Groups of special orders
Every group of prime order \(p\) is cyclic, and every group of order \(p^2\) is abelian, hence isomorphic to \(\ZZ/p^2\) or \(\ZZ/p\times\ZZ/p\). For primes \(p<q\), a group of order \(pq\) has a normal Sylow \(q\)-subgroup \(Q\cong\ZZ/q\), and is isomorphic to \(\ZZ/q\semidirect_\psi\ZZ/p\) for a homomorphism \(\psi\colon\ZZ/p\to\Aut(\ZZ/q)\cong(\ZZ/q)^\times\); a nontrivial \(\psi\) exists if and only if \(p\divides q-1\). For orders with more prime factors, Frattini's argument factors \(G\) as \(N_G(P)\,N\) for a normal subgroup \(N\) and a Sylow subgroup \(P\) of \(N\), every finite $p$-group is solvable, and a group of order \(pqr\) for distinct primes \(p<q<r\) is solvable.
Classification of groups of order $p$ (Tag PR-LFGHA)
Let \(p\) be a prime. Every group \(G\) of order \(p\) is cyclic, and hence \(G \cong \ZZ/p\ZZ\).
Classification of groups of order $p^2$ (Tag PR-IGLFV)
Let \(p\) be a prime. Every group \(G\) of order \(p^2\) is abelian, and hence \(G \cong C_{p^2}\) or \(G \cong C_p \times C_p\).
Classification of groups of order $pq$ (Tag PR-SLWTB)
Let \(p\) and \(q\) be primes with \(q<p\), and let \(G\) be a group of order \(pq\).
If \(q\notdivides p-1\), then \(G\cong C_{pq}\).
If \(q\divides p-1\), then either \(G\cong C_{pq}\), or \(G\) is nonabelian and \[ G\cong C_p\semidirect_\psi C_q \cong \gens{a, b \suchthat a^p=b^q=1,\ bab^{-1} = a^\ell}, \] where \(\psi\colon C_q\to\Aut(C_p)\) is nontrivial and \(\ell\) is any integer with \(\ell \not\equiv 1 \pmod p\) and \(\ell^q \equiv 1 \pmod p\). Up to isomorphism, the nonabelian group does not depend on the choice of \(\psi\) or \(\ell\).
Let \(P\) and \(Q\) be Sylow $p$- and $q$-subgroups of \(G\), so \(P\cong C_p\) and \(Q\cong C_q\). By the Sylow theorems \(n_p \equiv 1 \pmod p\) and \(n_p \divides q\); since \(q < p\), \(n_p = 1\) and \(P\normal G\). As \(P\cap Q=1\), \(\abs{PQ}=pq\), so \(G=PQ\cong P\semidirect_\psi Q\) for the conjugation action \(\psi\colon Q\to\Aut(P)\cong(\ZZ/p\ZZ)^\times\cong C_{p-1}\).
If \(q\notdivides p-1\), then \(\psi\) is trivial, so \(G\cong C_p\times C_q\cong C_{pq}\).
If \(q\divides p-1\) and \(\psi\) is trivial, again \(G\cong C_{pq}\). If \(\psi\) is nontrivial, it is injective because \(\abs{Q}=q\) is prime, and its image is the unique subgroup of order \(q\) of the cyclic group \(\Aut(P)\). Let \(a\) generate \(P\). Two nontrivial homomorphisms \(Q\to\Aut(P)\) with the same image differ by precomposition with an automorphism of \(Q\), which gives isomorphic semidirect products. Choosing a generator \(b\) of \(Q\) with \(\psi(b)(a)=a^\ell\) gives the presentation, and the conditions on \(\ell\) say exactly that \(a\mapsto a^\ell\) has order \(q\) in \(\Aut(P)\). This group is nonabelian because \(bab^{-1}=a^\ell\neq a\).
Frattini's argument (Tag L-6QBOJ)
Let \(G\) be a finite group, let \(N\) be a normal subgroup of \(G\), let \(p\) be a prime, and let \(P\) be a Sylow $p$-subgroup of \(N\). Then \(G=N_G(P)N\), where \(N_G(P)\) is the normalizer of \(P\) in \(G\).
Let \(g\in G\). Since \(N\) is normal, \(gPg^{-1}\subseteq N\), and it is a Sylow \(p\)-subgroup of \(N\). By Sylow's second theorem in \(N\), there is \(n\in N\) with \(ngPg^{-1}n^{-1}=P\), so \(ng\in N_G(P)\). Hence \(g=n^{-1}(ng)\in NN_G(P)=N_G(P)N\), the last equality because \(N\) is normal.
Finite $p$-groups are solvable (Tag L-Q44MT)
Let \(p\) be a prime. Every finite $p$-group is solvable.
Induct on \(\abs G\); the trivial group is solvable. A nontrivial finite \(p\)-group \(G\) has nontrivial center \(Z(G)\) by the class equation. The quotient \(G/Z(G)\) is a \(p\)-group of smaller order, hence solvable by induction, and \(Z(G)\) is an abelian normal subgroup. Pulling back a normal series of \(G/Z(G)\) with abelian factors to \(G\) and appending \(Z(G)\trianglerighteq\theset{e}\) gives a normal series of \(G\) with abelian factors.
A group of order $pq$ with $p<q$ has a normal Sylow $q$-subgroup (Tag L-WROBY)
Let \(p<q\) be primes and let \(G\) be a group with \(\abs G=pq\). Then \(G\) has a normal subgroup of order \(q\).
Let \(n_q\) be the number of Sylow \(q\)-subgroups of \(G\). By Sylow's third theorem, \(n_q\equiv1\pmod q\) and \(n_q\divides p\). The divisors of \(p\) are \(1\) and \(p\), and \(1<p<q\) gives \(p\not\equiv1\pmod q\), so \(n_q=1\) and \(G\) has a unique Sylow \(q\)-subgroup \(Q\). For every \(g\in G\), the conjugate \(gQg^{-1}\) is again a Sylow \(q\)-subgroup, so \(gQg^{-1}=Q\). Therefore \(Q\trianglelefteq G\), and \(\abs Q=q\).
Groups of order $pqr$ are solvable (Tag PR-2ZOAV)
Let \(p < q < r\) be primes and let \(G\) be a group with \(\abs{G} = pqr\). Then \(G\) is solvable.
The small orders
The abelian groups of each order are given by the structure theorem for finite abelian groups (Number of abelian groups of order $n$ as a product of partition numbers).
The smallest nonabelian group has order 6 (Tag FF-VAKLJ)
Every group of order at most \(5\) is abelian, and up to isomorphism the only nonabelian group of order \(6\) is the dihedral group \(D_3\cong S_3\).
Groups of order \(1\), \(2\), \(3\), and \(5\) are cyclic, groups of order \(4\) are abelian by FF-TSZNT, and the groups of order \(6\) are classified in FF-TEGRU.
Groups of order 4 (Tag FF-TSZNT)
Up to isomorphism, the groups of order \(4\) are \(\ZZ/4\ZZ\) and \((\ZZ/2\ZZ)^2\). Both are abelian.
Let \(\abs G=4\). If \(G\) has an element of order \(4\), then \(G\cong\ZZ/4\ZZ\). Otherwise every nonidentity element has order \(2\), so \(g^{-1}=g\) for all \(g\), and \(ab=(ab)^{-1}=b^{-1}a^{-1}=ba\) for all \(a,b\in G\). Then \(G\) is an abelian group of exponent \(2\), that is, written additively, a \(2\)-dimensional \(\FF_2\)-vector space, and \(G\cong(\ZZ/2\ZZ)^2\).
The cycle graphs of \(\ZZ/4\ZZ = \gens{g}\) and of \((\ZZ/2\ZZ)^2 = \theset{e, a, b, ab}\), with identity \(e\):
Groups of order 6 (Tag FF-TEGRU)
Up to isomorphism, the groups of order \(6\) are the cyclic group \(\ZZ/6\ZZ\), which is abelian, and the dihedral group \(D_3\cong S_3\) of order \(6\), which is nonabelian.
Let \(\abs G=6\). By Sylow's theorems, the number of Sylow \(3\)-subgroups divides \(2\) and is congruent to \(1\) modulo \(3\), so there is a unique, hence normal, Sylow \(3\)-subgroup \(N\cong\ZZ/3\ZZ\). A Sylow \(2\)-subgroup \(H\cong\ZZ/2\ZZ\) meets \(N\) trivially and \(NH=G\), so \(G\cong N\rtimes_\theta H\) for a homomorphism \(\theta\colon H\to\Aut(N)\cong\theset{\pm1}\). If \(\theta\) is trivial, \(G\cong\ZZ/3\ZZ\times\ZZ/2\ZZ\cong\ZZ/6\ZZ\); otherwise the generator of \(H\) acts by inversion and \(G\cong D_3\).
Groups of order 8 (Tag FF-TDXQT)
Up to isomorphism there are five groups of order \(8\). The abelian ones are \[ \ZZ/8\ZZ,\qquad\ZZ/4\ZZ\times\ZZ/2\ZZ,\qquad(\ZZ/2\ZZ)^3, \] and the nonabelian ones are the dihedral group \(D_4\) of order \(8\) and the quaternion group \(Q_8\).
Groups of order 9 (Tag FF-C3SN4)
Up to isomorphism, the groups of order \(9\) are \(\ZZ/9\ZZ\) and \((\ZZ/3\ZZ)^2\). Both are abelian.
A group \(G\) of order \(p^2\), \(p\) prime, has nontrivial center \(Z(G)\) by the class equation. If \(Z(G)\ne G\), then \(G/Z(G)\) has order \(p\) and is cyclic, which forces \(G\) to be abelian, a contradiction. So \(G\) is abelian, and the classification of finite abelian groups gives \(\ZZ/p^2\ZZ\) or \((\ZZ/p\ZZ)^2\).
Groups of order 10 (Tag FF-2NPCE)
Up to isomorphism, the groups of order \(10\) are the cyclic group \(\ZZ/10\ZZ\), which is abelian, and the dihedral group \(D_5\) of order \(10\), which is nonabelian.
Let \(\abs G=10\). By Sylow's theorems, the number of Sylow \(5\)-subgroups divides \(2\) and is congruent to \(1\) modulo \(5\), so there is a unique, hence normal, Sylow \(5\)-subgroup \(N\cong\ZZ/5\ZZ\). A Sylow \(2\)-subgroup \(H\cong\ZZ/2\ZZ\) meets \(N\) trivially and \(NH=G\), so \(G\cong N\rtimes_\theta H\) for a homomorphism \(\theta\colon H\to\Aut(N)\cong(\ZZ/5\ZZ)^\times\). The elements of order dividing \(2\) in \((\ZZ/5\ZZ)^\times\) are \(\pm1\). If \(\theta\) is trivial, \(G\cong\ZZ/5\ZZ\times\ZZ/2\ZZ\cong\ZZ/10\ZZ\); if the generator of \(H\) acts by inversion, \(G\cong D_5\).
Groups of order 12 (Tag FF-I6XN6)
Up to isomorphism there are five groups of order \(12\). The abelian ones are \[ \ZZ/4\ZZ\times\ZZ/3\ZZ\cong\ZZ/12\ZZ\qquad\text{and}\qquad(\ZZ/2\ZZ)^2\times\ZZ/3\ZZ, \] and the nonabelian ones are
the alternating group \(A_4\),
the dihedral group \(D_6\) of order \(12\),
the dicyclic group of order \(12\), \[ \gens{a,b\suchthat a^6=1,\ b^2=a^3,\ bab^{-1}=a^{-1}}\cong\ZZ/3\ZZ\rtimes\ZZ/4\ZZ, \] where a generator of \(\ZZ/4\ZZ\) acts on \(\ZZ/3\ZZ\) by inversion.
Groups of order 14 (Tag FF-GWCXC)
Up to isomorphism, the groups of order \(14\) are the cyclic group \(\ZZ/14\ZZ\), which is abelian, and the dihedral group \(D_7\) of order \(14\), which is nonabelian.
Let \(\abs G=14\). By Sylow's theorems, the number of Sylow \(7\)-subgroups divides \(2\) and is congruent to \(1\) modulo \(7\), so there is a unique, hence normal, Sylow \(7\)-subgroup \(N\cong\ZZ/7\ZZ\). A Sylow \(2\)-subgroup \(H\cong\ZZ/2\ZZ\) meets \(N\) trivially and \(NH=G\), so \(G\cong N\rtimes_\theta H\) for a homomorphism \(\theta\colon H\to\Aut(N)\cong(\ZZ/7\ZZ)^\times\). The elements of order dividing \(2\) in \((\ZZ/7\ZZ)^\times\) are \(\pm1\). If \(\theta\) is trivial, \(G\cong\ZZ/7\ZZ\times\ZZ/2\ZZ\cong\ZZ/14\ZZ\); if the generator of \(H\) acts by inversion, \(G\cong D_7\).
Groups of order 15 (Tag FF-Q6XCL)
Every group of order \(15\) is isomorphic to the cyclic group \(\ZZ/15\ZZ\).
Let \(\abs G=15\). By Sylow's theorems, the number of Sylow \(3\)-subgroups divides \(5\) and is congruent to \(1\) modulo \(3\), and the number of Sylow \(5\)-subgroups divides \(3\) and is congruent to \(1\) modulo \(5\); both numbers are \(1\). So \(G\) has normal subgroups \(P\cong\ZZ/3\ZZ\) and \(Q\cong\ZZ/5\ZZ\) with \(P\cap Q=\theset{e}\) and \(\abs{PQ}=15\), and FT-7NMQR gives \(G\cong\ZZ/3\ZZ\times\ZZ/5\ZZ\cong\ZZ/15\ZZ\).
Groups of order 16 (Tag FF-IRHG3)
Up to isomorphism there are fourteen groups of order \(16\). The five abelian ones are \[ \ZZ/16\ZZ,\quad(\ZZ/4\ZZ)^2,\quad\ZZ/8\ZZ\times\ZZ/2\ZZ,\quad\ZZ/4\ZZ\times(\ZZ/2\ZZ)^2,\quad(\ZZ/2\ZZ)^4, \] and the nine nonabelian ones are
\((\ZZ/4\ZZ\times\ZZ/2\ZZ)\rtimes\ZZ/2\ZZ=\gens{a,b,c\suchthat a^4=b^2=c^2=1,\ ab=ba,\ cac^{-1}=ab,\ cbc^{-1}=b}\),
\(\ZZ/4\ZZ\rtimes\ZZ/4\ZZ\), with a generator of the second factor acting by inversion,
the modular group \(\ZZ/8\ZZ\rtimes\ZZ/2\ZZ=\gens{r,s\suchthat r^8=s^2=1,\ srs^{-1}=r^5}\),
the dihedral group \(D_8\) of order \(16\),
\(D_4\times\ZZ/2\ZZ\), with \(D_4\) the dihedral group of order \(8\),
\(Q_8\times\ZZ/2\ZZ\), with \(Q_8\) the quaternion group,
the generalized quaternion group \(Q_{16}=\gens{a,b,c\suchthat a^4=b^2=c^2=abc}\),
the semidihedral group \(QD_{16}=\gens{r,s\suchthat r^8=s^2=1,\ srs^{-1}=r^3}\),
the Pauli group \((\ZZ/4\ZZ\times\ZZ/2\ZZ)\rtimes\ZZ/2\ZZ=\gens{a,b,c\suchthat a^4=b^2=c^2=1,\ ab=ba,\ cac^{-1}=a,\ cbc^{-1}=a^2b}\), generated by the Pauli matrices.
The groups in items 1 and 9 are both semidirect products \((\ZZ/4\ZZ\times\ZZ/2\ZZ)\rtimes\ZZ/2\ZZ\) and are not isomorphic: the center of the group in item 1 is \(\gens{a^2,b}\cong(\ZZ/2\ZZ)^2\), and the center of the Pauli group is \(\gens{a}\cong\ZZ/4\ZZ\).
Groups of order 18 (Tag FF-MBUKG)
Up to isomorphism there are five groups of order \(18\). The abelian ones are \(\ZZ/18\ZZ\) and \((\ZZ/3\ZZ)^2\times\ZZ/2\ZZ\), and the nonabelian ones are
the dihedral group \(D_9\) of order \(18\),
\(S_3\times\ZZ/3\ZZ\),
the generalized dihedral group \((\ZZ/3\ZZ)^2\rtimes\ZZ/2\ZZ\), with the generator of \(\ZZ/2\ZZ\) acting by inversion.
Groups of order 20 (Tag FF-JKCAM)
Up to isomorphism there are five groups of order \(20\). The abelian ones are \(\ZZ/20\ZZ\) and \((\ZZ/2\ZZ)^2\times\ZZ/5\ZZ\), and the nonabelian ones are
the dihedral group \(D_{10}\) of order \(20\),
\(\ZZ/5\ZZ\rtimes\ZZ/4\ZZ\) with \(\ZZ/4\ZZ\) acting faithfully, which is the affine group \(\operatorname{AGL}_1(\FF_5)\),
the dicyclic group of order \(20\), \[ \gens{a,b\suchthat a^{10}=1,\ b^2=a^5,\ bab^{-1}=a^{-1}}. \]
There are five groups of order \(8\) up to isomorphism, of which \(D_4\) and \(Q_8\) are nonabelian, and five groups of order \(12\), of which \(A_4\), \(D_6\), and \(\ZZ/3\semidirect\ZZ/4\) are nonabelian.