Young diagram for $S_5$

Question

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.

Answer 1

Although late in the party, I post an answer with images of two .pdf pages produced by $\LaTeX$ since MathJaX doesn't support the Young Tableaux package.

The symmetric group $\mathrm S_{5}$ has $5!\boldsymbol{=}120$ elements in 7 conjugacy classes. The corresponding partitions, Young Diagrams (YD), $\lambda$'s, $\nu$'s and the cardinalities $h_{\boldsymbol{\lambda}}$ are shown in Table I :

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Table I

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^{Note that \begin{equation} \!\!\!\!\sum\limits_{j\boldsymbol{=}1}^{j\boldsymbol{=}7}h_{j}\boldsymbol{=}h_{1}\boldsymbol{+}h_{2}\boldsymbol{+}h_{3}\boldsymbol{+}h_{4}\boldsymbol{+}h_{5}\boldsymbol{+}h_{6}\boldsymbol{+}h_{7}\boldsymbol{=}1\boldsymbol{+}10\boldsymbol{+}15\boldsymbol{+}20\boldsymbol{+}20\boldsymbol{+}30\boldsymbol{+}24\boldsymbol{=}120\boldsymbol{=}5! \tag{01}\label{01} \end{equation}}

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The degrees $\,f_{\boldsymbol{\lambda}}\,$ (dimensions) of irreducible representations of the symmetric group $\mathrm S_{5}$ from the number of Standard Young Tableaux (SYT) per partition-conjugacy class using the hook length formula \begin{equation} \boxed{\:\:f_{\boldsymbol{\lambda}}\boldsymbol{=}\dfrac{n!}{\prod_{\imath\jmath}\mathrm{h}_{\imath\jmath}}\,,\quad \boldsymbol{\lambda}\boldsymbol{\vdash}n \vphantom{\dfrac{\frac{a}{b}}{\frac{c}{d}}}\:\:} \tag{02}\label{02} \end{equation} are shown in Table II :

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Table II

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^{Note that \begin{equation} \sum\limits_{j\boldsymbol{=}1}^{j\boldsymbol{=}7}f^2_{j}\boldsymbol{=}f^2_{1}\boldsymbol{+}f^2_{2}\boldsymbol{+}f^2_{3}\boldsymbol{+}f^2_{4}\boldsymbol{+}f^2_{5}\boldsymbol{+}f^2_{6}\boldsymbol{+}f^2_{7}\boldsymbol{=}1^2\boldsymbol{+}4^2\boldsymbol{+}5^2\boldsymbol{+}6^2\boldsymbol{+}5^2\boldsymbol{+}4^2\boldsymbol{+}1^2\boldsymbol{=}120\boldsymbol{=}5! \tag{03}\label{03} \end{equation}}

Answer 2

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 $$ \left( \begin{array}{ccccccc} 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ -1 & 0 & -1 & 1 & 0 & 2 & 4 \\ 0 & -1 & 1 & -1 & 1 & 1 & 5 \\ 1 & 0 & 0 & 0 & -2 & 0 & 6 \\ 0 & 1 & -1 & -1 & 1 & -1 & 5 \\ -1 & 0 & 1 & 1 & 0 & -2 & 4 \\ 1 & -1 & -1 & 1 & 1 & -1 & 1 \\ \end{array} \right)\, . $$