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Let $S_n$ act on $S_n/H=\{ H,t_2 H,...,t_e H\}$ from the left and let $n\geq 5$, $e \geq 3$. Then the kernel of this action is trivial.

This supposedly follows because $e \geq 3$, but I haven't the slightest clue how? Now, the kernel is a normal subgroup of $S_n$, so it must be one of $S_n,A_n,\langle \mathrm{id} \rangle$. I'm actually not sure about the definition of the kernel of an action; Dummit/Foote defines the kernel of this particular action as $$\{ \sigma \in S_n \mid \sigma s_i H = s_i H, \forall s_i \in S_n \} = \cap_{s_i \in S_n} s_i H s_i^{-1}$$

I figure this is equivalent to $$\{ \sigma \in S_n \mid \sigma t_i H = t_i H, \forall t_i \in S_n//H \} = \cap_{t_i \in S_n//H} t_i H t_i^{-1}$$

In other words, those permutations stabilizing all the elements of $S_n/H$. When $e=2$ it can be checked "manually" that the kernel $A_n$, but showing that takes into account odd- and evenness and not the actual number $2$. How do I use the assumption $e\geq 3$ in proving the above assertion?

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    $\begingroup$ The kernel is contained in $H$. $e=[G:H]$. What is the index of $A_n$ in $S_n$ and how does the index of the kernel relate to the index of $[G:H]$? $\endgroup$ – Jack Schmidt Aug 3 '13 at 4:16
  • $\begingroup$ @JackSchmidt $[S_n : A_n]=2$ but as far as the rest goes I haven't a clue what you're getting at :) $\endgroup$ – Erik Vesterlund Aug 3 '13 at 4:23
  • $\begingroup$ @JackSchmidt Hold on a sec... The kernel is contained in $H$ of index $\geq 3$, and must be normal. $H$ isn't normal, so the kernel must be trivial. Is that it? $\endgroup$ – Erik Vesterlund Aug 3 '13 at 4:30
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    $\begingroup$ yup, that is correct. $\endgroup$ – Jack Schmidt Aug 3 '13 at 14:08
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Hint: Denote that kernel of this action by $K$. Then $K, t_2K,\cdots, t_eK$ are pairwise different.

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  • $\begingroup$ How do you deduce that? $\endgroup$ – Erik Vesterlund Aug 3 '13 at 4:24
  • $\begingroup$ @ErikVesterlund: Elements in $t_iK$ send $H$ to $t_iH$. $\endgroup$ – 23rd Aug 3 '13 at 4:39
  • $\begingroup$ @ErikVesterlund: Is everything clear to you now? $\endgroup$ – 23rd Aug 3 '13 at 5:53
  • $\begingroup$ I'd have to look into it, as is it doesn't provide any insights, sorry. Thanks for helping out though! $\endgroup$ – Erik Vesterlund Aug 3 '13 at 21:35
  • $\begingroup$ @ErikVesterlund: You are welcome. I think you have got the idea in your reply to Jack Schmidt's comment, so could you explain what is unclear to you in my answer? $\endgroup$ – 23rd Aug 4 '13 at 3:04

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