This is exercise 4.1.2 from Dummit and Foote:

Let $G$ be a permutation group on the set $A$ (i.e., $G \le S_A$), let $\sigma \in G$, and let $a\in A$. Prove that $\sigma G_a \sigma^{-1} = G_{\sigma(a)}$, where $G_a$ is the stabilizer of $a$ in $G$. Deduce that if $G$ acts transitively on $A$ then $\bigcap_{\sigma\in G} \sigma G_a \sigma^{-1} = \{\text{id}\}$.

I'm having trouble with the second part. From the exercise that precedes this one, I know that $\bigcap_{\sigma\in G} \sigma G_a \sigma^{-1}$ is the kernel of the action. After a bit of agonizing, I checked the solution on Project Crazy Project, but the key line doesn't make sense to me: Apparently, because $G \le S_A$, the homomorphism inducing the action is injective.

I thought I might try proving that if a set is acted upon by a subgroup of its group of permutations, then the inducing homomorphism must be injective. First, though, I made up an example: Let $A=\{1,2,3,4\}$, so that $S_A = S_4$, and let $G=\left< \sigma \right>$, such that $\sigma = (1\;2\;3\;4)$. If $\theta : G \to S_A$ such that $\sigma \mapsto \sigma^2$, then $\theta$ is a homomorphism that induces a group action (I think these two are equivalent, but I'm not sure), but its kernel is $\{\text{id}, \sigma^2\}$. So it doesn't seem as though I should try proving the statement. (I just realized this action is not transitive, but then the statement I'm interested in proving doesn't mention anything about transitivity either...)

Having typed the above, I noticed that a few useful links had appeared in the sidebar. The small portion of this question that deals with the second half of the exercise seems a bit garbled, but I'm working on sorting it out. Furthermore, this answer says the homomorphism induced by a permutation group must be injective but doesn't say why.

So I guess that's my question: Why must the homomorphism be injective? What in the example I gave was wrong or irrelevant to the question at hand? (I've left out the tedious computation, but I was pretty careful about it.) I realize there may be some other question in here that I don't realize I have; I'd appreciate anything anyone thinks related to what I've written.

  • $\begingroup$ You are just confusing yourself! In this question, your action map $\theta$ maps $\sigma$ to $\sigma$, which is clearly injective. $\endgroup$ – Derek Holt Dec 30 '15 at 18:23
  • $\begingroup$ @DerekHolt That I'm confusing myself seems totally plausible. However, I don't see how $\theta$ maps $\sigma$ to itself; I defined it as $\sigma \mapsto \sigma^2$. Unless I'm missing something -- like I said: totally plausible -- $\theta(\sigma^2)=\text{id}$. $\endgroup$ – dmk Dec 30 '15 at 18:33
  • $\begingroup$ In the solution, when it refers to the "homomorphism inducing the action", it means the map $\theta: \sigma \mapsto \sigma$. $\endgroup$ – Derek Holt Dec 30 '15 at 18:41
  • $\begingroup$ @DerekHolt So is my error in the terminology or in the computation? $\endgroup$ – dmk Dec 30 '15 at 19:19
  • $\begingroup$ @DerekHolt Just looked this up: D&F (p. 114) write: "If $G$ is a group, a permutation representation of $G$ is any homomorphism of $G$ into the symmetric group $S_A$ [...]. We shall say a given action of $G$ on $A$ affords or induces the associated permutation representation of $G$." $\endgroup$ – dmk Dec 30 '15 at 19:22

Firstly, the action of $S_A$ on $A$ obviously has trivial kernel (the only element of $S_A$ which acts as the identity on $A$ is the identity). Since here $G$ is assumed to be a subgroup of $S_A$, anything in $G$ which acts as the identity on $A$ must also be an element of $S_A$ which acts as the identity, hence is the identity of $G$.

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  • $\begingroup$ By "the" action, I'm assuming you're referring to the one used in the solution on PCP. I think at one point I may have assumed that was the identity map -- in which case, yes, the action clearly has a trivial kernel. But why are we allowed to choose how $G$ acts on $A$? That is, if each action is in one-to-one correspondence with a homomorphism from $G$ to $S_A$, what's to stop me from choosing something like $\endgroup$ – dmk Dec 30 '15 at 22:33

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