I Using group action to prove a set is a subgroup

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Problem: Let ##G=S_n##, fix ##i \in \{1,2, \dots, n \}## and let ##G_i = \{ \sigma \in G ~|~ \sigma (i) = i \}##. Use group actions to prove that ##G_i## is a subgroup of G. Find ##|G_i|##.

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?
 

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Problem: Let ##G=S_n##, fix ##i \in \{1,2, \dots, n \}## and let ##G_i = \{ \sigma \in G ~|~ \sigma (i) = i \}##. Use group actions to prove that ##G_i## is a subgroup of G. Find ##|G_i|##.

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?
True. I also would have considered the direct way easier to see. The point of this exercise is to prepare for the - now I don't know the exact English term - we call it orbit formula. It is of similar importance as Lagrange's theorem for finite groups. It says:

Let ##(G,\cdot )## be a group which operates on a set ##M##, say ##(g,m) \longmapsto g.m \in M\,.##
Then we have for every ##m \in M## a bijection $$G/G_m \longleftrightarrow G.m = \{\,g.m\, : \,g \in G\,\} \subseteq M$$ where ##G_m## is the stabilizer and ##G.m## the orbit of ##m## under the operation of ##G##. Note that ##|G/G_m|## isn't necessarily a group, as the stabilizer may not be a normal subgroup. However, one can still build the equivalences classes ##g \,\cdot \,G_m##, it's just not necessarily a group again.

This yields for finite sets and groups ##|G/G_m| = |G.m|## resp. ##|G| = |G_m|\,\cdot \,|G.m|\,.##
 

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