Let \(H,K\le G\) with \(H\cap K=1\) and \(HK=G\). If \(H\) and \(K\) are both normal, then \(G\cong H\times K\); if \(H\) is normal, then \(G\cong H\rtimes_\varphi K\) with \(\varphi(k)(h)=khk^{-1}\). For \(K\) cyclic and homomorphisms \(\varphi_1,\varphi_2\colon K\to\Aut(H)\) whose images are conjugate in \(\Aut(H)\), \(H\rtimes_{\varphi_1}K\cong H\rtimes_{\varphi_2}K\); and \(\Aut(\ZZ/n\ZZ)\cong(\ZZ/n\ZZ)^\times\). The fundamental theorem of finite abelian groups writes a finite abelian group uniquely as \(\ZZ/d_1\ZZ\times\cdots\times\ZZ/d_r\ZZ\) with \(1<d_1\mid d_2\mid\cdots\mid d_r\) (invariant factors), and uniquely up to order as a product of cyclic groups of prime-power order (elementary divisors). A presentation \(\langle S\mid R\rangle\) is the quotient of the free group on \(S\) by the normal subgroup generated by \(R\).
Group presentation (Tag D-V6UJ7)
Let \(S\) be a set, let \(F[S]\) be the free group on \(S\), and let \(R\subseteq F[S]\) be a set of words in \(S\). The group presented by generators \(S\) and relations \(R\) is \[ \gens{S \st R} \coloneqq F[S] / \cl_n(R), \] where \(\cl_n(R)\) is the normal closure of \(R\) in \(F[S]\), the smallest normal subgroup of \(F[S]\) containing \(R\). A presentation of a group \(G\) is an isomorphism \(G\cong\gens{S \st R}\).
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, \(ji=-k\), \(kj=-i\), \(ik=-j\):
Recognizing internal direct products (Tag T-SVJUN)
Let \(G\) be a group with subgroups \(H\) and \(K\) such that
\(H, K \normal G\),
\(G = HK\), and
\(H\intersect K = \theset{e}\).
Then the map \(H \times K\to G\), \((h,k)\mapsto hk\), is an isomorphism, so \(G \cong H \times K\).
Hypothesis 1 may be replaced by the condition that \(hk=kh\) for all \(h\in H\) and \(k\in K\): together with \(G=HK\), this condition implies that \(H\) and \(K\) are normal in \(G\).
Recognition theorem for internal direct products (Tag FT-7NMQR)
Let \(G\) be a group and let \(H,K\le G\) be normal subgroups with \(H\cap K=\theset{e}\) and \(HK=G\). Then the map \[ H\times K\to G,\qquad(h,k)\mapsto hk, \] is an isomorphism of groups.
For \(h\in H\) and \(k\in K\), the commutator \(hkh^{-1}k^{-1}\) lies in \(K\) because \(K\) is normal and in \(H\) because \(H\) is normal, so it is \(e\) and \(hk=kh\). Hence \((h,k)(h',k')=(hh',kk')\mapsto hh'kk'=hkh'k'\), and the map is a homomorphism. Its kernel consists of the pairs with \(hk=e\), so \(h=k^{-1}\in H\cap K=\theset{e}\), and the map is injective. It is surjective because \(HK=G\).
Recognizing internal direct products of finitely many subgroups (Tag T-YNKCZ)
Let \(G\) be a group with subgroups \(H_1,\ldots,H_n\) such that
\(H_i \normal G\) for all \(i\),
\(G = H_1 \cdots H_n\), and
for each \(k\), \(H_k \intersect \left(H_1 \cdots H_{k-1}H_{k+1} \cdots H_n\right) = \theset{e}\).
Then the map \(\prod_{i=1}^n H_i\to G\), \((h_1,\ldots,h_n)\mapsto h_1\cdots h_n\), is an isomorphism, so \(G \cong \prod_{i=1}^n H_i\).
Recognizing internal semidirect products (Tag T-SB6AV)
Let \(G\) be a group with subgroups \(N \normal G\) and \(H\leq G\) such that \(G = NH\) and \(N\intersect H=\theset{e}\). Let \(\psi\colon H \to \Aut(N)\) be the conjugation action, \(\psi(h)=h(\wait)h^{-1}\). Then \(G \cong N \semidirect_\psi H\).
If moreover \(H\normal G\), then \(hnh^{-1}n^{-1}\in N\intersect H=\theset{e}\) for all \(n\in N\) and \(h\in H\), so \(\psi\) is trivial and \(G\cong N\times H\).
Automorphism groups of cyclic groups and related counts (Tag PR-N6S6P)
Let \(\varphi\) be Euler's totient function.
For \(n \geq 1\), a cyclic group \(C_n\) has exactly \(\varphi(n)\) generators, and for primes \(p \neq q\) and \(k, \ell \geq 1\), \[ \begin{aligned} \varphi(p) &= p-1, \\ \varphi(p^k) &= p^{k-1}(p - 1), \\ \varphi(p^k q^\ell) &= \varphi(p^k)\,\varphi(q^\ell). \end{aligned} \]
\(\Aut(C_n) \cong \unitsof{(\ZZ/n\ZZ)}\), a group of order \(\varphi(n)\). It is cyclic if and only if \(n = 1, 2, 4, p^k\), or \(2p^k\) with \(p\) an odd prime and \(k \geq 1\).
For \(p\) an odd prime and \(k \geq 1\), \(\Aut(C_{p^k}) \cong C_{\varphi(p^k)}\); in particular \(\Aut(C_p) \cong C_{p-1}\).
\(\Aut(C_2) = 1\), \(\Aut(C_4) \cong C_2\), and for \(k \geq 3\), \(\Aut(C_{2^k}) \cong C_2 \times C_{2^{k-2}}\).
If \(G\) and \(H\) are finite groups of coprime orders, then \(\Aut(G \times H) \cong \Aut(G) \times \Aut(H)\). Consequently, if \(n = \prod_{k=1}^\ell p_k^{n_k}\) with distinct primes \(p_k\), \(p_1 = 2\), and \(n_1 \geq 3\), then \(C_n \cong \prod_{k=1}^{\ell} C_{p_k^{n_k}}\) and \[ \begin{aligned} \Aut(C_n) &\cong \prod_{k=1}^\ell \Aut\qty{C_{p_k^{n_k}}} \\ &\cong \prod_{k=1}^\ell \unitsof{(\ZZ/p_k^{n_k}\ZZ)} \\ &\cong \qty{C_2 \times C_{2^{n_1-2}} } \times \prod_{k=2}^\ell C_{m_k}, \qquad m_k \coloneqq \varphi(p_k^{n_k}) = p_k^{n_k-1}(p_k-1). \end{aligned} \]
For \(m \geq 2\) and \(n \geq 1\), \(\Aut(C_m^n) \cong \GL_n(\ZZ/m\ZZ)\). For \(p\) prime, \(\Aut(C_p^n) \cong \GL_n(\FF_p)\), and \[ \abs{\GL_n(\FF_p)} = \prod_{k=0}^{n-1}(p^n-p^k) = (p^n-1)(p^n-p)(p^n-p^2)\cdots(p^n-p^{n-1}). \]
For \(m, n \geq 1\), \(\abs{\Hom(C_n, C_m)} = \gcd(n, m)\).
Let \(N, H\) be groups, \(\psi\colon H \to \Aut(N)\) a homomorphism, \(\sigma \in \Aut(H)\), and \(\tau \in \Aut(N)\), and let \(c_\tau\colon \Aut(N) \to \Aut(N)\) be conjugation \(\alpha \mapsto \tau\alpha\inverseof{\tau}\). Then \(N \semidirect_\psi H \cong N \semidirect_{c_\tau \circ \psi \circ \sigma} H\).
For every group \(G\), \(\Inn(G) \cong G/Z(G)\).
\(\Aut(C_8) \cong \unitsof{(\ZZ/8\ZZ)} = \theset{1, 3, 5, 7}\) has every nonidentity element of order \(2\), so \(\Aut(C_8) \cong C_2 \times C_2 \not\cong C_4\).
By (h), for \(N = C_p^n\) and \(H = C_k = \gens{h}\), a homomorphism \(\psi\colon C_k \to \GL_n(\FF_p)\) is determined by the matrix \(A = \psi(h)\), and replacing \(A\) by a similar matrix \(PA\inverseof{P}\) gives an isomorphic semidirect product. So it suffices to take \(A\) in a canonical form, such as the rational canonical form.
N/C theorem: $N_G(H)/C_G(H)$ embeds in $\Aut(H)$ (Tag PR-NESS4)
Let \(G\) be a group and \(H\leq G\) a subgroup, with normalizer \(N_G(H)\) and centralizer \(C_G(H)\). The map \(c\colon N_G(H)\to\Aut(H)\), \(c(g)(h)=ghg^{-1}\), is a group homomorphism with kernel \(C_G(H)\). Hence \(C_G(H)\normal N_G(H)\) and \(c\) induces an injective homomorphism \[ N_G(H)/C_G(H)\hookrightarrow\Aut(H). \]
For \(g\in N_G(H)\) we have \(gHg^{-1}=H\), so \(c(g)\) is a bijection \(H\to H\), and \(c(g)(hh')=ghg^{-1}gh'g^{-1}=c(g)(h)\,c(g)(h')\); thus \(c(g)\in\Aut(H)\). Since \(c(gg')(h)=gg'h(gg')^{-1}=c(g)(c(g')(h))\), the map \(c\) is a homomorphism. Its kernel is \(\theset{g\in N_G(H)\suchthat ghg^{-1}=h\ \forall h\in H}=C_G(H)\). The first isomorphism theorem gives \(N_G(H)/C_G(H)\cong c(N_G(H))\leq\Aut(H)\).
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 $pq$ (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 \(bab^{-1}=a^\ell\neq a\).
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 \(20\), \[ \gens{a,b\suchthat a^{10}=1,\ b^2=a^5,\ bab^{-1}=a^{-1}}. \]
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.