- #1

Mr Davis 97

- 1,462

- 44

So here is what I did. Let ##A = \{1,2, \dots, n \}##. I claim that ##G## acts on ##A## by the group action ##\sigma \cdot i = \sigma (i)##. Proof: Let ##I_p## is the identity permutation. Then ##I_p \cdot i = I_p (i) = i##. Also, if ##\sigma_1, \sigma_2 \in G##, then it can quickly be checked that ##\sigma_1 \cdot (\sigma_2 \cdot i) = (\sigma_1 \circ \sigma_2) \cdot i##. This shows that we have a group action, which means that ##G_i##, the stabilizer of ##i## in ##G##, is automatically a subgroup. Also, by simple counting, we see that ##|G_i| = (n-1)!##.

Here is my real question: what was the point of this exercise? What's the utility of using the language of group actions when it seems I could have easily accomplished the same thing without it?