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I am attempting to understand the representation of the induced module $\mathrm{Ind}_{S_2}^{S_3}V = \Bbbk S_3 \otimes_{\Bbbk S_2} V$, where $V = \{v \in \Bbbk^2: v_1+v_2=0\} = \Bbbk(e_1 - e_2)$, and what the actual image of a given element in this representation. I know that $\Bbbk S_3$ is a free $\Bbbk S_2$-module with basis given by $\{(1),(13),(23)\}$, so computing the image of an element should just boil down to figuring out what the action of said element is on each member of the basis for $S_3 \otimes_{\Bbbk S_3}V$. The issue I'm having is I can't figure out how this action works.

For example, suppose I want to compute the image of $(123)$ under this representation. Then I must figure out the action of $(123)$ on $1 \otimes (e_1 - e_2)$, $(13) \otimes (e_1 - e_2)$, and $(23) \otimes (e_1 - e_2)$. I know in the end, I get a representation $S_3 \rightarrow GL_3(\Bbbk)$ since $\mathrm{dim}(S_3 \otimes_{\Bbbk S_3} V) = [S_3:S_2] \cdot \dim(V) = 3$, but I just don't see how to get this matrix, let alone compute the group action. I have a misunderstanding somewhere, I'm just not sure where it is.

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  • $\begingroup$ does $(1, 2, 3)\otimes (e_1-e_2) = (1,3)(1,2)\otimes(e_1-e_2) =(1,3)\otimes(1,2)(e_1-e_2)= (1,3) \otimes (e_2-e_1) = - (1,3)\otimes(e_1-e_2)$ help? It's not quite clear to me what you don't understand. $\endgroup$ Feb 16 at 20:10
  • $\begingroup$ @MatthewTowers Yes, that helps! I think I just stared at it too long and confused myself. Thanks! $\endgroup$
    – Red Sleuth
    Feb 16 at 20:16
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To apply $(123)$ to e.g. $1\otimes(e_2-e_3)$, you just apply $(123)$ to the first factor $1$, to get $(123)\otimes(e_2-e_3)$. This may not appear to be one of your basis elements for $\mathrm{Ind}_{S_2}^{S_3}V$, but it actually is. You just need to slide the element of $(12)\in S_2$ across the $\otimes$ symbol:

$$ \begin{array}{ll} (123)\otimes(e_1-e_2) & =(123)(12)^{-1}(12)\otimes (e_1-e_2) \\ & = (123)(12)^{-1}\otimes(12)(e_1-e_2) \\ & = (123)(12)\otimes(e_2-e_1) \\ & = (13)\otimes(-(e_1-e_2)). \end{array} $$

Thus, you get $-(13)\otimes(e_1-e_2)$. I'll let you try your hand at computing other group elements applied to other basis elements to write down the matrices.

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