Characters and the Orthogonality Relations — Examples

Since $g^n=1$, a homomorphism $\phi :G\to {\mathbb{C}}^{\times }$ sends $g$ to an $n$-th root of unity: $\phi (g{)}^n=\phi (g^n)=\phi (1)=1$. By [F4] the $n$-th roots of unity in $\mathbb{C}$ are exactly the $n$ distinct powers ${\zeta }^0,\dots ,{\zeta }^{n-1}$ of a primitive root $\zeta$ of [F3].

For each $k$, the formula ${\chi }_k(g^m)={\zeta }^{km}$ is well defined by [F1] and is a homomorphism, since ${\chi }_k(g^a{g}^b)={\zeta }^{k(a+b)}={\zeta }^{ka}{\zeta }^{kb}$. They are pairwise distinct because ${\zeta }^{k-k^{\mathrm{\prime }}}\mathrm{\ne }1$ for $0<\mathrm{\mid }k-k^{\mathrm{\prime }}\mathrm{\mid }<n$ by primitivity of [F3].

By [F5] and [F6], $\mathbb{C}$ is a splitting field for the finite abelian group $G$; then [F7] shows every irreducible representation of $G$ has degree $1$. By [F2] it corresponds to a homomorphism $\phi :G\to {\mathbb{C}}^{\times }$, and step 1.1 shows $\phi (g)={\zeta }^k$ for some $k$, so $\phi ={\chi }_k$. Hence the irreducible characters are exactly the ${\chi }_k$.

By [F1] the entry in row ${\chi }_k$ and column $g^m$ is ${\zeta }^{km}$, so the table has the stated shape.

If $n=1$, then $G$ is trivial and ${\sum }_{m=0}^{n-1}{\zeta }^{km}=1$ for the unique value $k=0$. Assume now that $n>1$. Then the polynomial $t^n-1$ has coefficient $0$ at $t^{n-1}$, so [A1] and [F2] give ${\sum }_{m=0}^{n-1}{\zeta }^m=0$. More generally, if $k\equiv 0(\mathrm{m}\mathrm{o}\mathrm{d}n)$ then ${\sum }_{m=0}^{n-1}{\zeta }^{km}=n$; if $0<k<n$, the element $\eta ={\zeta }^k$ has order $n\mathrm{/}d$, where $d=\gcd (k,n)$, so ${\sum }_{m=0}^{n-1}{\zeta }^{km}=d{\sum }_{r=0}^{n\mathrm{/}d-1}{\eta }^r\mathrm{.}$ Here $n\mathrm{/}d>1$, and the inner sum vanishes by the same Vieta argument applied to $t^{n\mathrm{/}d}-1$. Thus ${\sum }_{m=0}^{n-1}{\zeta }^{km}$ is $n$ when $k\equiv 0(\mathrm{m}\mathrm{o}\mathrm{d}n)$ and $0$ otherwise.

Row orthogonality: by [F3] and step 1.1, ${\delta }_{kl}=⟨{\chi }_k,{\chi }_l⟩=\frac{1}{n}{\sum }_m{\zeta }^{km}\overline{{\zeta }^{lm}}=\frac{1}{n}{\sum }_m{\zeta }^{(k-l)m}$, and the last sum is $n$ when $k\equiv l(\mathrm{m}\mathrm{o}\mathrm{d}n)$ and $0$ otherwise by step 1.2, which matches ${\delta }_{kl}$.

Column orthogonality: by [F4] and step 1.1, ${\sum }_k{\chi }_k(g^m)\stackrel{‾}{{\chi }_k(g^{m^{\mathrm{\prime }}})}={\sum }_k{\zeta }^{k(m-m^{\mathrm{\prime }})}$, which is $n$ when $m\equiv m^{\mathrm{\prime }}(\mathrm{m}\mathrm{o}\mathrm{d}n)$ and $0$ otherwise by step 1.2; since $G$ is abelian every centralizer is $G$, so this is exactly $\mathrm{\mid }{C}_G(g)\mathrm{\mid }{\delta }_{m{m}^{\mathrm{\prime }}}$.

All $n$ degrees are $1$, so ${\sum }_k{n}_k^2=n=\mathrm{\mid }G\mathrm{\mid }$.

The line $L$ of constant vectors is fixed pointwise by every permutation, so it is a trivial subrepresentation, and the sum map $(x_1,\dots ,x_n)\mapsto {\sum }_i{x}_i$ is preserved by every permutation, so its kernel $V_{\mathrm{s}\mathrm{t}\mathrm{d}}$ is a subrepresentation. Every vector decomposes uniquely as a constant vector plus a vector of sum zero, which is [A1].

By [F2] applied to [A1], the permutation character ${\chi }_{{\mathbb{C}}^n}$ satisfies ${\chi }_{{\mathbb{C}}^n}={\chi }_{\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{v}}+{\chi }_{\mathrm{s}\mathrm{t}\mathrm{d}}$. The trivial character is the constant $1$, so ${\chi }_{\mathrm{s}\mathrm{t}\mathrm{d}}(\sigma )={\chi }_{{\mathbb{C}}^n}(\sigma )-1$.

By [F1], ${\chi }_{{\mathbb{C}}^n}(\sigma )=\mathrm{fix}(\sigma )$, the number of fixed points. Combining with step 1.2 gives ${\chi }_{\mathrm{s}\mathrm{t}\mathrm{d}}(\sigma )=\mathrm{fix}(\sigma )-1$.

By [F1] the cycle types of $S_3$ are the identity type, the type of a transposition, and the type of a $3$-cycle, so $S_3$ has three conjugacy classes; a direct count of each type gives class sizes $1$, $3$, and $2$.

By [F3] the trivial and sign characters are distinct one-dimensional characters, hence irreducible: a one-dimensional space has no proper nonzero subspaces. Their degrees are $1$ and $1$.

By [F4], the standard character has values $2$, $0$, and $-1$ on the three classes of step 1.1 (fixed points $3$, $1$, and $0$ minus $1$). Using [A1], $⟨{\chi }_{\mathrm{s}\mathrm{t}\mathrm{d}},{\chi }_{\mathrm{s}\mathrm{t}\mathrm{d}}⟩=\frac{1}{6}(4+3\cdot 0+2\cdot 1)=1$, so by [F5] the standard character is irreducible, of degree ${\chi }_{\mathrm{s}\mathrm{t}\mathrm{d}}(1)=2$.

The three irreducible characters of steps 1.2 and 2.1 have squared degrees $1+1+4=6=\mathrm{\mid }{S}_3\mathrm{\mid }$. By [F2] the sum over all irreducible characters is also $6$, so no further irreducible character exists.

By [F1] there are three irreducible characters and three classes; [A1] and [A2] supply the second and third rows, and the trivial character is the constant $1$. Hence the displayed table is the character table.

Row orthogonality: the rows $1$ and $\mathrm{sgn}$ have inner product $\frac{1}{6}(1-3+2)=0$, the row $1$ with ${\chi }_{\mathrm{s}\mathrm{t}\mathrm{d}}$ gives $\frac{1}{6}(2+0-2)=0$, $\mathrm{sgn}$ with ${\chi }_{\mathrm{s}\mathrm{t}\mathrm{d}}$ gives $\frac{1}{6}(2-0-2)=0$, and each row has self-inner-product $1$; this matches [F2].

Column orthogonality: the first column has squared norm $1+1+4=6=\mathrm{\mid }{C}_{S_3}(1)\mathrm{\mid }$; the second has $1+1+0=2=\mathrm{\mid }{C}_{S_3}((12))\mathrm{\mid }$; the third has $1+1+1=3=\mathrm{\mid }{C}_{S_3}((123))\mathrm{\mid }$; and each pair of distinct columns has inner product $1-1+0=0$, $1+1-2=0$, $1-1+0=0$, matching [F3].

By [F2], the tensor square of the two-dimensional representation has character ${\chi }_2^2$; from [F1] its values are $(2,0,-1{)}^2=(4,0,1)$ on the three classes.

Using [A1] and the values of step 1.1, the multiplicities are $⟨{\chi }_2^2,1⟩=\frac{1}{6}(4+0+2)=1$, $⟨{\chi }_2^2,\mathrm{sgn}⟩=\frac{1}{6}(4+0+2)=1$, and $⟨{\chi }_2^2,{\chi }_2⟩=\frac{1}{6}(8+0-2)=1$. By [F3] these are the coefficients of $1$, $\mathrm{sgn}$, and ${\chi }_2$.

The value at $1$ checks: ${\chi }_2^2(1)=4=1\cdot 1+1\cdot 1+1\cdot 2$, the sum of the degrees with the multiplicities of step 2.1; hence ${\chi }_2^2=1+\mathrm{sgn}+{\chi }_2$.

By [A1] the three nonidentity elements of $V_4$ are even permutations and multiply pairwise to the third, so $V_4$ is a subgroup of $A_4$ of order $4$.

For the $3$-cycle $(123)$, the cycle lengths are $3$ and $1$, odd and distinct, so [F2] applies: its $S_4$-class, of size $8$, splits into two $A_4$-classes of equal size $4$. These two classes account for all eight $3$-cycles of $A_4$.

By [A2], any conjugate of a double transposition is again a double transposition, and there are exactly three of them; by [F1] they all lie in the single $S_4$-class of cycle type $(2,2)$. Hence every conjugate of every element of $V_4$ lies in $V_4$, so $V_4⊴A_4$.

By [F3], $\mathrm{\mid }{A}_4\mathrm{/}{V}_4\mathrm{\mid }=\mathrm{\mid }{A}_4\mathrm{\mid }\mathrm{/}\mathrm{\mid }{V}_4\mathrm{\mid }=12\mathrm{/}4=3$.

The identity class of $A_4$ is $\{1\}$. For the double transposition $(12)(34)$, the cycle type $(2,2)$ has two equal lengths, so [F2] does not apply and its $S_4$-class does not split: its $A_4$-class is the whole class of double transpositions, of size $3$ by step 2.1.

Every nonidentity element of $A_4$ is a double transposition or a $3$-cycle, so steps 3.1 and 1.2 list all classes: sizes $1$, $3$, $4$, $4$.

By [F1] and [F2] the quotient $A_4\mathrm{/}{V}_4$ has order $3$, hence is cyclic. Its three homomorphisms to ${\mathbb{C}}^{\times }$ send a generator to $1$, $\omega$, or ${\omega }^2$; by [F3] and [A1], they pull back to three one-dimensional characters of $A_4$ whose values are the first three rows. Conversely, every homomorphism $A_4\to {\mathbb{C}}^{\times }$ kills commutators, so [A2] makes it kill $(12)(34)$ and therefore, by conjugacy, every nonidentity element of $V_4$. Hence every degree-one character is trivial on $V_4$ and factors through $A_4\mathrm{/}{V}_4$ by [A1]. Thus these are exactly the three one-dimensional characters of $A_4$.

By [F4], the remaining irreducible degree $d$ satisfies $1+1+1+d^2=12$, so $d=3$.

By [F6], the column of $(12)(34)$ is orthogonal to the column of $1$: $1+1+1+\psi ((12)(34))=0$, so $\psi ((12)(34))=-1$.

By [F6], the column of $(123)$ is orthogonal to the column of $(12)(34)$: $1\cdot \bar{1}+1\cdot \bar{\omega }+1\cdot \overline{{\omega }^2}+(-1)\cdot \overline{\psi ((123))}=0$. Since $1+\bar{\omega }+\overline{{\omega }^2}=1+{\omega }^2+\omega =0$, this forces $\overline{\psi ((123))}=0$, hence $\psi ((123))=0$; the same argument gives $\psi ((132))=0$.

The four rows assembled in steps 1.1 through 4.1 form the displayed table. Row orthogonality holds by [F5]: the first three rows are orthonormal ($\frac{1}{12}(4+3\cdot 1+4+4)=1$ for the identity, with cross terms $1+3\cdot 1+4\omega +4{\omega }^2=1+3-4=0$), and $⟨\psi ,\psi ⟩=\frac{1}{12}(9+3\cdot 1)=1$, while $\psi$ is orthogonal to each of the first three rows.

Column orthogonality holds by [F6]: the squared norms are $12$, $4$, $3$, $3$, matching the centralizer sizes $12$, $4$, $3$, $3$ of the four classes, and distinct columns are orthogonal.

By [F1] and [A1], the classes of $S_4$ are exactly those five cycle types.

The class size of a cycle type $c$ is $\frac{4!}{{\prod }_k{k}^{c_k}{c}_k!}$ by [F2]. For the types of [A1] these sizes are $\frac{24}{4}=6$, $\frac{24}{3}=8$, $\frac{24}{2^2\cdot 2}=3$, $\frac{24}{2}=6$, and $1$.

Matching the representatives $1$, $(12)$, $(123)$, $(1234)$, $(12)(34)$ with their cycle types in [A1] and the corresponding values from step 1.2 gives class sizes $1$, $6$, $8$, $6$, $3$. The sum $1+6+8+6+3=24=\mathrm{\mid }{S}_4\mathrm{\mid }$, so the count is complete.

By [F2] the sign row is $(1,-1,1,-1,1)$ on the representatives: values $+1$ on even permutations and $-1$ on odd ones. By [F3], the permutation character of $S_4$ has values $4$, $2$, $1$, $0$, $0$ (fixed points), so the standard character has values $3$, $1$, $0$, $-1$, $-1$.

By [F4], the sign twist $\epsilon {\chi }_3$ has values $(3,-1,0,1,-1)$.

The trivial and sign characters are one-dimensional, hence irreducible. Using [A3] and the values from steps 1.1 and 2.1 gives $⟨{\chi }_3,{\chi }_3⟩=\frac{1}{24}(9+6+0+6+3)=1,⟨{\chi }_3,1⟩=\frac{1}{24}(3+6+0-6-3)=0.$ Because $\mathrm{\mid }\epsilon (g)\mathrm{\mid }=1$ for every $g\in S_4$, the same computation gives $⟨\epsilon {\chi }_3,\epsilon {\chi }_3⟩=1$, and multiplying one factor by $\epsilon$ preserves orthogonality with $1$ and with $\epsilon$. Therefore $1$, $\epsilon$, ${\chi }_3$, and $\epsilon {\chi }_3$ are four pairwise orthogonal irreducible characters, the last two by [F5].

By [F6], irreducible characters form an orthonormal basis of the $5$-dimensional class-function space of $S_4$, so after the four orthogonal irreducibles of step 3.1 there is exactly one remaining irreducible character, call it ${\chi }_2$. By [F7], its degree $d$ satisfies $1+1+9+9+d^2=24$, so $d=2$.

By [F8], each column is orthogonal to the first column $(1,1,3,3,2)$, so reading off the first four entries gives the fifth entry: at $(12)$, $1-1+3-3+2\overline{{\chi }_2((12))}=0$, so ${\chi }_2((12))=0$; at $(123)$, $1+1+0+0+2\overline{{\chi }_2((123))}=0$, so ${\chi }_2((123))=-1$; at $(1234)$, $1-1-3+3+2\overline{{\chi }_2((1234))}=0$, so ${\chi }_2((1234))=0$; at $(12)(34)$, $1+1-3-3+2\overline{{\chi }_2((12)(34))}=0$, so ${\chi }_2((12)(34))=2$.

The five rows from steps 1.1, 2.1, and 5.1 now form an orthonormal basis of the class functions: step 3.1 already handles the first four rows, and $⟨{\chi }_2,{\chi }_2⟩=\frac{1}{24}(4+0+8+0+12)=1.$ Since ${\chi }_2$ is orthogonal to the first four rows by construction from step 5.1, this is the displayed character table.

The kernels, by [A1]: $\ker 1=S_4$; $\ker \epsilon =A_4$ (the even permutations); $\ker {\chi }_3=\{1\}$, because ${\chi }_3(\sigma )=3$ means $\mathrm{fix}(\sigma )=4$, i.e. $\sigma =1$; $\ker \epsilon {\chi }_3=\{1\}$; $\ker {\chi }_2=V_4$, because ${\chi }_2(\sigma )=2$ exactly at the identity and the double transpositions, and [A2] identifies the class with exactly two fixed points as the transpositions.

By [F9], the normal subgroups are exactly the intersections of the five kernels of step 6.2; by [A2] these intersections are $\{1\}$, $V_4$, $A_4$, and $S_4$.