Show that there are at least two nonisomorphic nonabelian groups of order \(40\).
Solution
1. Let\[ G\coloneqq C_{20}\rtimes C_2, \]where the nonidentity element of \(C_2\) acts on \(C_{20}\) by inversion. Then\(G\) is a nonabelian group of order \(40\).
Write \[ C_{20}=\langle a\rangle, \qquad C_2=\langle s\rangle, \] with \[ sas^{-1}=a^{-1}. \] The inversion map is an automorphism of \(C_{20}\) of order \(2\), so the semidirect product is defined and has \[ \abs{G}=20\cdot2=40. \] Since \(a\) has order \(20\), one has \(a\neq a^{-1}\), and therefore \(sa\neq as\). Thus \(G\) is nonabelian.
2. The group \(G\) has exactly \(21\) elements of order \(2\).
Inside the cyclic subgroup \(\langle a\rangle\), the unique involution is \(a^{10}\). Every element outside that subgroup has the form \(a^is\), and \[ (a^is)^2 = a^is a^is = a^ia^{-i} =1. \] The \(20\) elements \(a^is\) are distinct and nonidentity. Hence \(G\) has \(20+1=21\) involutions.
3. Let\[ H\coloneqq C_5\rtimes C_8, \]where a generator of \(C_8\) acts on \(C_5\) by inversion. Then \(H\) is anonabelian group of order \(40\).
Write \[ C_5=\langle c\rangle, \qquad C_8=\langle b\rangle, \] and define the action by \[ bcb^{-1}=c^{-1}. \] Since inversion has order \(2\), the assignment of the generator \(b\) to inversion defines a homomorphism \(C_8\to\Aut(C_5)\). Hence the semidirect product exists and has \[ \abs{H}=5\cdot8=40. \] Because \(c\neq c^{-1}\), the relation \(bcb^{-1}=c^{-1}\) shows that \(H\) is nonabelian.
4. The group \(H\) has exactly one element of order \(2\).
Every element of \(H\) has a unique form \[ c^i b^j, \qquad 0\leq i<5, \quad 0\leq j<8. \] The relation \(bcb^{-1}=c^{-1}\) gives \[ (c^ib^j)^2 = c^{\,i(1+(-1)^j)}b^{2j}. \] If \(j\) is odd, then \(b^{2j}\neq1\), so the square is not \(1\). If \(j\) is even, the square equals \[ c^{2i}b^{2j}. \] For this to be \(1\), one must have \(j=0\) or \(j=4\) and \(i=0\). The case \((i,j)=(0,0)\) is the identity, while \((i,j)=(0,4)\) gives the involution \(b^4\). Thus \(b^4\) is the unique involution in \(H\).
5. The groups \(G\) and \(H\) are not isomorphic.
An isomorphism preserves element orders and hence preserves the number of involutions. By step 2, \(G\) has \(21\) involutions, whereas by step 4, \(H\) has exactly one. Therefore \(G\not\cong H\).
6. Q.E.D.
Steps 1 and 3 construct two nonabelian groups of order \(40\), and step 5 proves that they are nonisomorphic.