# Finding Unitary Irreducible Representations

• Mar 16th 2010, 07:54 PM
redsoxfan325
Finding Unitary Irreducible Representations
We are studying representation theory and we were asked to find the unirreps (unitary irreducible representations) for a group $\displaystyle G$ given that $\displaystyle |G|=p^3$, $\displaystyle p$ a prime.

We have the theorem that states that the number of unirreps is equal to the number of conjugacy classes of the group $\displaystyle G$.

There are five cases, three of which are abelian and therefore trivial. The other two groups, which we figured out in class, are:

1.) Let $\displaystyle z=xyx^{-1}y^{-1}$. Then $\displaystyle G=\langle x,y~|~x^p=y^p=z^p=1,~xz=zx,~yz=zy\rangle$

2.) $\displaystyle G=\langle x,y~|~x^{p^2}=y^p=1,~yxy^{-1}=x^{p+1}\rangle$

How do I go about finding the conjugacy classes of these groups, or is there a better way to find the number of unirreps for each group?

Thanks.
• Mar 17th 2010, 07:32 AM
Opalg
Quote:

Originally Posted by redsoxfan325
We are studying representation theory and we were asked to find the unirreps (unitary irreducible representations) for a group $\displaystyle G$ given that $\displaystyle |G|=p^3$, $\displaystyle p$ a prime.

We have the theorem that states that the number of unirreps is equal to the number of conjugacy classes of the group $\displaystyle G$.

There are five cases, three of which are abelian and therefore trivial. The other two groups, which we figured out in class, are:

1.) Let $\displaystyle z=xyx^{-1}y^{-1}$. Then $\displaystyle G=\langle x,y~|~x^p=y^p=z^p=1,~xz=zx,~yz=zy\rangle$

2.) $\displaystyle G=\langle x,y~|~x^{p^2}=y^p=1,~yxy^{-1}=x^{p+1}\rangle$

How do I go about finding the conjugacy classes of these groups, or is there a better way to find the number of unirreps for each group?

For 1), notice that z commutes with x and y, so that the cyclic subgroup Z generated by z is in the centre of G. The quotient group G/Z is the abelian group $\displaystyle \mathbb{Z}_p\times\mathbb{Z}_p$ (generated by the images of x and y), so it has p^2 one-dimensional unirreps, corresponding to the conjugacy classes $\displaystyle x^jy^kZ$ ($\displaystyle 0\leqslant j,k<p$; j, k not both zero) and the conjugacy class of the identity element.

That leaves the elements $\displaystyle z^k\ (1\leqslant k<p)$, each of which is central and therefore forms a singleton conjugacy class. The correponding unirreps must (I guess) have dimension p. Then we have $\displaystyle p^2$ one-dimensional unirreps, and $\displaystyle (p-1)$ p-dimensional unirreps, and the squares of their dimensions add up to the order of the group, as they should: $\displaystyle p^3 = p^2\times1^2 + (p-1)\times p^2$.

Edit. The analysis for group 2) looks very similar. Here, the centre is the cyclic subgroup Z generated by $\displaystyle x^p$. There are again $\displaystyle p^2$ one-dimensional unirreps corresponding to representations of the quotient group G/Z, and $\displaystyle p-1$ p-dimensional unirreps corresponding to the elements of Z (other than the identity).