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I am not sure if Sage can be used to find the irreducible representations of the symmetric group.

For example:

For $g = (123)$, we have $$D(a)=\begin{bmatrix}0&0&1\\ 1 & 0 & 0\\0 & 1 & 0\end{bmatrix}$$

One decomposition is the following:
$$D(a)=\begin{bmatrix}1&0&0\\ 0 & -\frac{1}{2} & -\frac{\sqrt{3}}{2}\\0 & \frac{\sqrt{3}}{2} & -\frac{1}{2}\end{bmatrix}$$

so we get two irreducible representations: $$1,\ \ \begin{bmatrix} -\frac{1}{2} & -\frac{\sqrt{3}}{2}\\ \frac{\sqrt{3}}{2} & -\frac{1}{2}\end{bmatrix}$$

The answer is not unique since we can pick another similar transformation with different basis.

Can Sage be used to do this for the general case $S_n$? If not, any software can make it?

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    $\begingroup$ I am not sure what "this" refers to. You seem to just have a single matrix rather than a representation of a symmetric group. And I am not sure what the uniqueness is relevant for. $\endgroup$ Commented Jun 11, 2018 at 4:33
  • $\begingroup$ @TobiasKildetoft What I mean is for $S_3$, input is a matrix representation $D(a)$ of an element, say $(123)$ of $S_3$. The output is $F(a)$. Of course the order of $S_3$ is $6$. "This" refers to "if the input is an permutation matrix, the output is the block matrix such that each block is irreducible representation" $\endgroup$ Commented Jun 11, 2018 at 5:04
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    $\begingroup$ A block of a matrix is not a representation. You need to do this simultaneously for all group elements to get a representation. $\endgroup$ Commented Jun 11, 2018 at 6:31

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Symmetric group representations are implemented in Sage.

See the documentation:

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