4 character table

A table is like: row - conjugacy class column - irreducible character we call this table “Character table”.

For a given group, we hope to find the character table of this group. Using the properties as following helps us a lot:

• orthogonality
• conjugacy of irreducible character also irreducible
• normal subgroups and lifted characters, especially $G/G^{\prime }$ and linear character
• product of linear character and irreducible character
• decomposing ${\chi }^2$ as ${\chi }_A$ and ${\chi }_S$
• for a permutation module, $\#\mathrm{fix}(g)-1$ is a character (may not irreducible)

Character table is square

It suffices to show $\#$conjugate classes = $\#$irreducible characters.

All class functions form a vector space and so its dimension is bigger than the number of irreducible characters.

Conversely, since $\#$conjugate classes equals the dimension of $Z(\mathbb{C}G)$, consider an element $z\in Z(\mathbb{C}G)$.

Suppose $V=W_1\oplus ...\oplus W_k$ and $1=f_1+...+f_k$ where $f_i\in W_i$. For each $\mathbb{C}G$-map ${\theta }_i(v)=vz$ from $W_i$ to $W_i$, $vz={\lambda }_{z}v$ by Schur’s lemma and so $z={\sum }_{i=1}^k{\lambda }_i{f}_i$.

Thus $\#$irreducible characters equals to $k$ is bigger than $\#$conjugacy classes.

Cor. Suppose $\psi$ is a class function, then $\psi ={\sum }_{i=1}^k{\lambda }_i{\chi }_i$, where ${\lambda }_i=⟨{\chi }_i,\psi ⟩$.

Orthogonality relations

We have known that $⟨\chi ,\psi ⟩=0$, and guess that there are similar conclusion in column.

Thm. $⟨{\chi }_r,{\chi }_s⟩={\delta }_{rs}$, and $$ \sum_{i=1}^k \chi_i(g_r)\overline{\chi_i(g_s)}=\delta_{rs}|C_G(g_r)|^{1/2}|C_G(g_s)|^{1/2}.

Let $M_{ij}=\chi_i(g_j)/(C_G(g_j))^{1/2}$. Then $M$ is a unitary matrix. ### Lifted characters Assume $N\lhd G$ and $\widetilde{\chi}$ a character of $G/N$. Then $$ \chi:G\to \mathbb C,\chi(g)=\widetilde{\chi}(Ng) $$is a character of $G$, which is called the lift of $\widetilde{\chi}$ to $G$. **Thm.** Assume $N\lhd G$. By associating each character of $G/N$ with its lift to $G$, we obtain a bijective correspondence between the set of characters of $G/N$ and the set of character $\chi$ of $G$ which satisfy $N\leq \mathrm{Ker}\chi$. Also, irreducible characters of $G/N$ correspond to irreducible characters of $G$. This theorem provides us a new way to find irreducible characters: consider normal subgroup $N$ and an easier irreducible character of $N$. Also, a character table can help us to find normal subgroup of $G$: **Prop.** The group is not simple iff $\chi(g)=\chi(1)$ for some non-trivial irreducible character $\chi$ and some non-identity element $g$. **Prop.** If $N\lhd G$ then there exist irreducible characters $\chi_1,\cdots,\chi_s$ of $G$ such that$$ N=\bigcap_{i=1}^s\mathrm{Ker}\chi_i.

We can use this property to prove $A_5$ is simple.

In particular, $G^{ab}=G/G^{\prime }$ is always useful to find all linear characters of $G$. Furthermore, the product of linear character and irreducible characters also yields new character.

And the next part answers the question: “What happens if we product two arbitrary characters (not necessary linear)?”

Using the theory of tensor product we know it is still a character, and the most useful conclusion is following:

Properties of ${\chi }^n$

Thm. Let $\chi$ be a faithful character of $G$, and suppose that $\chi (g)$ takes precisely $r$ different values as $g$ varies over all the elements of $G$. Then every irreducible character of $G$ is a constituent of one of the powers ${\chi }^0,{\chi }^1,\cdots ,{\chi }^{r-1}$.

A quickly corollary: consider the regular character $\chi$ that takes just two values. We have known that every irreducible character of $G$ is a constituent of $1_G$ or $\chi$.

Proof. Use the Vandermonde determinant.

This property is helpful to find irreducible characters:

Prop. For $g\in G$, we have two characters:$$ \begin{aligned} \chi_S(g)=&\frac{1}{2}(\chi^2(g)+\chi(g^2))\ \chi_A(g)=&\frac{1}{2}(\chi^2(g)-\chi(g^2)) . \end{aligned}

If one of $\chi_S$ or $\chi_A$ is irreducible, we can repeat the procedure and may find more.