k$ is nilpotent by the results of Group Actions and Structure.

Class Equations

The class equations of the small groups display the arithmetic of the conjugacy classes, and each is a sum of the sizes recorded above.

The dihedral column is the case $n = 4$ of the formula $(n+6)/2$ for even $n$, giving $5$ classes, and the alternating column is $A_5$, whose five classes have the sizes $1, 12, 12, 15, 20$ computed above. In the class equation of a group of order $p^k$ the terms equal to $1$ are exactly the central elements, so there are $|Z(G)|$ of them and the remaining terms are divisible by $p$; since the equation sums to $p^k$, a nontrivial finite $p$-group has nontrivial centre. This is the argument used in Group Actions and Structure and again in the classification of order $8$ above, where it forces $|Z(G)| = 2$ and $G/Z(G) \cong V_4$.

Automorphism Groups of the Small Groups

Theorem. The automorphism groups of the small groups are as follows, where $\operatorname{Hol}(C_n) = C_n \rtimes (\mathbb{Z}/n\mathbb{Z})^\times$ is the holomorph of $C_n$.

Proof. The cyclic case is the proposition on $C_n$. For $D_n$ with $n \geq 3$ an automorphism is determined by the images of $r$ and $s$: the image of $r$ is an element of order $n$, that is, $r^k$ with $\gcd(k,n) = 1$, and once $r \mapsto r^k$ is chosen the image of $s$ is any element of the coset $s\langle r\rangle$, giving at most $n\varphi(n)$ automorphisms. Each such assignment preserves the relations, and the automorphisms with $s \mapsto s$ form the group $(\mathbb{Z}/n\mathbb{Z})^\times$ acting on the normal $C_n$, so $\operatorname{Aut}(D_n) = \operatorname{Hol}(C_n)$ of order $n\varphi(n)$; for $n = 3, 4, 6$ this is $6, 8, 12$. Every permutation of the three non-identity elements of $V_4$ is an automorphism, so $\operatorname{Aut}(V_4) \cong S_3$. An automorphism of $Q_8$ permutes the three subgroups $\langle e_k \rangle$ of order $4$, giving a homomorphism $\operatorname{Aut}(Q_8) \to S_3$ which is onto because $e_1 \mapsto e_2$, $e_2 \mapsto e_1$, $e_3 \mapsto -e_3$ realises a transposition, while $e_1 \mapsto e_2$, $e_2 \mapsto e_3$, $e_3 \mapsto e_1$ realises a $3$-cycle; the kernel consists of the maps fixing each $e_k$ up to sign, whose three signs are constrained by $e_1 e_2 = e_3$ and hence amount to $4$ choices. So $|\operatorname{Aut}(Q_8)| = 4 \cdot 6 = 24$, and $\operatorname{Aut}(Q_8) \cong S_4$, with $\operatorname{Inn}(Q_8) \cong Q_8/\{\pm 1\} \cong V_4$ and $\operatorname{Out}(Q_8) \cong S_4/V_4 \cong S_3$. An automorphism of $A_4$ preserves the characteristic subgroup $V_4$ and permutes the four Sylow $3$-subgroups, giving an injective homomorphism $\operatorname{Aut}(A_4) \to S_4$, so $|\operatorname{Aut}(A_4)| \leq 24$; the inner automorphisms already realise $A_4$ of order $12$ in the image, and conjugation by an odd permutation of $S_4$ realises an odd element of the image, so the image has order $24$ and $\operatorname{Aut}(A_4) \cong S_4$, with $\operatorname{Out}(A_4) \cong C_2$. Finally every automorphism of $S_4$ is inner and $\operatorname{Aut}(S_4) \cong S_4$; more generally $\operatorname{Aut}(S_n) \cong S_n$ for $n \neq 2, 6$, the exception being $\operatorname{Out}(S_6) \cong C_2$, the classical exception of Hölder.

Remark. The quotient $\operatorname{Out}(G) = \operatorname{Aut}(G)/\operatorname{Inn}(G)$, with $\operatorname{Inn}(G) \cong G/Z(G)$, measures the failure of the automorphisms to be inner. Here $\operatorname{Inn}(D_4) \cong D_4/Z(D_4)$ has order $4$, so $\operatorname{Out}(D_4) \cong C_2$; $\operatorname{Out}(A_4) \cong C_2$; and $\operatorname{Out}(S_4) = 1$. In general $\operatorname{Out}(D_n)$ has order $\varphi(n)/2$ for odd $n$ and $\varphi(n)$ for even $n$, and $\operatorname{Out}(A_n) \cong C_2$ for $n \geq 4$ with $n \neq 6$, the exceptional case $n = 6$ having $\operatorname{Out}(A_6) \cong C_2 \times C_2$. A group with trivial centre all of whose automorphisms are inner is called complete; the groups $S_n$ for $n \neq 2, 6$ are complete, while $A_4$ is not, its outer automorphisms expressing the symmetry between the three subgroups of order $2$ in $V_4$.

Sylow Data of the Small Groups

The Sylow numbers of the small groups are obtained by counting elements of each order; the tables of the preceding sections supply the counts.

Example. In $A_5$ there are $15$ elements of order $2$, $20$ of order $3$ and $24$ of order $5$. A Sylow $2$-subgroup is a $V_4$, so it contains three of the $15$ involutions and $n_2 = 15/3 = 5$; a Sylow $3$-subgroup is a $C_3$, containing two of the $20$ elements of order $3$, so $n_3 = 20/2 = 10$; and a Sylow $5$-subgroup is a $C_5$, containing four of the $24$ five-cycles, so $n_5 = 24/4 = 6$. The values satisfy the congruence $n_p \equiv 1 \pmod p$ and the divisibility $n_p \mid |G|$ of the Sylow theorems. In the same way, $S_4$ has $9$ involutions and $8$ elements of order $3$, giving $n_3 = 8/2 = 4$, while its Sylow $2$-subgroups are the three copies of $D_4$, containing two elements of order $4$ each out of the six that $S_4$ possesses; and $A_4$ has $n_2 = 1$ because its Klein four group $V_4$ is normal, which is exactly the feature that makes $A_4$ solvable and that $A_5$ does not possess.

Summary

Finite groups are groups of transformations, and the small ones are classified concretely. A cyclic group $C_n$ has exactly one subgroup of each order dividing $n$, $\varphi(n)$ generators and automorphism group $(\mathbb{Z}/n\mathbb{Z})^\times$; the dihedral group $D_n$ of order $2n$ is $C_n \rtimes C_2$ with the inversion action, its conjugacy classes are the pairs $\{r^k, r^{-k}\}$ together with one or two classes of the elements outside $\langle r \rangle$, and every subgroup is cyclic or dihedral. The symmetric group $S_n$ has conjugacy classes indexed by the partitions of $n$, with class sizes $n!/(\prod_k k^{m_k} m_k!)$, and the classes of $A_n$ either persist or split in two according as the cycle type fails or succeeds in having distinct odd parts alone; $A_n$ is simple for $n \geq 5$, so $S_n$ is solvable exactly for $n \leq 4$. The quaternion group $Q_8$ is the smallest Hamiltonian group, with centre $\{\pm 1\}$, five classes and six subgroups, all normal, and the dicyclic groups $\operatorname{Dic}_n$ are the finite groups with a unique involution apart from the cyclic groups of even order.

The groups of order $n \leq 16$ are counted by $1, 1, 1, 2, 1, 2, 1, 5, 2, 2, 1, 5, 1, 2, 1, 14$, the entries for $n \leq 15$ being derived from the structure of $p$-groups, of groups of order $pq$ and of groups of order $2p$. The automorphism groups of the small groups are $(\mathbb{Z}/n\mathbb{Z})^\times$ for $C_n$, the holomorph $\operatorname{Hol}(C_n) = C_n \rtimes (\mathbb{Z}/n\mathbb{Z})^\times$ of order $n\varphi(n)$ for $D_n$, and $S_3, D_4, S_4, S_4, S_4$ for $V_4, D_4, Q_8, A_4, S_4$; the Sylow numbers of the small groups are recovered by counting elements of each order, giving $n_2 = 5, n_3 = 10, n_5 = 6$ for $A_5$ and showing the normality of the Klein four group in $A_4$ that separates the two alternating groups.

Summary of Notation

Further Reading

• Harold S. M. Coxeter and William O. J. Moser, Generators and Relations for Discrete Groups (Springer, 4th ed. 1980), for presentations of the cyclic and dihedral groups.
• Joseph J. Rotman, An Introduction to the Theory of Groups (Springer, 4th ed. 1995), for the classification of the small orders and the simplicity of $A_n$.
• Jean-Pierre Serre, Cours d'arithmétique (Presses Universitaires de France, 1970), for the groups with a unique involution.