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Only the character ${\chi }_2$ of the two-dimensional irreducible representation remains unknown. Write it as $(2,a,b,c,d)$. The regular representation has character zero off the identity and decomposes with multiplicities equal to the irreducible representation dimensions. Thus, for a nonidentity element $g$,For the four nonidentity conjugacy classes this gives $a=0$, $b=2$, $c=-1$, $d=0$. The complete character table is consequentlyFor example, $⟨{\chi }_2,{\chi }_2⟩=(4+12+8)/24=1$, agreeing with character orthogonality.

There is one complex irreducible representation for each conjugacy class, hence three. The trivial representation $1$ and sign representation $\epsilon$ are distinct and one-dimensional. The sum of squares of irreducible representation dimensions leavesCall the remaining irreducible representation $2$. The permutation representation on ${\mathbb{C}}^3$ has character equal to the number of fixed objects, namely $(3,1,0)$ on the ordered conjugacy classes $(e,\text{transpositions},\text{three-cycles})$. Removing its constant-vector trivial representation gives $(2,0,-1)$, whose character inner product with itself is one. This constructs $2$ and gives the character tableTo obtain every Clebsch-Gordan decomposition of group representations, multiply the characters pointwise. In particular, ${\chi }_{\epsilon }^2={\chi }_1$, ${\chi }_{\epsilon }{\chi }_2={\chi }_2$ andThus all ordered pairs of irreducible representations are covered by

Write $1$ for the trivial representation, $\epsilon$ for the sign representation, and $2$ for the given two-dimensional group representation. Their matrix representations on the generators of a group arewith $\alpha =e^{2\pi i/3}$. The two eigenvalues of $r$ in $2$ are distinct, and $s$ exchanges its two eigenlines; no line is invariant under both. Thus $2$ is an irreducible representation. The dimensions already exhaust the sum of squares of irreducible representation dimensions, so this is the complete list.

Since $\alpha +{\alpha }^{-1}=-1$, the character table isMultiplying characters and using character orthogonality gives every tensor product of group representations:For example ${\chi }_2^2=(4,1,0)={\chi }_1+{\chi }_{\epsilon }+{\chi }_2$. Explicit Clebsch-Gordan coefficients can be read from two doublets $x=(x_1,x_2{)}^T$, $y=(y_1,y_2{)}^T$:The first two combinations are invariant under $r$ and respectively even and odd under $s$; the last transforms by the displayed doublet matrices.

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.

A Frobenius group is a finite transitive group action on more than one point in which a nonidentity element fixes at most one point and a point stabilizer is nontrivial. Equivalently it has a nontrivial proper subgroup $H$ with $H\cap H^g=1$ for every $g\notin H$. The Frobenius complement is such a point stabilizer $H$.

The Frobenius kernel is the identity together with all fixed-point-free elements in this action, equivalentlyThe Frobenius kernel theorem asserts that this set is a normal subgroup, acts regularly on the points, and gives $G=N⋊H$. In particular $C_N(h)=1$ for $1\ne h\in H$.

Brauer's permutation lemma states that an automorphism of a finite group fixes equally many conjugacy classes and irreducible characters. More generally a group of automorphisms has equal permutation characters on those two sets. Indeed, the character table is invertible and intertwines the two permutation matrices: simultaneous permutation of rows and columns preserves character evaluation. Similar matrices have equal traces, and a permutation-matrix trace counts its fixed points. The fixed objects on the group side are classes, not individual elements.

Put $w_j=|C_j|/|G|$. The character orthogonality for the five supplied rows say ${\sum }_j{w}_j{\chi }_r(C_j){\chi }_s(C_j)={\delta }_{rs}$. Solving these linear equations givesFor example orthogonality of the two degree-four rows gives $16w_1+w_3+w_4-4w_5-w_6=0$, and their norms give $16w_1+w_3+w_4+4w_5+w_6=1$; the remaining row products determine the other weights. Since $C_1=\{e\}$, $w_1=1/|G|$, so $|G|=120$ and the class sizes are the displayed numerators.

Tensoring the fifth irreducible representation with the second, one-dimensional one gives another irreducible character, with row $(5,1,-1,0,-1,-1,1)$. It differs from all five known rows. There are seven conjugacy classes, hence seven irreducible characters. The sum of squared degrees is $|G|$, so the final degree is $\sqrt{120-(1+1+16+16+25+25)}=6$.

The regular representation has character zero off the identity and is the sum of irreducible characters weighted by their degrees. Applying this column by column determines the last row. The completed character table, in the supplied class order, isNo identification of the group with a familiar group is needed.

Directly tracking the supplied permutations givesThus every group element reduces to $a^k$ or $b{a}^k$, $0\le k<6$. These are twelve distinct elements: $a$ has order six, while $b$ is not a power of $a$ because it exchanges the two six-element supports that every power of $a$ preserves. Therefore $|G|=12$. This is the dicyclic group with parameter three; the calculations give its order-twelve dicyclic character table.

Conjugation by $b$ inverts powers of $a$, while conjugation by $a$ changes the exponent in $b{a}^k$ by two modulo six. Conjugation by $b$ preserves that exponent's parity. The complete conjugacy classes areTheir sizes are $1,1,2,2,3,3$.

The normal subgroup $N=⟨a^2⟩$ has order three. The quotient is cyclic of order four, generated by $bN$, since $(bN{)}^2=aN$. Inflating its characters gives ${\chi }_j(b)=i^j$, ${\chi }_j(a)=(-1{)}^j$, for $j=0,1,2,3$. These are all the linear characters: in any one-dimensional representation, the relation $ba{b}^{-1}=a^{-1}$ forces the image of $a$ to have square one, so $a^2$ is in the kernel.

In the displayed class order, the full character table isTo construct the last two rows, put $\zeta =e^{i\pi /3}$ and, for $k=1,2$, setThe matrices obey all three relations. The distinct eigenlines of ${\rho }_k(a)$ are exchanged by ${\rho }_k(b)$, so the representation is irreducible. Its traces on $a^m$ are $2\cos (km\pi /3)$ and on $b{a}^m$ are zero, producing the rows above. Their weighted inner products are orthonormal, and the degree-square sum $4\cdot 1^2+2\cdot 2^2=12$ proves completeness.

Each $H$-conjugacy class represented by $x_i$ has $|H|/|C_H(x_i)|$ elements. For every element of that class, the number of conjugators from $g$ to it is $|C_G(g)|$. Grouping the induced character formula by these classes therefore gives

In the example, $a$ has order four, $b$ has order two, and $bab=a^{-1}$, so $H\cong D_8$. Its five conjugacy classes are $1$, $a^2$, $\{a,a^3\}$, $\{b,a^{2}b\}$ and $\{ab,a^{3}b\}$, of sizes $1,1,2,2,2$. The four one-dimensional irreducible characters are determined by $\psi (a)=\epsilon$ and $\psi (b)=\eta$, each sign independently chosen. The remaining two-dimensional irreducible character comes from the plane action of the square:Their squared degrees sum to eight, so this is the full character table.

In $S_4$, use class representatives $1,(12),(12)(34),(123),(1234)$. Their intersections with $H$ are respectively $\{1\}$, $\{b,a^{2}b\}$, $\{a^2\}\cup \{ab,a^{3}b\}$, the empty set, and $\{a,a^3\}$. The corresponding $S_4$ centralizer orders are $24,4,8,3,4$, while the noncentral $H$ centralizers have order four. The induced values are consequentlyIn the same $S_4$ class order, all five induced irreducible characters are

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.

Representations. A complex group representation is a group homomorphism $\rho :G\to \mathrm{GL}(W)$ to the invertible linear maps on a vector space. It is an irreducible representation when $W\ne 0$ and the only invariant subspaces are $0$ and $W$.

Group table. The supplied matrices satisfy $A=-I$, $C=-B$, and $A^2=B^2=C^2=I$. Their multiplication table isThe table proves closure and gives identity $I$ and an inverse for every element; associativity is inherited from matrix multiplication. The table is symmetric, so the group is abelian. It is the Klein four-group $V\cong C_2\times C_2$.

Dimensions and conjugacy. The sum of squares of irreducible representation dimensions is $|V|=4$. A two-dimensional irreducible would alone contribute four, leaving no room for the ever-present one-dimensional trivial representation. Hence the given two-dimensional representation is reducible. Since the group is abelian, every conjugacy class is a singleton. There are therefore four inequivalent complex irreducible representations, each of dimension one:Taking the trace of the given matrices gives their character of a representation:

Irreducible characters. In a one-dimensional representation, the values on $A$ and $B$ must each be $\pm 1$, since both elements square to $I$. Their values determine the value on $C=AB$. The four choices give the complete character table:Each row is a character because it respects the generating relations, and the rows have unit norm and pairwise zero character inner product. The multiplicities of these rows in $(2,-2,0,0)$ are respectively $0,0,1,1$, by character orthogonality. Thus

Diagonal form. The common invariant lines are spanned by $(1,1{)}^T$ and $(1,-1{)}^T$. Use their normalized vectors as columns ofThen simultaneous diagonalization givesReading the first and second diagonal entries across the four matrices recovers ${\chi }_3$ and ${\chi }_4$, respectively, verifying the decomposition explicitly.