# Math Help - Finding Unitary Irreducible Representations

1. ## Finding Unitary Irreducible Representations

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

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

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

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

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.

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

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

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

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

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 $\mathbb{Z}_p\times\mathbb{Z}_p$ (generated by the images of x and y), so it has p^2 one-dimensional unirreps, corresponding to the conjugacy classes $x^jy^kZ$ ( $0\leqslant j,k; j, k not both zero) and the conjugacy class of the identity element.

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

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