= Solution
Write the <nonabelian group of order 21> as
$$
G=\langle r,s\mid r^7=s^3=1,\ srs^{-1}=r^2\rangle
\cong C_7\rtimes C_3;
$$
replacing $2$ by $4$ gives the same group. Conjugation by $s$ splits the six nonidentity elements of the normal subgroup $\langle r\rangle$ into the two classes
$$
C_1=\{r,r^2,r^4\},\qquad
C_2=\{r^3,r^5,r^6\}.
$$
Each of the two other cosets is one conjugacy class:
$$
C_3=\langle r\rangle s,\qquad C_4=\langle r\rangle s^2.
$$
Thus the class sizes are $1,3,3,7,7$, so there are five <irreducible representation>[irreducible characters].
The <abelianization> is $C_3$, giving three <linear character>[linear characters]. Put $\omega=e^{2\pi i/3}$ and $\zeta=e^{2\pi i/7}$, and define
$$
\alpha=\zeta+\zeta^2+\zeta^4=\frac{-1+i\sqrt7}{2},
\qquad
\beta=\zeta^3+\zeta^5+\zeta^6=\frac{-1-i\sqrt7}{2}.
$$
The two nontrivial orbits of characters of the normal $C_7$ induce two irreducible characters of degree three. Their values vanish outside $C_7$, and summing each orbit gives $\alpha,\beta$ on the two nonidentity classes. The resulting <character table> is
$$
\begin{array}{c|ccccc}
&1&C_1&C_2&C_3&C_4\\
\hline
|C|&1&3&3&7&7\\
\chi_1&1&1&1&1&1\\
\chi_2&1&1&1&\omega&\omega^2\\
\chi_3&1&1&1&\omega^2&\omega\\
\chi_4&3&\alpha&\beta&0&0\\
\chi_5&3&\beta&\alpha&0&0
\end{array}.
$$
The degree check
$$
1^2+1^2+1^2+3^2+3^2=21
$$
and <character orthogonality>[character orthogonality relations] verify that this is the complete table.
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