More than 1 character for a conjugacy class

[ChrisVer]

Hi,
I am trying to find the characters for the $S_4$ group, in the conjugacy class $K= (..)(..)$ and the two 3 dimensional and one 2 dimensional representations of the group.

I use the relation obtained from the classes' algebra, which after some steps becomes::
$|K|^2 \chi^{\nu} (K) \chi^\nu (K) = d_\nu \Big[ 3 d_\nu + 2 |K| \chi^\nu (K) \Big]$
after inserting the numbers ($|K|=3, d=3$) I find for the two 3-dimensional the same equation, and give for those $X$'s:
$X(K)^2 - 2 X(K) -3 =0 \Rightarrow X=-1, X=3$

and for the 2-dimensional ($|K|=3, d=2$), $x$:
$3 x(K)^2 - 4 x(K) -4=0 \Rightarrow X=2, X= -\frac{2}{3}$

Obviously I'm getting the same result as the bibliography but I also get the $X=3$ and $x= - \frac{2}{3}$... How can I discard them?

 

[jvicens]

Hi there,

First of all, great job on using the relation obtained from the classes' algebra to find the characters for the S_4 group. Let's go through your calculations to see if we can figure out why you are getting the extra solutions of X=3 and x= - \frac{2}{3}.

Starting with the two 3-dimensional representations, we have the equation X(K)^2 - 2 X(K) -3 =0. This is a quadratic equation, and we can use the quadratic formula to solve for X:
X = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
where a=1, b=-2, and c=-3. Plugging in these values, we get:
X = \frac{2 \pm \sqrt{4 - 4(-3)}}{2}
= \frac{2 \pm \sqrt{16}}{2}
= \frac{2 \pm 4}{2}
= -1 or 3
So, we can see that both solutions are valid for this equation. However, since we are looking for the characters for the S_4 group, we can eliminate the solution of X=3 because it is not a valid character. The characters for a group must be integers, and 3 is not an integer.

Moving on to the 2-dimensional representation, we have the equation 3 x(K)^2 - 4 x(K) -4=0. Again, this is a quadratic equation and we can solve for x using the quadratic formula:
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
where a=3, b=-4, and c=-4. Plugging in these values, we get:
x = \frac{4 \pm \sqrt{16 - 4(3)(-4)}}{2(3)}
= \frac{4 \pm \sqrt{64}}{6}
= \frac{4 \pm 8}{6}
= 2 or -\frac{2}{3}
So, both solutions are once again valid for this equation. However, we can eliminate the solution of x=2 because it is not a valid character for the S_4 group. The characters for a group must sum to the order of the group, which for S_4 is 24. The character x