# Understanding how the induced representation of $S_3$ under $S_2$ with the standard representation works

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.

• 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. Feb 16 at 20:10
• @MatthewTowers Yes, that helps! I think I just stared at it too long and confused myself. Thanks! Feb 16 at 20:16

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.