Representation of $S_4$

Is there a general method to work out all irreducible complex representation of a group?

Describe all the the irreducible complex representation of the group $S_4$.

$S_4$ is the symmetric group on four letter.

• If you have a semester free, you can take a course and learn a general method. If you don't have that much time, there are tricks that work for this group, and tricks that work for that one. The more you know about a group, the more tricks you can find. What do you know about $S_4$? Do you know its conjugacy classes, their orders, its normal subgroups? and what do you know about group representations? If you're ready for it, it's done in section 8.7 of rutherglen.science.mq.edu.au/ccooper/Groups/… – Gerry Myerson Feb 16 '14 at 3:09
• What do you know already? what do we need to know about representation of a group? what are conjugacy classes of $S_4$? you heard about character table? – user87543 Feb 16 '14 at 3:12
• Just occurred to me that the Cooper link I gave gives the irreducible characters, not the representations, so there will still be some work to do. – Gerry Myerson Feb 16 '14 at 4:28
• math.stackexchange.com/questions/914545/… – k73586 Aug 31 '14 at 4:34

In some generality, the irreducible complex representations of $S_n$ are naturally indexed by the partitions of $n$. The irreducible representation associated to a partition $\lambda$ is called the Specht module $S^{\lambda}$. It has a basis indexed by the standard Young tableaux of shape $\lambda$.

In principle, the Specht modules of $S_n$ can be described explicitly. I think the case of $n=4$ should be reasonably straightforward. I would suggest that you look in "Young Tableaux" by Fulton.

Just in case Peter Crooks' excellent answer is more general than you really want, here is a specific argument for $S_4$.

Presumably you can find the representations of $S_3$, or you wouldn't be attempting $S_4$.

Since $S_3$ is a quotient of $S_4$, you have the representations of degrees $1,1$ and $2$ of $S_3$.

Then you have the $3$-dimensional representation $\rho$ which is a constituent of the standard permutation representation, so that is easily calculated. (The permutation representation of any 2-transitive permutation group decomposes as the trivial rep plus an irreducible).

Finally, you get $\rho \otimes -1$, which equals $\rho$ on $A_4$ and $-\rho$ on $S_4 \setminus A_4$. Since $1^2+1^2+2^2+3^2+3^2=24$ that's the lot!