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Write $N=⟨a⟩$. The relation $ba{b}^{-1}=a^2$ lets every element be expressed as $a^i{b}^j$ with $0\le i<7$ and $0\le j<3$. There are 21 such expressions, so the assumption $|G|=21$ makes them distinct. In particular $a$ has order seven, $b$ has order three, and $N$ is a normal subgroup. This is the nonabelian group of order 21, the semidirect product $C_7⋊C_3$ with the action given by multiplication by two modulo seven.

Conjugation by $a$ fixes $N$, while conjugation by $b$ sends $a^i$ to $a^{2i}$. The nonidentity conjugacy classes inside $N$ are thereforeFor the other elements, direct calculation givesWhen $j=1$ or $2$, the coefficient $1-2^j$ is nonzero modulo seven, so letting $t$ vary gives every member of the coset $N{b}^j$. These cosets cannot merge under conjugation, since $G/N$ is an abelian group and conjugation acts trivially on that quotient group. Thus the five conjugacy classes have representatives and sizes

To find the linear characters, note that the commutator subgroup $G^{\prime }$ contains $ba{b}^{-1}{a}^{-1}=a$. The quotient group $G/N\cong C_3$ is abelian, so $G^{\prime }\subseteq N$; hence $G^{\prime }=N$. The abelianization is $C_3$. With $\omega =e^{2\pi i/3}$, its three linear characters areThis exhausts the linear characters because one-dimensional characters factor through the abelianization.

For the remaining irreducible characters, put $\zeta =e^{2\pi i/7}$ and take the linear characters ${\theta }_t$ of $N$ defined by ${\theta }_t(a)={\zeta }^t$. The six nontrivial linear characters form two orbits under conjugation, with exponent sets $\{1,2,4\}$ and $\{3,5,6\}$. Define the induced charactersFor $a^i\in N$, the character formula for an induced representation givesBoth vanish outside $N$, since no element outside a normal subgroup can be conjugate into it. This is induction from an abelian normal subgroup with a free character orbit. For completeness, its irreducibility here follows directly from Frobenius reciprocity and character orthogonality on the cyclic group $N$:The same calculation with ${\theta }_3,{\theta }_5,{\theta }_6$ proves that $\overline{\psi }$ is an irreducible character.

LetThe sum of the nontrivial seventh roots of unity gives $\alpha +\beta =-1$. Squaring and reducing exponents modulo seven givesThus $\alpha ,\beta$ are the two polynomial roots of $T^2+T+2$, andTo fix the signs with our specified $\zeta$, observe that $\mathrm{Im}\alpha =\sin (2\pi /7)+\sin (3\pi /7)-\sin (\pi /7)>0$, using the increase of sine on $(0,\pi /2)$. These Gaussian periods for seventh roots of unity make the answer explicit.

The resulting character table isThe two degree-three irreducible characters are distinct because $\alpha \ne \beta$. Together with the three linear characters, they exhaust the irreducible characters: the sum of squares of irreducible degrees is $3\cdot 1^2+2\cdot 3^2=21=|G|$. Equivalently, we have five distinct irreducible characters for the five conjugacy classes.

As a check on part (b), inversion interchanges $K_1$ and $K_2$, and also interchanges the two conjugacy classes outside $N$. Only the identity conjugacy class is real. Correspondingly ${\lambda }_0$ is the only real-valued character among the irreducible characters; ${\lambda }_1,{\lambda }_2$ and $\psi ,\overline{\psi }$ are two conjugate pairs.

Write a matrix as $(r,b)$, where $r=a^2$ is a nonzero quadratic residue modulo $11$. The square subgroup isMatrix multiplication becomesThus $H$ is the affine semidirect product of cyclic groups of orders eleven and fiveThere are five choices for $r$ and eleven for $b$, so $|H|=55$. It is nonabelian, since

LetConjugation by the complement givesHence the ten nonidentity translations split into the two orbitseach of size five. For $r\ne 1$,Since $1-r\ne 0$, varying $t$ gives every $b\in {\mathbb{F}}_{11}$. Conjugation cannot change $r$ in the abelian quotient $H/N$. Therefore, withthe seven conjugacy classes areof sizes $1,5,5,11,11,11,11$, agreeing with the conjugacy classes in the affine semidirect product of orders eleven and five.

The commutators with $(4,0)$ generate every translation because multiplication by $4-1=3$ is invertible in ${\mathbb{F}}_{11}$. Hence the commutator subgroup is $N$, and the abelianization isPut $\zeta =e^{2\pi i/5}$. The one-dimensional characters factor through the abelianization, giving five characters

For the remaining characters, put $\omega =e^{2\pi i/11}$ and define a character of $N$ byThe complement has two free orbits on the nontrivial ${\theta }_m$, indexed by $m\in Q$ and $m\in 2Q$. By induction from an abelian normal subgroup with a free character orbit, the charactersare irreducible of degree five and vanish outside $N$.

SetThe quadratic periods modulo eleven give the nonsquare sum $\overline{\eta }=(-1-i\sqrt{11})/2$. Multiplication by a square preserves $Q$ and multiplication by a nonsquare exchanges the two square classes, so the complete character table isFinally,and there are seven rows for the seven conjugacy classes. Thus these are all irreducible characters, as described by the irreducible characters of the affine semidirect product of orders eleven and five.