Inducing a representation from a subgroup. Find all the irreducible representations of the group given by:
$<x,y,z|x^2=y^2=(xy)^2=z^6=1,zxz^{-1}=y, zyz^{-1}=xy>$.
I have 8 conjugacy classes:
$\{1\}, \{z^3\}, \{x,y,xy\}, \{z,xyz,xz,yz\}, \{z^2,xz^2,yz^2,xyz^2\}, \{yz^3,xz^3,xyz^3\}, \{z^4, xz^4, yz^4\}, \{z^5,xz^5,yz^5,xyz^5\}$
and hence 8 irreducible representations. The abelianization of the group has 6 elements. Hence there are 6 irreducible representations of dimension 1.  The sum of the squares of the dimensions of the irreducible representations is equal to the order of the group. Hence we have two irreducible representations with the square of the dimensions summing to 18. Hence we have two representations of dimension 3 each. 
If I find one of these representations then by tensoring by a well chosen representation of dimension 1 I can hopefully find the other one.
I know that I can find a 3 dimensional representation by inducing from a subgroup of order 8. How do I do this?
 A: A subgroup $H$ of order 8 is generated by $x,y,z^3$. We easily see that it is elementary abelian, and also (as a union of conjugcay classes of $G$) a normal subgroup.
You want to induce such a character $\chi$ of $H$ that is not a restriction of a character of $G$ (I'm afraid I don't know/remember the general theory, but this requirement is surely needed for otherwise we would just multiply the character by the index). Let us try and induce the 1-dimensional character $\chi$ defined by $\chi(x)=-1$, $\chi(y)=\chi(z^3)=1$. This is not a restriction of a character of $G$, because $x$ and $y$ are conjugate in $G$. As $\chi:H\rightarrow\mathbf{C}^*$ is a homomorphism of groups we get that under $\chi$ the elements $xy,xz^3,xyz^3$ are all mapped to $-1$ but $1,yz^3\mapsto 1$.
Let $\psi=\mathrm{Ind}_H^G(\chi)$. From the formula of the induced character we see that
$\psi$ has the following values on conjugacy classes: $\psi(1)=\psi(z^3)=3$, $\psi(\{x,y,xy\})=-1+1-1=-1$, $\psi(\{yz^3,xyz^3,xz^3\})=-1$, and the rest are mapped to zero (trivially, as $H$ is normal in $G$). 
As a check for irreducibility let us compute the inner product:
$$
\langle\psi,\psi\rangle=\frac{1}{24}(1\cdot3^2+1\cdot3^2+3\cdot(-1)^2+3\cdot(-1)^2)=1.
$$
Looks like $\psi$ is, indeed, irreducible.
