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My current approach is to take some elements of $M^{(2, 2)}$ and examine the submodules generated by them, in the hopes of finding a basis for them. Each submodule will correspond to an irreducible representation (since it was generated by a single element). We can find the representations from the submodules using the relationship $\rho (\sigma)b = \sigma b$ for $b$ a basis vector.

For example ,if we start with the element of $M^{(2,2)}$ that is the sum of all the tabloids, then the submodule is spanned by said element, since it is invariant under the effects of (12), (23), (34). Hence $\rho (\sigma) = 1$ for all $\sigma \in \mathfrak{S}$

I am struggling to apply this method to other elements of $M^{(2, 2)}$. How will I know when I have all of the irreducible representations?

EDIT: My apologies, I thought the notation I'd used was more well known. I'm looking at representation theory of the symmetric group. $M^{(2, 2)}$ is the permutation module spanned by tabloids of shape $(2, 2)$. Here is a reference that covers some of the topic area: http://www-personal.umich.edu/~charchan/RepresentationTheorySymmetricGroupsNotes.pdf

I have now figured out the crux of the problem. There is a 1-d, a 2-d and a 3-d submodule. One can prove that there are no further submodules of these by showing that any single element that spans such a submodule but be 0.

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  • $\begingroup$ You should probably explain what $M^{(2,2)}$ is, for our mind reading machines are all out of order this week. (Giving information about what field you are working with would not hurt, probably!) $\endgroup$ – Mariano Suárez-Álvarez Feb 6 '18 at 3:36
  • $\begingroup$ Being generated by a single element is far from enough to guarantee being irreducible. $\endgroup$ – Tobias Kildetoft Feb 6 '18 at 4:47
  • $\begingroup$ @TobiasKildetoft Thank you! That clarified things for me. If a submodule is spanned by a single element then it is irreducible, but being generated by a single element is not the same. $\endgroup$ – simples123 Feb 6 '18 at 23:01

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