# Young diagram for $S_5$

I am trying to draw the Young diagram for $S_5$. I know the following pieces of information about $S_5$.

1. The order of the group is $120$.
2. The number of conjugacy classes and so partitions is $7$.
3. Degrees of irreducible representations $1,1,4,4,5,5,6$.
4. The partition is $1 + 10 + 15 + 20 + 20 + 24 + 30 = 120$.

I understand that the Young diagram should contain $30$ boxes in the first row, $24$ boxes in the second row, $20$ boxes in the third and fourth rows, $15$ boxes in the fifth row, $10$ boxes in the sixth row and $1$ box in the seventh row.

So, the Young diagram is as follows.

My question:

Am I doing it right? I understand that the next step is to fill up the boxes to make it a Young tableau.

UPDATE 1:

I was able to compute the partition as follows.

What should be my next step?

UPDATE 2:

I think I was able to draw the Young diagrams.

• It's not clear what you are trying to do here. Each irreducible representation of $S_5$ is associated with a partition and young diagram, but what is the "young diagram of $S_5$"? Why do you need to make it into a young tableau? – Jair Taylor Dec 5 '15 at 0:42
• Is this a homework question? It might help if you state the problem. – Jair Taylor Dec 5 '15 at 0:44
• @JairTaylor, this is not a homework problem. I am trying to compute all the irreps of $S_5$. I am done with trivial, parity, standard and product of sign and standard representations. To compute the rest three, I am going through Young Tableaux and the Representations of the Symmetric Group by Jeremy Booher. It is suggested in that note that a more systematic approach is to work out the Young diagram and Young tableaux first. – Omar Shehab Dec 5 '15 at 0:49
• A Young diagram for $S_5$ should correspond to a partition of 5, not 5! – Max Dec 5 '15 at 1:05
• Each irrep will correspond to a partition of 5. So the $7$ associated partitions are $5$, $41$,$32$,$311$,$221$,$2111$,$11111$. If you count the number of standard young tableaux for each of these shapes you will get the dimension of each irrep. – Jair Taylor Dec 5 '15 at 1:12

I do not think your first attempt is right, but it seems your updates are much more succesful. The partitions of $$5$$ are $$\{\{5\},\{4,1\},\{3,2\},\{3,1,1\},\{2,2,1\},\{2,1,1,1\},\{1,1,1,1,1\}\}$$ so your Young diagram will have $$5$$ boxes. If $$\{\lambda_1,\lambda_2,\ldots,\lambda_p\}$$ is one of the above partitions, then the corresponding Young diagram will have $$\lambda_1$$ boxes on row $$1$$, $$\lambda_2$$ boxes on row $$2$$ and so forth, up to $$\lambda_p$$ boxes on row $$p$$, with $$p\le 5$$, in agreement with your second update.
One can use instead of Young diagrams so-called Ferrers diagrams, which are basically Young diagrams with boxes replaced by dots, and more amenable to MathJax than boxes. Those would be (I don't have much control over the spacings forgive me) $$\bullet\bullet\bullet\bullet\bullet\, \qquad \stackrel{\Large\bullet\,\bullet\,\bullet\,\bullet}{\bullet\phantom{E\cdot \cdot\cdot}} \, ,\qquad \stackrel{\Large\bullet\,\bullet\,\bullet}{\bullet\bullet\phantom{\cdot}} \qquad \stackrel{\Large\bullet\bullet}{\stackrel{\Large\bullet\phantom{e}}{\bullet\phantom{e}}}$$ etc. for the partitions $$\{5\}, \{4,1\}, \{3,2\} ,\{3,1,1\}$$ respectively. The character table is $$\left( \begin{array}{ccccccc} 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ -1 & 0 & -1 & 1 & 0 & 2 & 4 \\ 0 & -1 & 1 & -1 & 1 & 1 & 5 \\ 1 & 0 & 0 & 0 & -2 & 0 & 6 \\ 0 & 1 & -1 & -1 & 1 & -1 & 5 \\ -1 & 0 & 1 & 1 & 0 & -2 & 4 \\ 1 & -1 & -1 & 1 & 1 & -1 & 1 \\ \end{array} \right)\, .$$