irreducible characters of a group

I am currently attempting a past exam paper and am stuck on the following question for part a) $\mu$ is an irreducible character iff it is equal to the character of an irreducible representation, which in my notes is a representation afforded by a $\mathbb{C}G-$module that only has 2 submodules, itself and $0$

but i cannot make the connection

assuming part a i am still unsure of how to do the remaining parts of the questions and would appreciate some hints and tips as the mark schemes for these exam papers have not been provided meaning i have no way to check my answers or point myself in the right direction of the solution.

thanks in advance for the help

• A character of degree one is the same thing as a group homomorphism $G\rightarrow \mathbb{C}^\times$. And all characters of degree $1$ are irreducible for obvious reasons (i.e. you're not going to be able to split a vector space of dimension $1$ into two proper non-zero subspaces). – Hayden May 17 '15 at 20:19

a) Check that $\mu$ is the character attached to the representation $\det \rho: G \to \mathbb{C}^* = GL_1(\mathbb{C})$.

b) Put $k = \#G$. Since $\mu(g)^k = 1$ for all $g$, the group $G/\ker \mu = \mu(G)$ is a subgroup of the cyclic group $\{z\in \mathbb{C}:\, z^k = 1\}$ and thus itself cyclic.

c) Here $\mu:G \twoheadrightarrow \mathbb{Z}_d$ with $d$ even by the fact that some $\mu(g) = -1$. Take the preimage of an index-$2$ subgroup of $\mathbb{Z}_d$.

d) Since the representation $\rho$ is over a field of characteristic $0$, the matrix $\rho(g)$ is diagonalizable with $\rho(g)^2 = 1$. The eigenvalues $\lambda_1, \dots, \lambda_n$ of $\rho(g)$ are thus $\pm 1$, so $\chi(g) = \operatorname{tr} \rho(g) = \lambda_1 + \cdots + \lambda_n\in \mathbb{Z}$ with $\chi(g)\equiv n\pmod{2}$.

e) By part (c), $\det \rho(g) = \lambda_1 \cdots \lambda_n = 1$ for all $g\in G$ (in the notation of part (d)). Thus $$\#\{i:\lambda_i = -1\} \equiv 0 \pmod{2}.$$ Since $$\chi(g) = \lambda_1 + \cdots + \lambda_n = n - 2\#\{i:\lambda_i = -1\},$$ the result follows.

For $a$, It is enough to show that $det$ is an character as its degree is $1$ , it is automatically irreducable.

For $b$, $G/Ker(\mu)$ is ismorphic to a subgroup of $\mathbb C^*$. Thus, $G/Ker(\mu)$ is abelian.

$c)$ if $\mu (g)=-1$ then the order of $g$ is even. Hence, $G/Ker(\mu)$ is an abelian group of even order. Then, $G/Ker(\mu)$ has an subgroup of index $2$. By correspondonce theorem, $G$ has a subgroup of index $2$.

$d)$ Let $g\to A$ where $A\in Gl(n,\mathbb C)$. We can find a base $B$ such that $[A]_B$ is diagonizible and $[A_{ii}]_B$ is a root of unity $2$, Which means $[A_{ii}]_B\in \{1,-1\}$. Thus, $\chi(g)=r-s$ where $r$ is the number of the one and $s$ is the number of the $-1$. We can also see that $r+s=n$. Hence, $n-\chi(g)=2s$ which is even.

$e)$ If $G$ has no subgroup of index $2$ then $\mu(g)\neq -1$. Bu we have $\mu(g^2)=\mu(g)^2=1\implies \mu(g)=1=det([A]_B)$. Hence $s$ is even. Thus, $n-\chi(g)=2s$ is divisible by $4$.