1. ## Group Representation Theory

(1) Let G:= <g|g^4 = 1> and F be the field of real numbers. Show that the representation

p:g -> (0 -1)
(-1 0)

is irreducible over F.

(2) Let G be a finite group, x & y two characters of G

(i) If x is linear, show the x.y is irreducible <=> y is irreducible

(ii) If y(1) > 1, show that y .y is always irreducible.

(iii)Let x be an element of Irreducible G and no other character in Irreducible G have the same degree as x. Suppose that there is a linear character y such that for some g, y(G) is not equal to 1. Show that x(g) = 0.

2. Originally Posted by Turloughmack
(1) Let G:= <g|g^4 = 1> and F be the field of real numbers. Show that the representation

p:g -> (0 -1)
(-1 0)

is irreducible over F.
I think you may be over thinking this. So, I assume that you mean $\displaystyle \rho:\mathbb{Z}_4\to\text{GL}_2(\mathbb{R})$ and it is the homomorphism generated by $\displaystyle 1\mapsto \begin{pmatrix}0 &1\\-1 & 0\end{pmatrix}$ (since as you can quick check, for the matrix you put the space $\displaystyle \text{span}(1,1)$ is a non-trivial $\displaystyle \rho$-invariant subspace). Assume then that $\displaystyle \mathbb{R}^2$ had a non-trivial invariant subspace $\displaystyle S$, then by a dimension argument we see that $\displaystyle \dim_\mathbb{R}(S)=1$ and thus $\displaystyle S=\text{span}(x)$ for some common eigenvector $\displaystyle x$ of $\displaystyle \rho_0,\cdots,\rho_3$. That said, this is impossible since both the eigenvectors for $\displaystyle \rho(1)$ are non-real. Get the idea?

(2) Let G be a finite group, x & y two characters of G

(i) If x is linear, show the x.y is irreducible <=> y is irreducible

What field are you working over here? If it's $\displaystyle \mathbb{C}$ then everything's easy because you can check easily that $\displaystyle \left\langle \lambda\chi,\lambda\chi\right\rangle=1\left\langle \chi,\chi\right\rangle=1$ (this is because $\displaystyle \lambda$ [what I'm calling the linear character] is irrelevant since $\displaystyle |\lambda(g)|=1$ and so won't affect the inner product).

If the field is $\displaystyle \mathbb{R}$ then this is simple since by definition an irreducible character $\displaystyle \chi$ is a character of an irrep $\displaystyle \rho$ and a linear character $\displaystyle \lambda$ is the character of a degree one irrep $\displaystyle \psi$ and thus $\displaystyle \lambda\chi$ is the character of $\displaystyle \psi\otimes\rho$. Thus, it suffices to show that $\displaystyle \psi\otimes\rho$ is irreducible if and only if $\displaystyle \rho$ is irreducible. But, by assumption you can assume the representation space of $\displaystyle \psi\otimes\rho$ is (up to equivalence) $\displaystyle \mathbb{R}\otimes\mathbb{R}^m$ etc. You can take it from there?

(ii) If y(1) > 1, show that y .y is always irreducible.
Is this really what you want to say here?

(iii)Let x be an element of Irreducible G and no other character in Irreducible G have the same degree as x. Suppose that there is a linear character y such that for some g, y(G) is not equal to 1. Show that x(g) = 0.
[/quote]

You know (from i) that $\displaystyle \lambda\chi$ is an irreducible character of $\displaystyle G$ with the same degree as $\displaystyle \chi$. Thus, by assumption you have that $\displaystyle \lambda\chi=\chi$ and thus, in particular $\displaystyle \lambda(g)\chi(g)=\chi(g)$ since $\displaystyle \lambda(g)\ne 1$ you may conclude.

3. Thanks Drexel28, this helps alot!
For (ii) I actually mean "If y(1) > 1, show that y .y(BAR) is always irreducible"