Let \(G\) be the group of order \(21\) defined by the relations \[ a^7=b^3=1\quad\text{and}\quad b^{-1}ab=a^2. \]
Verify that the conjugacy classes of \(G\) are \[ \begin{align*} C_1&=\{1\},\\ C_2&=\{a,a^2,a^4\},\\ C_3&=\{a^3,a^5,a^6\},\\ C_4&=\{a^kb:k=0,\ldots,6\},\\ C_5&=\{a^kb^2:k=0,\ldots,6\}. \end{align*} \]
Show that \(H=\{a^k:k=0,\ldots,6\}\) is a normal subgroup of \(G\) for which \(G/H\) is isomorphic to \(\mathbb{Z}_3\). Give the character table for the lifting of the \(3\) linear characters of \(G/H\) to \(G\).
Let \(\chi\) be the linear character of \(H\) given by \[ \chi(a^k)=\eta^k\quad\text{for }k=0,\ldots,6 \] where \(\eta=e^{2\pi i/7}\). Show that \(\chi\uparrow_H^G\) is an irreducible character of \(G\).
Use parts (b) and (c) to give a complete character table for \(G\).
Solution
Let \(H=\langle a\rangle\cong C_7\). From \(b^{-1}ab=a^2\) one obtains \[ b^{-1}a^k b=a^{2k},\qquad ba^kb^{-1}=a^{4k}. \]
1. The conjugacy classes contained in \(H\) are\[ C_1=\{1\},\qquad C_2=\{a,a^2,a^4\},\qquad C_3=\{a^3,a^5,a^6\}. \]
Conjugation by powers of \(a\) fixes \(H\), while conjugation by \(b^{-1}\) multiplies exponents modulo \(7\) by \(2\). The nonzero orbits are \(\{1,2,4\}\) and \(\{3,6,5\}\).
2. The remaining conjugacy classes are\[ C_4=\{a^kb:0\le k<7\},\qquad C_5=\{a^kb^2:0\le k<7\}. \]
Since \(ba^j=a^{4j}b\), \[ a^{-j}ba^j=a^{3j}b. \] Multiplication by \(3\) permutes \(\mathbb Z/7\mathbb Z\), so the conjugates of \(b\) by \(H\) are exactly the seven elements of \(Hb\). Likewise \(b^2a^j=a^{2j}b^2\), hence \[ a^{-j}b^2a^j=a^jb^2, \] so the conjugates of \(b^2\) by \(H\) are exactly \(Hb^2\). The displayed classes have total size \(1+3+3+7+7=21\), so they exhaust \(G\).
3. The subgroup \(H\) is normal and \(G/H\cong C_3\).
The relation \(b^{-1}ab=a^2\) shows that \(b\) normalizes \(H\), and \(a\) does as well. Thus \(H\trianglelefteq G\). The quotient is generated by \(bH\), which has order \(3\), so \(G/H\cong C_3\).
4. If \(\omega=e^{2\pi i/3}\), the three linear characters inflated from \(G/H\) are\[ \begin{array}{c|rrrrr} &C_1&C_2&C_3&C_4&C_5\\ \hline \rho_0&1&1&1&1&1\\ \rho_1&1&1&1&\omega&\omega^2\\ \rho_2&1&1&1&\omega^2&\omega \end{array}. \]
The three characters of \(C_3=\langle bH\rangle\) send \(bH\) to \(1,\omega,\omega^2\). Their inflations are trivial on \(H\), and on \(Hb^2\) take the square of their value on \(Hb\).
Let \(\eta=e^{2\pi i/7}\), let \(\chi(a^k)=\eta^k\), and put \(\theta=\operatorname{Ind}_H^G\chi\).
5. The values of \(\theta\) are\[ \theta(C_1)=3,\qquad \theta(C_2)=\alpha,\qquad \theta(C_3)=\beta,\qquad \theta(C_4)=\theta(C_5)=0, \]where\[ \alpha=\eta+\eta^2+\eta^4,\qquad \beta=\eta^3+\eta^5+\eta^6. \]
Because \(H\) is normal, the induced character vanishes on \(G\setminus H\). For \(a^k\in H\), using coset representatives \(1,b,b^2\) gives \[ \theta(a^k)=\eta^k+\eta^{2k}+\eta^{4k}. \] This is \(\alpha\) on \(C_2\), \(\beta\) on \(C_3\), and \(3\) at the identity.
6. One has \(\alpha+\beta=-1\) and \(\alpha\beta=2\), hence \(|\alpha|^2=|\beta|^2=2\).
The first identity follows from \(1+\eta+\cdots+\eta^6=0\). Also \[ \alpha\beta =(\eta+\eta^2+\eta^4)(\eta^3+\eta^5+\eta^6) =3+(\eta+\eta^2+\eta^3+\eta^4+\eta^5+\eta^6)=2. \] Complex conjugation interchanges the two exponent sets, so \(\beta=\overline\alpha\).
7. The character \(\theta\) is irreducible.
Using the class sizes, \[ \langle\theta,\theta\rangle_G =\frac1{21}\left(9+3|\alpha|^2+3|\beta|^2\right) =\frac{9+6+6}{21}=1. \] Thus \(\theta\) is irreducible.
8. If \(\chi_3(a^k)=\eta^{3k}\) and \(\theta'=\operatorname{Ind}_H^G\chi_3\), then \(\theta'\) is irreducible with values\[ 3,\beta,\alpha,0,0 \]on \(C_1,\ldots,C_5\).
The nontrivial characters of \(H\) split into the two conjugation orbits with exponent sets \(\{1,2,4\}\) and \(\{3,5,6\}\). Repeating <1>5–<1>7 for the second orbit gives the stated values and norm \(1\).
9. Therefore the complete irreducible character table is\[ \begin{array}{c|rrrrr} &C_1&C_2&C_3&C_4&C_5\\ \text{class size}&1&3&3&7&7\\ \hline \rho_0&1&1&1&1&1\\ \rho_1&1&1&1&\omega&\omega^2\\ \rho_2&1&1&1&\omega^2&\omega\\ \theta&3&\alpha&\beta&0&0\\ \theta'&3&\beta&\alpha&0&0 \end{array}. \]
8. Their degree squares sum to\[ 1+1+1+9+9=21=|G|, \]so there are no further irreducible characters.