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The conjugacy classes are $\{1\}$, $\{a^n\}$, $\{a^r,a^{-r}\}$ for $1\le r\le n-1$, and two classes containing respectively $b{a}^{2s}$ and $b{a}^{2s+1}$, each of size $n$. Conjugation by $b$ inverts powers of $a$, while conjugation by $a$ changes the exponent of $b{a}^r$ by two; these facts prove the list and sizes. The sizes total $4n$.

The complete character table is specified by the following array, with one column for every $r$ and one row for every $j$ in the indicated ranges:Here $1\le r,j\le n-1$; the four pairs $(\alpha ,\beta )$ are those in part (ii). In particular, this formula displays both parity cases without concealing the different values $\pm i$ when $n$ is odd. The zero entries follow because the two-dimensional matrices representing $b{a}^r$ are off-diagonal.

The identity is one conjugacy class. In $N\setminus \{1\}$, conjugation by $c$ pairs each element with its inverse, giving four classes represented by $a,b,ab,a{b}^2$, each of size two. All nine elements of $Nc$ are conjugate, since conjugation by $n\in N$ changes the reflection component by $n^2$ and squaring is a bijection of $N$.

There are two linear characters: the trivial character and the character equal to one on $N$ and minus one on $Nc$. Write $\omega =e^{2\pi i/3}$. The remaining four irreducible characters of a representation are induced from the nontrivial characters $a^i{b}^j\mapsto {\omega }^{ri+sj}$ of $N$, pairing $(r,s)$ with $(-r,-s)$. They are exactly the two-dimensional representations already proved irreducible. Taking representatives $(1,0),(0,1),(1,1),(1,2)$ yields the character tableIndeed the induced value on $a^i{b}^j$ is ${\omega }^{ri+sj}+{\omega }^{-(ri+sj)}$, equal to two or minus one, and is zero off $N$. The squared dimensions sum to $1+1+4\cdot 4=18$, so this is the complete table.

An element of $N$ centralizes $c$ only if it equals its inverse, hence only if it is the identity. No reflection centralizes $a$. Thus $Z(E)=\{1\}$ is cyclic, but the two linear representations kill $N$ and each two-dimensional irreducible has a nontrivial kernel of a group homomorphism of order three. Therefore a cyclic centre does not imply a faithful irreducible representation.

Conjugation by $P$ partitions $H$ into orbits. Each non-singleton orbit has size divisible by $p$, by the orbit-stabilizer theorem. The singleton orbits are exactly $H\cap Z(P)$. Since $|H|$ is divisible by $p$, $|H\cap Z(P)|\equiv 0(\mathrm{mod}p)$. This set contains $1$, so $\boxed{|H\cap Z(P)|\ge p}$.

For the deduction we also give the representation-theoretic ingredient behind Burnside's theorem: a nonabelian finite simple group has no nontrivial conjugacy class of prime power size. Here is a proof, so no stronger theorem is being assumed without justification. Suppose $g\ne 1$ has class size $p^b$. For an irreducible character $\chi$ of degree $d$ coprime to $p$, the class sum acts as the scalar $p^b\chi (g)/d$, which is an algebraic integer: the class sum has an integer-entry matrix in the regular representation, which contains every irreducible representation, so this scalar is an eigenvalue of an integer-entry matrix. Also $\chi (g)$ is an algebraic integer. Bezout's identity then makes $\chi (g)/d$ an algebraic integer. All its algebraic conjugates have modulus at most $1$, because character values of a representation are sums of $d$ roots of unity. An algebraic integer with this property is zero or a root of unity: the monic polynomials of all its powers have uniformly bounded integer coefficients, so two powers must coincide unless it is zero.

Every nontrivial irreducible representation of a simple group is faithful. If $\chi (g)/d$ were a root of unity, equality in the triangle inequality would force its representing matrix to be scalar, hence $g$ central, impossible. Thus $\chi (g)=0$ whenever $p\nmid d$, except for the trivial character of a representation. Orthogonality of the identity column and the $g$ column in the character table now givesThis would make $1/p$ an algebraic integer, contradicting the fact that rational algebraic integers are integers. The same proof covers class size $1$ directly through triviality of the center.

If an abelian subgroup $A$ had index $p^a$, every $g\in A\setminus \{1\}$ would have $A\subseteq C_G(g)$, making its class size a divisor of $p^a$, a contradiction. Such a $g$ exists: if $A=1$, then $G$ is a nontrivial p-group, whose center is nontrivial by the first argument with $H=P$. Therefore a nonabelian simple group has no abelian subgroup of prime power index.

Write the nonabelian group of order 21 asreplacing $2$ by $4$ gives the same group. Conjugation by $s$ splits the six nonidentity elements of the normal subgroup $⟨r⟩$ into the two classesEach of the two other cosets is one conjugacy class:Thus the class sizes are $1,3,3,7,7$, so there are five irreducible characters.

The abelianization is $C_3$, giving three linear characters. Put $\omega =e^{2\pi i/3}$ and $\zeta =e^{2\pi i/7}$, and defineThe 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 isThe degree checkand character orthogonality relations verify that this is the complete table.

Irreducible degrees divide the order of a finite $2$-group and satisfyThe number of linear characters is $|G/G^{\prime }|$. For a nontrivial finite $2$-group this is at least $4$ unless the group is abelian, and every nonlinear degree is at least $2$. A degree $4$ constituent is therefore impossible in the nonabelian case. If there are $\ell$ linear and $m$ degree-two characters, thenThus the only degree collections arewith respectively $r=16,10,7$.

For $1^{16}$ take the cyclic group $G=C_{16}=⟨g⟩$. Every conjugacy class is a singleton, and its character table iswhere ${\zeta }_{16}=e^{2\pi i/16}$.

For $1^8,2^2$, take $G=D_8\times C_2$, where
$D_8=⟨r,s:r^4=s^2=1,srs=r^{-1}⟩$ and
$C_2=⟨z⟩$. On the five classesthe four linear characters of $D_8$, indexed by
$\epsilon ,\delta \in \{\pm 1\}$, and its degree-two character areThe ten classes of $G$ are $(C,1)$ and $(C,z)$. Its full table consists ofwhere $\chi$ runs over the above five rows and
$\eta \in \{1,-1\}$ runs over the two characters of $C_2$. This explicitly gives eight linear and two degree-two rows.

For $1^4,2^3$, take the dihedral groupOrder its seven classes asThe four linear rows areThe three degree-two rows, for $k=1,2,3$, areThe row norms and mutual inner products, weighted by the displayed class sizes, verify irreducibility and completeness.

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.