For each of the three orders \(n = 15, 20, 21\), either find a non-abelian group of order \(n\) or prove that all groups of order \(n\) are abelian.
Solution
1. Every group of order \(15\) is cyclic, hence abelian.
Let \(|G|=15\). Sylow’s theorems give \(n_5\mid3\), \(n_5\equiv1\pmod5\), and \(n_3\mid5\), \(n_3\equiv1\pmod3\) [@DF04]. The first pair forces \(n_5=1\); the second forces \(n_3=1\) because \(5\not\equiv1\pmod3\). Thus the Sylow subgroups \(P\) and \(Q\) of orders \(3\) and \(5\) are both normal. Their intersection is trivial by Lagrange’s theorem.
For \(u\in P\), \(v\in Q\), the commutator \(uvu^{-1}v^{-1}\) lies in both \(P\) and \(Q\) by their normality, so is one. Hence the multiplication map \(P\times Q\to G\) is a homomorphism. Its kernel is trivial, since \(uv=1\) implies \(u=v^{-1}\in P\cap Q\). Domain and codomain both have order \(15\), so this is an isomorphism. Both prime-order factors are cyclic, and a pair of generators has order \(\operatorname{lcm}(3,5)=15\). Therefore \(G\) is cyclic.
2. There is a nonabelian group of order \(20\).
On \(\mathbb Z/10\mathbb Z\), consider the permutations \[ f_{\varepsilon,b}(x)=\varepsilon x+b, \qquad \varepsilon\in\{1,-1\},\quad b\in\mathbb Z/10\mathbb Z. \] They form a group under composition, because \[ f_{\varepsilon,b}\circ f_{\delta,c} =f_{\varepsilon\delta,\,b+\varepsilon c}, \qquad f_{\varepsilon,b}^{-1}=f_{\varepsilon,\,-\varepsilon b}, \] and \(f_{1,0}\) is the identity. Associativity is inherited from composition of functions. Evaluation at \(0\) determines \(b\), and evaluation at \(1\) then determines \(\varepsilon\), since \(1\ne-1\) modulo \(10\). Thus these are exactly twenty distinct permutations.
Let \(r(x)=x+1\) and \(s(x)=-x\). Then \(srs^{-1}(x)=x-1\), which differs from \(r(x)=x+1\) modulo \(10\). Hence \(r,s\) do not commute. This gives the required nonabelian group, the dihedral group of order \(20\).
3. There is a nonabelian group of order \(21\).
The subset \(U=\{1,2,4\}\subseteq\mathbb F_7^\times\) is a subgroup: it consists of the powers of \(2\), with \(2^3=1\) and \(2,2^2\ne1\). On \(\mathbb F_7\) take the maps \[ g_{a,b}(x)=ax+b, \qquad a\in U,\quad b\in\mathbb F_7. \] Composition and inverse are \[ g_{a,b}\circ g_{c,d}=g_{ac,\,b+ad}, \qquad g_{a,b}^{-1}=g_{a^{-1},\,-a^{-1}b}. \] These formulas and the identity \(g_{1,0}\) prove that the maps form a group. They are distinct because evaluation at \(0\) recovers \(b\), and evaluation at \(1\) then recovers \(a\). The group therefore has \(3\cdot7=21\) elements.
For \(r(x)=x+1\) and \(t(x)=2x\), one has \(trt^{-1}(x)=x+2\ne x+1=r(x)\). Thus the group is nonabelian. This proves existence for the final order.