1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Abelian Permutation Group

  1. May 26, 2017 #1
    1. The problem statement, all variables and given/known data
    Assume that ##G## is an abelian transitive subgroup of ##S_A## that acts on the set ##A##. Show that ##\sigma(a) \neq a## for all ##\sigma \in G - \{1\}## and all ##a \in A##. Deduce that ##|G| = |A|##.

    2. Relevant equations

    A group is said to act transitively on a set if there is only one orbit.

    Fact: Let ##G## be a permutation group on the set ##A## (i.e., ##G \le S_A##), and let ##\sigma \in G## and let ##a \in A##. Then ##\sigma G_a \sigma^{-1} = G_{\sigma(a)}## (where ##G_a## denotes the stabilizer of ##a \in A##), and if ##G## acts transitively, then

    $$\bigcap_{\sigma \in G} \sigma G_a \sigma^{-1} = \{e\}$$

    3. The attempt at a solution

    If ##G## acts transitively on ##A##, then by the above fact we see that the kernel of the action is faithful. This means that ##\sigma (a) = a## if and only if ##\sigma = e##, which is to say that ##\sigma(a) \neq a## for all ##\sigma \in G - \{e\}## and ##a \in A##. I am now going to switch from the function notation ##\sigma(a)## to the customary group action notation ##g \cdot a##.

    Now we construct a bijection between ##A## and ##G## to show they have the same cardinality. Consider ##f : G \to A## defined by ##f(g) = g \cdot a##. First recall that the action has one orbit since it is transitive. This implies that there exists a single ##a \in A## such that ##A = \cal{O}_a##. Clearly our function is surjective, since if ##b \in A = \cal{O}_a##, there exists a ##g \in G## such that ##b = g \cdot a## or ##b = f(g)##. Now we show that it is injective. Suppose that ##f(g) = f(h)##. Then ##(h^{-1}g) \cdot a = a##. We will show this implies that ##(h^{-1}g)\cdot x = x## for all ##x \in A##. If ##x \in A##, then there exists a ##k \in G## such that ##x = k \cdot a##. Hence,

    $$(h^{-1}g) \cdot x = (h^{-1}g) \cdot (k \cdot a) = (h^{-1}k) \cdot (g \cdot a) = (k h^{-1}h) \cdot a = k \cdot a = x$$

    In the above we used commutativity several times. Since the kernel is trivial, ##h^{-1}g = e## or ##g = h##, thereby proving injectivity, and hence bijectivity.

    Does this sound right?
     
  2. jcsd
  3. May 28, 2017 #2

    fresh_42

    Staff: Mentor

    If ##G## is Abelian, then ##\bigcap_{\sigma \in G} \sigma G_a \sigma^{-1} = G_a## and your assumption, that this equals ##\{e\}## is what you want to prove.
     
  4. May 29, 2017 #3
    I am not sure I understand this remark. Have I not proven what I have been asked to prove?
     
  5. May 29, 2017 #4

    fresh_42

    Staff: Mentor

    ##\sigma G_a \sigma^{-1}=\sigma \sigma^{-1}G_a = G_a## for Abelian groups and therefore ##\bigcap_{\sigma \in G} \sigma G_a \sigma^{-1} = G_a##. Now you claimed, that transitivity of the action implies ##\bigcap_{\sigma \in G} \sigma G_a \sigma^{-1}=\{e_G\}##. Combining it, it reads ##G_a=\{g\in G\,\vert \,g.a=a\}=\{e_G\}## for all ##a \in A##, that is ##e_G## is the only operation with fix points.

    Nothing to prove anymore.
     
  6. May 29, 2017 #5
    On second thought, I think I may have made a mistake when proving this fact. Let ##g \in \sigma G_a \sigma^{-1}##. This occurs if and only if there exists a ##\tau \in G_a## such that ##g = \sigma \tau \sigma^{-1}## iff and only if ##\sigma^{-1} g = \tau \sigma^{-1}##...No matter how much fiddling around with the equation, I can seem to get ##g(\sigma(a)) = \sigma(a)##, which would complete the proof since this implies ##g## stablizes the element ##\sigma(a)##. Here is what makes me even more nervous: ##g = \sigma \tau \sigma^{-1}## seems to imply that ##\sigma^{-1} g \sigma = \tau \in G_a##, which would suggest that the stabilizer is always a normal subgroup...But this isn't always the case...what am I misunderstanding!?!?!

    Similarly, ##G_{\sigma(a)} = \{g \in G ~|~ g(\sigma(a)) = \sigma(a) \} = ##\{g \in G , but I am having trouble justifying why the last set in this string of set equalities is the set ##\sigma G_a \sigma^{-1}##.

    No matter which way I attack it, it seems that I cannot finish the proof.
     
  7. May 29, 2017 #6

    fresh_42

    Staff: Mentor

    Or might it be the case, that what you stated under section 2 is actually what you want to show instead of what you want to use? I stopped reading it, as it was already the solution there.
     
  8. May 29, 2017 #7
    No, what I wrote in section 3 was not intended to be used to proof what I mentioned in part 2. However, I just now realized that I incorrectly proved the fact I alluded in section 2 (see my post above).

    WAIT! I may have a proof. Give me a few seconds!

    ##g \in G_{\sigma(a)}## if and only ##g(\sigma(a)) = \sigma(a)## if and only if ## (\sigma^{-1}g \sigma)(a) = a## if and only if ## \sigma^{-1} g \sigma \in G_a## if and only if there exists a ##\tau \in G_a## such that ##\sigma^{-1} g \sigma = \tau## if and only if ##g = \sigma \tau \sigma^{-1} \in \sigma G_a \sigma^{-1}##.

    Does this seem right?
     
    Last edited: May 29, 2017
  9. May 29, 2017 #8

    fresh_42

    Staff: Mentor

    I don't see the point with the conjugation. In general ##\sigma G_a \sigma^{-1}.\sigma(a)=\sigma(a)##, i.e. ##\sigma G_a \sigma^{-1} =G_{\sigma(a)}##.
    On the other hand, your very first sentence was
    so all conjugations are already the identity: ##\sigma g \sigma^{-1}=g\,##.

    Therefore my question is: Why is ##\bigcap_{\sigma \in G}\sigma G_a \sigma^{-1}=\{e_G\}\;##?
     
  10. May 29, 2017 #9
    Yes, but the "fact" I alluded to in section 2 of my OP holds whether or not ##G## is abelian. So the proof I gave in post #7 proceeded without this assumption.
     
  11. May 29, 2017 #10

    fresh_42

    Staff: Mentor

    I'm completely confused now, sorry. Let me summarize. We have by commutativity and transitivity
    $$G_a= \bigcap_{\sigma \in G}\sigma G_a \sigma^{-1} = \bigcap_{\sigma \in G} G_{\sigma(a)}=\bigcap_{b\in A}G_b$$
    This means, every ##\sigma \in G_a## acts as identity on ##A##. Since all elements of ##G## are defined as elements of ##S_A##, it means ##\sigma = e_G## which proves the first part.

    Now we proceed by
    the orbit of ##a##
    It would be a good habit to learn to avoid such words: clearly, obviously and similar. What might be obvious to you doesn't have to be obvious to others. I remember an occasion an "obviously" took me two days of search and three substitutions of complex variables and written twenty lines of calculations ...
    I think you can stop here. The first part already showed that a fix point for one element is automatically a fix point for all elements and ##h^{-1}g=e_G##. And I think this has been meant by (my flag):
     
    Last edited: May 29, 2017
  12. Jun 4, 2017 #11

    WWGD

    User Avatar
    Science Advisor
    Gold Member

    I find your notation unusual: when you mean that the kernel is faithful. I have heard of faithful actions, but not faithful kernels, do you mean that the kernel ( when the action is seen as an injection of ## G ## into the permutation group ## S_{|X|}## , where ## X## is the set you are acting on ) is trivial?.EDIT: And you want to show that if the associated map to the permutation group is injective, then the action is faithful ?
     
  13. Jun 4, 2017 #12

    fresh_42

    Staff: Mentor

    Those kernels ##\{e\}## are especially faithful in group theory. You can factor them out, separate them in direct factors, but you do not get rid of them, never ever. :biggrin:

    (Sorry @Bashyboy, couldn't resist.)
     
  14. Jun 4, 2017 #13

    WWGD

    User Avatar
    Science Advisor
    Gold Member

    And she has your phone number and she will call you even after you break up with her...., er, I mean, I agree, kernels, yes, kernels ;).
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Abelian Permutation Group
  1. Abelian group (Replies: 9)

  2. Abelian group (Replies: 1)

  3. Abelian Groups (Replies: 5)

  4. Abelian Groups (Replies: 9)

  5. Abelian group (Replies: 2)

Loading...