Using group action to prove a set is a subgroup

[Mr Davis 97]

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?

 

[fresh_42]

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. $|G| = |G_m|\,\cdot \,|G.m|\,.$