3.4 Group Characters

Definition

A character $\chi$ of a group $G$ with values in a field is a group homomorphism

$$\chi :G\to L^{\times }.$$

Example. ${\mathrm{GL}}_n(\mathbb{C})\to {\mathbb{C}}^{\times },M\to \det M$ is a character.

Definition

Let ${\chi }_1,\cdots ,{\chi }_n:G\to L^{\times }$ be characters. We say they are linearly independent over $L$, if they are linearly independent as functions on $G$; i.e., there does not exist non-trivial relation $a_1{\chi }_1+\cdots +a_n{\chi }_n=0$ as function on $G$.

Theorem

Suppose ${\chi }_1,\cdots ,{\chi }_n:G\to L^{\times }$ are distinct characters, then they are linearly independent.

\begin{proof} If not, there exist minimal $m⩾2$ and $a_i$ not all zero, such that ${\sum }_{i=1}^m{a}_i{\chi }_i=0$. Since ${\chi }_1\ne {\chi }_m$, there exists $g_0\in G$ such that ${\chi }_1(g_0)\ne {\chi }_m(g_0)$. Consider

$$\sum _{i=1}^m{a}_i{\chi }_i(g)=\sum _{i=1}^m{a}_i{\chi }_i(g_{0}g)=0,$$

then it deduces that

$$\sum _{i=2}^m({\chi }_1(g_0)-{\chi }_i(g_0))a_i{\chi }_i(g)=0,$$

which contradicts with minimality of $m$. \end{proof}

Recall that if $K,L$ are fields, then field homomorphism $f:K\to L$ is always injective. Therefore, $f:K^{\times }\to L^{\times }$ is a character.

Corollary

Let ${\sigma }_i:K\hookrightarrow L$ be the different embedding, then they are linear independent over $L$.

Theorem

Let $F$ be a field, and let $G<\mathrm{Aut}(F)$ be a finite group with $G=\{{\sigma }_1,\cdots ,{\sigma }_n\}$. Let $K=F^G$, then $[F:K]=n=|G|$.

\begin{proof} Firstly, we show $[F:K]=m⩾n$. Suppose otherwise, $m<n$. Let $w_1,\cdots ,w_m$ be the basis of $F/K$. Consider the following system of linear equation with indeterminateness’ $x_1,\cdots ,x_n$

$$\begin{array}{rl}& {\sigma }_1(w_1)x_1+{\sigma }_2(w_1)x_2+\cdots +{\sigma }_n(w_1)x_n=0\\ & \cdots \\ & {\sigma }_1(w_m)x_1+{\sigma }_2(w_m)x_2+\cdots +{\sigma }_n(w_m)x_n=0\end{array}$$

and by $n>m$ there exists ${\beta }_i\in F$ such that ${\sum }_{i=1}^n{\sigma }_i(w_j){\beta }_i=0$ for all $j\in \{1,\cdots ,m\}$. It deduces that $({\sum }_{i=1}^n{\beta }_i{\sigma }_i)(w_j)=0$ for all $j\in \{1,\cdots ,m\}$. Note that ${\sigma }_i{|}_K=\mathrm{id}{|}_K$, then for any $a_j\in K$, we have

$$(\sum _{i=1}^n{\beta }_i{\sigma }_i)(\sum _{j=1}^m{a}_j{m}_j)=0$$

and so ${\sum }_{i=1}^n{\beta }_i{\sigma }_i=0$ as functions on $F$. It is impossible by ^a23929.

Then we show $[F:K]⩽n$. If not, then $[F:K]>n$. Take ${\alpha }_1,\cdots ,{\alpha }_{n+1}\in F$ linearly independent over $K$, and consider the following system of linear equation

$$\begin{array}{rl}& {\sigma }_1({\alpha }_1)x_1+{\sigma }_1({\alpha }_2)x_2+\cdots +{\sigma }_1({\alpha }_{n+1})x_{n+1}=0\\ & \cdots \\ & {\sigma }_n({\alpha }_1)x_1+{\sigma }_n({\alpha }_2)x_2+\cdots +{\sigma }_n({\alpha }_{n+1})x_{n+1}=0.\end{array}$$

There exists nonzero solution $({\beta }_1,\cdots ,{\beta }_{n+1})\in F$. WLOG assume ${\beta }_1,\cdots ,{\beta }_r$ are not all zero, while ${\beta }_{r+1},\cdots ,{\beta }_{n+1}$ are zero. By multiplying $1/{\beta }_r$ we can assume ${\beta }_r=1$. Take ${\sigma }_1=\mathrm{id}$, then ${\sum }_{i=1}^r{\alpha }_i{\beta }_i={\alpha }_1{\beta }_1+\cdots +{\alpha }_{r-1}{\beta }_{r-1}+{\alpha }_r=0$. Since ${\alpha }_i$ are linearly independent, not all ${\beta }_i\in K$ and WLOG we assume ${\beta }_1\notin K$. Hence, there exists ${\sigma }_{k_0}({\beta }_1)\ne {\beta }_1$. Notice that

$${\sigma }_i({\alpha }_1){\beta }_1+\cdots +{\sigma }_i({\alpha }_{r-1}){\beta }_{r-1}+{\sigma }_i({\alpha }_r)=0,i\in \{1,\cdots ,n\}\text{\hspace{1em}(1)}$$

deduces that

$${\sigma }_{k_0}{\sigma }_i({\alpha }_1){\sigma }_{k_0}({\beta }_1)+\cdots +{\sigma }_{k_0}{\sigma }_i({\alpha }_{r-1}){\sigma }_{k_0}({\beta }_{r-1})+{\sigma }_{k_0}{\sigma }_i({\alpha }_r)=0,i\in \{1,\cdots ,n\}.\text{\hspace{1em}(*)}$$

As $\{{\sigma }_{k_0}{\sigma }_1,\cdots ,{\sigma }_{k_0}{\sigma }_n\}=G$, after reordering $(\ast )$ we get

$${\sigma }_i({\alpha }_1){\sigma }_{k_0}({\beta }_1)+\cdots +{\sigma }_i({\alpha }_{r-1}){\sigma }_{k_0}({\beta }_{r-1})+{\sigma }_i({\alpha }_r)=0,i\in \{1,\cdots ,n\}.\text{\hspace{1em}(2)}$$

With $(1)$ and $(2)$ we can get a smaller $r$, which is a contradiction. \end{proof}

Corollary

Suppose $[F:K]<\infty$. Then $|{\mathrm{Aut}}_{K}F|⩽[F:K]$. The equality holds iff $F/K$ is Galois.

\begin{proof} Since $[F:K]<\infty$, ${\mathrm{Aut}}_{K}F$ is a finite group. Define $E=F^{{\mathrm{Aut}}_{K}F}$, then $E\supseteq K$ and so $|{\mathrm{Aut}}_{K}F|=[F:E]⩽[F:K]$.

Furthermore, the equality holds iff $E=K$ iff $F^{{\mathrm{Aut}}_{K}F}=K$ iff $F/K$ is Galois. \end{proof}

Corollary

Let $F$ be a field, and let $G<\mathrm{Aut}(F)$ be a finite group. Let $E=F^G$, then $G={\mathrm{Aut}}_{E}F$ and $F/E$ is Galois.

\begin{proof} Since $E=F^G$, we have $G<{\mathrm{Aut}}_{E}F$. By ^ehefrs there is $[F:E]=|G|⩽|{\mathrm{Aut}}_{E}F|$ and by ^m1looz there is $|{\mathrm{Aut}}_{E}F|⩽[F:E]$. Hence $[F:E]=|G|=|{\mathrm{Aut}}_{E}F|$ and so $F/E$ is Galois. \end{proof}

Corollary

Let $F$ be a field, and let $G_1,G_2<\mathrm{Aut}F$ be two different finite subgroups. Let $E_i=F^{G_i}$, then $E_1\ne E_2$.

\begin{proof} By ^c7e515, $G_i={\mathrm{Aut}}_{E_i}F$ are distinct and yields $E_i$ are distinct. \end{proof}