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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?

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Since the abelianization of the group has $6$ elements, you know $6$ lines out of $8$ of the character table of this group. Why don't you just try using character orthogonality relations to fill the table? 2 lines can't be that hard to fill. If there are many zeros in there you might find the rest of the table quite easily. Another tactic would be to find a character of this group which might not be irreducible but contains an irreducible component of degree $3$. –  Patrick Da Silva Jun 2 '12 at 16:52
    
(By the looks of the group, I don't think there is a subgroup of order $8$... note that your group is isomorphic to $(\mathbb Z/2\mathbb Z \times \mathbb Z / 2 \mathbb Z) \rtimes \mathbb Z/ 6 \mathbb Z)$.) But I agree with you that the abelianization of the group has $6$ elements ; it gives you $x=y=xy$, hence $x=y=1$, but $z$ remains "untouched"). –  Patrick Da Silva Jun 2 '12 at 16:52
    
Well, you have an elementary Abelian subgroup $A,$ say of order $8$ generated by $x, y$ and $z^{3}.$ I suggest that you consider a $1$-deimensional representation of $A$ which does not contain $z^{3}$ in its kernel and induce that representation. –  Geoff Robinson Jun 2 '12 at 16:56
    
It loooks to me as if your group is the direct product of $A_{4}$ with a cyclic group of order $2.$ –  Geoff Robinson Jun 2 '12 at 17:05
    
@Geoff, you know this better. Is the idea not to select a character $\chi$ of a normal subgroup $H$ in such a way that the inertia group is equal to $N$? As $H$ is maximal in this case, this translates to requiring that $\chi$ is not a restriction of a character of $G$? –  Jyrki Lahtonen Jun 2 '12 at 17:32

1 Answer 1

up vote 1 down vote accepted

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.

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Clifford's theorem is the relevant theory here. –  Geoff Robinson Jun 2 '12 at 18:41

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