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    Is this group abelian?

    Contents

    Throughout, \(G\) is a finite group and \(p<q\) are primes.

    From the order alone

    \(\size G\) Conclusion
    \(p\) cyclic, hence abelian
    \(p^2\) abelian
    \(pq\) with \(p \notdivides q-1\) cyclic, hence abelian
    \(pq\) with \(p \divides q-1\) there is exactly one nonabelian group of this order up to isomorphism, \(\ZZ/q\semidirect\ZZ/p\)
    \(p^3\) not necessarily abelian: \(D_4\) and \(Q_8\) have order \(8\)

    Classification of groups of order $p$ (Tag PR-LFGHA)

    Every group of prime order is cyclic.Suggest a slogan

    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)

    Every group of order $p^2$ is abelian: cyclic or $C_p\times C_p$.Suggest a slogan

    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)

    Groups of order $pq$ are cyclic unless the smaller prime divides the larger minus one, when one nonabelian semidirect product also appears.Suggest a slogan

    Let \(p\) and \(q\) be primes with \(q<p\), and let \(G\) be a group of order \(pq\).

    1. If \(q\notdivides p-1\), then \(G\cong C_{pq}\).

    2. 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 \(bab^{-1}=a^\ell\neq a\).

    For \(\size G = p^2\), the center \(Z(G)\) is nontrivial by the class equation; if \(\size{Z(G)} = p\), then \(G/Z(G)\) has order \(p\) and is cyclic, and a group whose quotient by its center is cyclic is abelian, so \(Z(G)=G\), a contradiction. Hence \(Z(G)=G\).

    From a quotient

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

    Let \(gZ(G)\) generate \(G/Z(G)\). Every element of \(G\) has the form \(g^iz\) with \(i\in\ZZ\) and \(z\in Z(G)\), and \(g^iz\cdot g^jw = g^{i+j}zw = g^jw\cdot g^iz\) for \(z,w\in Z(G)\).

    $G/Z(G)$ cyclic implies $G$ abelian; $G/N$ abelian iff $[G,G]\le N$ (Tag E-O73XQ)

    • Show that if \(G/Z(G)\) is cyclic then \(G\) is abelian.

    • Show that \(G/N\) is abelian iff \([G, G] \leq N\).

    • Show that normal subgroups are not necessarily contained in \(Z(G)\).

      • Hint: consider the order 3 subgroup of \(S_3\).
    Solution

    If \(G/Z(G)\) is cyclic, then \(G\) is abelian:

    • Write \(H\da Z(G)\) and \(G/H = \gens{xH}\) as a cyclic quotient.

    • Fix \(a, b\in G\), then \(aH = x^n H\) and \(bH = x^m H\).

    • So \(ax^{-n} = h_1, bx^{-m} = h_2\) where the \(h_i\) are now central.

    • Now write \(ab = (x^n h_1)(x^m h_2) = ba\) by commuting everything.

    If \(G\) is nonabelian, then \(G/Z(G)\) is not cyclic, so \([G:Z(G)]\) is not prime. In particular \(\size{Z(G)} \neq \size G/p\) for every prime \(p\).

    From the structure theorem

    A finite abelian group is a direct product of cyclic groups of prime-power order, and the isomorphism classes of abelian groups of order \(\prod_i p_i^{e_i}\) correspond to tuples of partitions of the exponents \(e_i\). For example, there are \(2\cdot 1 = 2\) abelian groups of order \(p^2 q\), namely \(\ZZ/p^2\times\ZZ/q\) and \(\ZZ/p\times\ZZ/p\times\ZZ/q\). This is the classification of finitely generated modules over the principal ideal domain \(\ZZ\); see Classify this module.

    Number of abelian groups of order $n$ as a product of partition numbers (Tag PR-2JG3F)

    Abelian groups of order $\prod p_k^{e_k}$ are counted by the product of the partition numbers $P(e_k)$.Suggest a slogan

    Let \(n = \prod_{k=1}^m p_k^{e_k}\), where \(p_1, \ldots, p_m\) are distinct primes and \(e_1, \ldots, e_m \geq 1\), and for an integer \(e \geq 1\) let \(P(e)\) be the number of partitions of \(e\). Up to isomorphism, there are exactly \(\prod_{k=1}^m P(e_k)\) abelian groups of order \(n\).

    Distinguishing nonabelian groups

    A semidirect product decomposition and the following isomorphism invariants separate nonabelian groups of the same order:

    • A normal Sylow subgroup \(N\) with a complement \(H\) exhibits \(G \cong N \semidirect H\); see Show $G$ is not simple and Quotients, products, and automorphisms.

    • The number of elements of each order: \(D_4\) has five elements of order \(2\) and \(Q_8\) has one.

    • The abelianization \(G/[G,G]\), computable from a presentation by adjoining the relations \([x,y]=1\) for all generators \(x,y\).

    • The center, the sizes of the conjugacy classes, and the automorphism group.

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