Classify the groups of order \(182 = 2 \cdot 7 \cdot 13\).
Solution
There are exactly four isomorphism classes: \[ C_{182},\qquad C_7\times D_{26},\qquad D_{14}\times C_{13},\qquad D_{182}, \] where \(C_m\) is cyclic of order \(m\) and \(D_{2m}\) is the dihedral group of order \(2m\).
1. Every group \(G\) of order \(182\) has a normal cyclicsubgroup \(N\) of order \(91\).
Sylow’s theorems give \(n_7\mid26\) and \(n_7\equiv1\pmod7\) [@DF04]. Among \(1,2,13,26\), only \(1\) has this congruence. Thus the Sylow \(7\)-subgroup \(P\) is normal. The quotient \(G/P\) has order \(26\). Its Sylow \(13\)-subgroup is unique, since its number divides \(2\) and is congruent to \(1\) modulo \(13\). Its inverse image \(N\) is normal in \(G\) and has order \(7\cdot13=91\).
In \(N\), the Sylow \(13\)-subgroup \(Q\) is unique because its number divides \(7\) and is \(1\) modulo \(13\). The Sylow \(7\)-subgroup is also unique, since its number divides \(13\) and \(13\not\equiv1\pmod7\). Both are cyclic and normal. Their intersection is trivial by Lagrange’s theorem. For \(u\in P\) and \(v\in Q\), normality puts \(uvu^{-1}v^{-1}\) in both subgroups, so they commute. Multiplication therefore identifies \(P\times Q\) with \(N\): it is a homomorphism with trivial kernel between groups of order \(91\). A pair of generators has order \(91\), proving that \(N\) is cyclic.
2. Every such \(G\) is obtained from \(C_{91}\) by anautomorphism whose square is the identity, and exactly four qualify.
Cauchy’s theorem gives an element \(s\in G\) of order two [@DF04]. Since \(N\) has odd order, \(s\notin N\). As \([G:N]=2\), every element has a unique form \(a^i s^e\), where \(a\) generates \(N\), \(i\in\mathbb Z/91\mathbb Z\), and \(e\in\{0,1\}\). Conjugation has the form \(sas^{-1}=a^u\) for a unit \(u\) modulo \(91\). The relation \(s^2=1\) forces \(u^2\equiv1\pmod{91}\).
Modulo each of the primes \(7\) and \(13\), the equation \((u-1)(u+1)=0\) forces \(u=1\) or \(u=-1\). The Chinese remainder theorem gives one residue modulo \(91\) for each pair of signs [@DF04]. Thus the possibilities are precisely \[ (u\bmod7,u\bmod13)=(1,1),(1,-1),(-1,1),(-1,-1). \] Each is a unit and satisfies the required square equation.
Conversely, for any such \(u\), define multiplication on \((\mathbb Z/91\mathbb Z)\times(\mathbb Z/2\mathbb Z)\) by \[ (i,e)(j,f)=(i+u^e j,e+f). \] This is well defined because \(u^2=1\). Both ways of associating a triple give first coordinate \(i+u^e j+u^{e+f}k\) and second coordinate \(e+f+h\). The identity is \((0,0)\) and the inverse of \((i,e)\) is \((-u^{-e}i,-e)\). Hence this is a group of order \(182\). The rule \((i,e)\mapsto a^i s^e\) is an isomorphism onto any \(G\) with the given conjugation action.
3. The four actions give the displayed four groups,and none of them are isomorphic to one another.
Identify \(C_{91}\) with \(C_7\times C_{13}\). The signs in step 2 specify whether the involution fixes or inverts each factor. Fixing both gives \(C_{91}\times C_2\cong C_{182}\). Inverting just one gives the corresponding dihedral factor and the other cyclic factor. Inverting both gives \(D_{182}\), since inversion on \(C_7\times C_{13}\) is inversion on \(C_{91}\).
To distinguish these groups intrinsically, observe that \(N\) is exactly the set of odd-order elements of \(G\). Elements of \(N\) have odd order by Lagrange’s theorem, whereas an element outside \(N\) has nontrivial image in \(G/N\cong C_2\) and therefore has even order. Thus any isomorphism \(G_u\to G_v\) takes \(N_u\) to \(N_v\). With the generators of step 2, it must take \(a\mapsto b^k\) for a unit \(k\) modulo \(91\), and \(s\mapsto b^j t\) for some \(j\). Since \(N_v\) is abelian, conjugation by \(b^j t\) acts on \(N_v\) in the same way as conjugation by \(t\). Applying the isomorphism to \(sas^{-1}=a^u\) therefore gives \(b^{kv}=b^{ku}\). Canceling the unit \(k\) modulo \(91\) yields \(u=v\). The four residues in step 2 are distinct, so the four groups are pairwise nonisomorphic. The preceding steps prove that the list is exhaustive.