Nonabelian models and \(p\)-groups
The dihedral group \(D_4\) and the quaternion group \(Q_8\) are the nonabelian groups of order \(8\). A finite $p$-group has nontrivial center, by the class equation, and has a normal subgroup of each order dividing its order. A transitive subgroup of \(S_n\) acts transitively on \(\ts{1,\ldots,n}\), so its order is divisible by \(n\).
Dihedral group (Tag D-4R2Z5)
Let \(n\geq 1\). The dihedral group of order \(2n\) is the group with presentation \[ D_n \coloneqq \gens{r, s \st r^n, s^2, srs^{-1}r } . \]
For \(n\geq 1\), \[ \gens{r, s \st r^n, s^2, srs^{-1}r } \cong \gens{r, s \st r^n, s^2, (rs)^2 }, \] by an isomorphism sending \(r\mapsto r\) and \(s\mapsto s\).
For \(n\geq 3\), \(D_n\) is the group of symmetries of a regular \(n\)-gon in the plane, with \(r\) the rotation by \(2\pi/n\) and \(s\) a reflection. Labeling the vertices \(1, \ldots, n\) in cyclic order identifies \(D_n\) with a subgroup of \(S_n\), in which \(r\) is an \(n\)-cycle.
Label the vertices of a square \(1,2,3,4\) in cyclic order. Then \(D_4 = \gens{(1,2,3,4), (1,3)} \leq S_4\), where \((1,2,3,4)\) is the rotation by \(2\pi/4\) and \((1,3)\) is the reflection across the diagonal through the vertices \(2\) and \(4\).
Label the vertices of a regular pentagon \(1,\ldots,5\) in cyclic order. Then \(D_5 = \gens{(1,2,3,4,5), (1,5)(2,4)} \leq S_5\), where \((1,2,3,4,5)\) is the rotation by \(2\pi/5\) and \((1,5)(2,4)\) is the reflection across the line through vertex \(3\) and the center.
Quaternion group $Q_8$ (Tag D-KRKV7)
The quaternion group is the group of order \(8\) \[ \begin{aligned} Q_8 &= \gens{-1, i, j, k \suchthat (-1)^2 = 1,\ i^2 = j^2 = k^2 = ijk = -1} \\ &\cong \gens{x, y \suchthat x^4 = y^4,\ x^2 = y^2,\ yxy^{-1} = x^{-1}}, \end{aligned} \] where the isomorphism sends \(x\mapsto i\) and \(y\mapsto j\). Its elements are \(\pm1,\pm i,\pm j,\pm k\).
Every element of \(\theset{i,j,k}\) squares to \(-1\), and \(ijk=-1\). Products of two distinct elements of \(\theset{i,j,k}\) taken along the cycle \(i\to j\to k\to i\) are positive, \(ij=k\), \(jk=i\), \(ki=j\), and products taken against the cycle are negative, \(ji=-k\), \(kj=-i\), \(ik=-j\):
Transitive subgroup of $S_n$ (Tag D-7UIPO)
Let \(n\geq 1\). A subgroup \(H\leq S_n\) is transitive if its action on \(\theset{1, \ldots, n}\) is transitive, that is, if for all \(i, j\in\theset{1,\ldots,n}\) there exists \(\sigma\in H\) with \(\sigma(i) = j\).
$p$-groups (Tag D-FIB7S)
Let \(p\) be a prime. A finite group \(G\) is a \(p\)-group if \(\abs{G} = p^k\) for some integer \(k\geq 0\).
$p$-group (Tag FD-OR3BW)
Let \(p\) be a prime. A \(p\)-group is a group of order \(p^n\) for some integer \(n\geq 1\).
Cyclic groups
Subgroups of cyclic groups (Tag T-QMKXQ)
Let \(G\) be a finite group of order \(n\). Then \(G\) is cyclic if and only if, for each positive divisor \(d\) of \(n\), \(G\) has a unique subgroup of order \(d\).
Cyclic groups, Euler's totient, and $\operatorname{Aut}(\mathbb{Z}/n\mathbb{Z})$ (Tag E-DAS3L)
Show that any cyclic group is abelian.
Show that every subgroup of a cyclic group is cyclic.
Show that \[\phi(n) = n \prod_{p\mid n}\qty{1 - {1\over p}}.\]
Compute \(\aut(\ZZ/n\ZZ)\) for \(n\) composite.
Compute \(\aut(\qty{\ZZ/p\ZZ}^n)\).
Solution
<1>1. A cyclic group \(G=\langle g\rangle\) is abelian.
Any two elements are \(g^a,g^b\), and \(g^ag^b=g^{a+b}=g^bg^a\).
<1>2. Every subgroup \(H\) of \(G=\langle g\rangle\) is cyclic.
If \(H=\{e\}\), then \(H=\langle e\rangle\). Otherwise \(H\) contains some \(g^k\) with \(k\neq0\), hence also \(g^{|k|}\), so let \(m\) be the least positive integer with \(g^m\in H\). For \(g^k\in H\), write \(k=qm+r\) with \(0\le r<m\); then \(g^r=g^k(g^m)^{-q}\in H\), so \(r=0\) by minimality. Hence \(H=\langle g^m\rangle\).
<1>3. \(\phi(n)=|(\ZZ/n\ZZ)^\times|=n\prod_{p\mid n}\left(1-\frac1p\right)\).
For \(n=\prod_i p_i^{a_i}\), the Chinese remainder theorem gives \((\ZZ/n\ZZ)^\times\cong\prod_i(\ZZ/p_i^{a_i}\ZZ)^\times\). The nonunits of \(\ZZ/p^a\ZZ\) are the \(p^{a-1}\) multiples of \(p\), so \(|(\ZZ/p^a\ZZ)^\times|=p^a-p^{a-1}=p^a(1-1/p)\). Multiplying over the prime powers dividing \(n\) gives the formula.
<1>4. \(\Aut(\ZZ/n\ZZ)\cong(\ZZ/n\ZZ)^\times\cong\prod_i(\ZZ/p_i^{a_i}\ZZ)^\times\), where \[(\ZZ/2^a\ZZ)^\times\cong\begin{cases}1,&a=1,\\ C_2,&a=2,\\ C_2\times C_{2^{a-2}},&a\ge3,\end{cases}\] and \((\ZZ/p^a\ZZ)^\times\) is cyclic of order \(p^{a-1}(p-1)\) for \(p\) odd.
An endomorphism of \(\ZZ/n\ZZ\) is multiplication by \(u=\varphi(1)\), and it is bijective exactly when \(u\) is a unit; composition corresponds to multiplication of the units. The product decomposition is step <1>3, For odd \(p\) there is a primitive root modulo \(p^a\); for \(a\ge3\), the class of \(5\) has order \(2^{a-2}\) modulo \(2^a\) and \((\ZZ/2^a\ZZ)^\times=\langle-1\rangle\times\langle5\rangle\). For composite \(n\) the group need not be cyclic, e.g. \(\Aut(\ZZ/8\ZZ)\cong C_2\times C_2\).
<1>5. \(\Aut((\ZZ/p\ZZ)^n)\cong\operatorname{GL}_n(\FF_p)\), of order \(\prod_{i=0}^{n-1}(p^n-p^i)\).
A group homomorphism of \((\ZZ/p\ZZ)^n=\FF_p^n\) is additive, hence \(\FF_p\)-linear, so the automorphisms are the invertible \(\FF_p\)-linear maps. An invertible matrix is a choice of columns \(v_1,\dots,v_n\) with \(v_{i+1}\notin\operatorname{span}(v_1,\dots,v_i)\), which leaves \(p^n-p^i\) choices for \(v_{i+1}\).
(Cyclic groups)
A cyclic group of order \(n\) is isomorphic to \(\ZZ/n\), has exactly one subgroup of each order \(d\) dividing \(n\), and each of its subgroups and quotients is cyclic. By the structure theorem, every finite abelian group is a direct product of cyclic groups of prime-power order.
Symmetric and alternating groups
Transposition presentation of the symmetric group (Tag D-6BTFJ)
Let \(n\geq 2\). The symmetric group \(S_n\) is the group with presentation \[ S_n \coloneqq \gens{ \sigma_1, \ldots, \sigma_{n-1} \st \sigma_i^2 \ (1\leq i\leq n-1),\ \sigma_i\sigma_j\sigma_i^{-1}\sigma_j^{-1} \ (\abs{i-j}\geq 2),\ \sigma_i \sigma_{i+1} \sigma_i \sigma_{i+1}^{-1} \sigma_i^{-1} \sigma_{i+1}^{-1} \ (1\leq i\leq n-2) }. \]
Sending \(\sigma_i\) to the transposition \((i, i+1)\) identifies this group with the group of bijections of \(\theset{1, \ldots, n}\).
Sign homomorphism (Tag D-BZ3KD)
Let \(n\geq 1\). The sign homomorphism is the map \[ \sgn\colon S_n \to (\ZZ/2, +), \qquad \sigma \mapsto k \bmod 2, \] where \(\sigma = \tau_1 \cdots \tau_k\) is any expression of \(\sigma\) as a product of \(k\geq 0\) transpositions.
Let \(n \geq 1\). For \(\sigma\in S_n\), all expressions of \(\sigma\) as a product of transpositions have the same number of factors modulo \(2\), so \(\sgn\) is well defined, and \(\sgn\) is a group homomorphism.
Even and odd permutations (Tag FD-PIVQZ)
Let \(\varepsilon\colon S_n\to\theset{\pm1}\) be the sign homomorphism and \(\sigma\in S_n\). The permutation \(\sigma\) is even if \(\varepsilon(\sigma) = 1\) and odd if \(\varepsilon(\sigma) = -1\).
Since every permutation is a product of transpositions and each transposition has sign \(-1\), a product of \(t\) transpositions has sign \((-1)^t\). So \(\sigma\) is even if and only if it is a product of an even number of transpositions, and odd if and only if it is a product of an odd number of transpositions.
Sign of a permutation from its cycle decomposition (Tag FD-GHDF2)
Let \(\sgn\colon S_n\to\theset{\pm1}\) be the sign homomorphism, and let \(\sigma\in S_n\) have disjoint cycle decomposition \(\sigma=c_1\cdots c_t\) with \(c_j\) a cycle of length \(k_j\). Then \[ \sgn(\sigma)=\prod_{j=1}^t(-1)^{k_j-1}. \] In particular, a cycle of length \(k\) is odd if and only if \(k\) is even, and \(\sigma\) is odd if and only if an odd number of the \(c_j\) have even length.
A \(k\)-cycle is a product of \(k-1\) transpositions: \((a_1\,a_2\,\cdots\,a_k)=(a_1\,a_k)(a_1\,a_{k-1})\cdots(a_1\,a_2)\). Since each transposition has sign \(-1\) and \(\sgn\) is a homomorphism, \(\sgn(c_j)=(-1)^{k_j-1}\), and the displayed formula follows by multiplying over \(j\). The factor \((-1)^{k_j-1}\) equals \(-1\) exactly when \(k_j\) is even, so the product equals \(-1\) exactly when the number of even-length cycles is odd.
Sign of $(1\,2\,3\,4\,5\,6)(7\,8\,9)(10\,11)(12\,13\,14\,15)(16\,17\,18)$ (Tag FF-6MNV6)
The permutation \(\sigma=(1\,2\,3\,4\,5\,6)(7\,8\,9)(10\,11)(12\,13\,14\,15)(16\,17\,18)\in S_{18}\) has sign \(\varepsilon(\sigma)=-1\).
A cycle of length \(\ell\) is a product of \(\ell-1\) transpositions, so its sign is \((-1)^{\ell-1}\) (FD-GHDF2). The disjoint cycles of \(\sigma\) have lengths \(6,3,2,4,3\), so \[ \varepsilon(\sigma)=(-1)^{5+2+1+3+2}=(-1)^{13}=-1. \] Equivalently, \(\sigma\) has three cycles of even length.
Alternating group (Tag D-TMME3)
Let \(n\geq 1\). A permutation \(\sigma\in S_n\) is even if \(\sgn(\sigma) = 0\), where \(\sgn\colon S_n\to\ZZ/2\) is the sign homomorphism. The alternating group is the subgroup of even permutations of the symmetric group: \[ A_n \coloneqq \theset{\sigma \in S_n \suchthat \sgn(\sigma) = 0} = \ker(\sgn). \]
A cycle of length \(\ell\) is a product of \(\ell-1\) transpositions, so a permutation is even if and only if its disjoint cycle decomposition has an even number of cycles of even length.
Alternating group (Tag FD-X4DBP)
Let \(n\geq 1\) and let \(\sgn\colon S_n\to\theset{\pm1}\) be the sign homomorphism. The alternating group is \[ A_n\coloneqq\ker(\sgn)=\theset{\sigma\in S_n \st \sgn(\sigma)=1}, \] the subgroup of even permutations of \(S_n\).
$A_n$ is generated by $3$-cycles (Tag PR-JY4PV)
For \(n\geq 3\), the alternating group \(A_n\) is generated by \(3\)-cycles.
(Conjugacy classes in $S_n$ and $A_n$)
Two permutations in \(S_n\) are conjugate if and only if they have the same cycle type, so the conjugacy classes of \(S_n\) correspond to the partitions of \(n\). For \(\sigma\in A_n\), the \(S_n\)-conjugacy class of \(\sigma\) is either one \(A_n\)-conjugacy class or the union of two of equal size; it splits into two if and only if the cycles of \(\sigma\), including fixed points as cycles of length \(1\), have distinct odd lengths.