Results 1 to 4 of 4

Math Help - Transitive Group Action on a Finite Set

  1. #1
    Newbie
    Joined
    Sep 2011
    Posts
    2

    Transitive Group Action on a Finite Set

    I am working on problem 4.1.9 in Dummit and Foote's Abstract Algebra. Thought I am in an algebra course, we don't use this book, and this isn't a homework assignment.

    G acts transitively on a finite set A and H is a normal subgroup of G. O_1,...,O_r are the distinct orbits of H on A.

    I have already done half of the problem, which is to show all orbits of H have the same cardinality and that G is transitive on the set {O_1,...,O_r} where G acts on this set via representative.

    My problem is this: Prove that if a\in O_1 then |O_1|=|H:H\cap G_a| and prove that r=|G:HG_a|.

    I cannot do the second part (the first isn't hard). A hint is to use the 2nd isomorphism theorem, but I don't see how it helps.

    Thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21

    Re: Transitive Group Action on a Finite Set

    Quote Originally Posted by EntryLevel View Post
    I am working on problem 4.1.9 in Dummit and Foote's Abstract Algebra. Thought I am in an algebra course, we don't use this book, and this isn't a homework assignment.

    G acts transitively on a finite set A and H is a normal subgroup of G. O_1,...,O_r are the distinct orbits of H on A.

    I have already done half of the problem, which is to show all orbits of H have the same cardinality and that G is transitive on the set {O_1,...,O_r} where G acts on this set via representative.

    My problem is this: Prove that if a\in O_1 then |O_1|=|H:H\cap G_a| and prove that r=|G:HG_a|.

    I cannot do the second part (the first isn't hard). A hint is to use the 2nd isomorphism theorem, but I don't see how it helps.

    Thanks
    What have you tried? Are the G_a the stabiliers in G or H?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Sep 2011
    Posts
    2

    Re: Transitive Group Action on a Finite Set

    Quote Originally Posted by Drexel28 View Post
    What have you tried? Are the G_a the stabiliers in G or H?
    They are the stabilizers in G.

    Actually, I decided to ignore the advice about the 2nd isomorphism theorem and I thought of the following map that I think is a bijection.

    \phi :G / HG_a \rightarrow \{O_1,...,O_r\}
    \phi (g)= the orbit containing g(a), where a is given in the problem.

    The map is definitely surjective. It is well defined because if you have two cosets gHG_a and g'HG_a then it must be true that g=g'hg_a for elements h\in H and g_a\in G_a

    g(a)=g'hg_a(a)=g'h(a)=g'(a)

    where the 2nd equality is because g_a is a stabilizer and the last equality because in part (a), it is shown that any element of G permutes the different orbits, taking all elements in one orbit to all elements in another...

    Finally it is injective because if g(a)=g'(a) then g^{-1}g'\in G_a\subset HG_a

    This seems to work... perhaps the hint was a red-herring (or perhaps I made a mistake).
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21

    Re: Transitive Group Action on a Finite Set

    Quote Originally Posted by EntryLevel View Post
    They are the stabilizers in G.

    Actually, I decided to ignore the advice about the 2nd isomorphism theorem and I thought of the following map that I think is a bijection.

    \phi :G / HG_a \rightarrow \{O_1,...,O_r\}
    \phi (g)= the orbit containing g(a), where a is given in the problem.

    The map is definitely surjective. It is well defined because if you have two cosets gHG_a and g'HG_a then it must be true that g=g'hg_a for elements h\in H and g_a\in G_a

    g(a)=g'hg_a(a)=g'h(a)=g'(a)

    where the 2nd equality is because g_a is a stabilizer and the last equality because in part (a), it is shown that any element of G permutes the different orbits, taking all elements in one orbit to all elements in another...

    Finally it is injective because if g(a)=g'(a) then g^{-1}g'\in G_a\subset HG_a

    This seems to work... perhaps the hint was a red-herring (or perhaps I made a mistake).
    Seems to work from what I see. I was going to just note the following. Since G acts transitively on A you have that \text{card}(A)=[G:G_a] for any a, right (this is just the orbit stabilizer theorem and using the fact that the orbit must be the whole set). But, from the orbit decomposition theorem and the fact that all the orbits of A under H are equipotent we have that \text{card}(A)=r\text{card}(O_1). So, we have that


    \displaystyle \begin{aligned}\frac{|G|}{|HG_a|} &=\frac{|G|}{\displaystyle \frac{|H||G_a|}{|H\cap G_a|}}\\ &=\frac{|G|}{\displaystyle \frac{|H|}{|H\cap G_a|}|G_a|}\\ &=\frac{\displaystyle \frac{|G|}{|G_a|}}{\text{card}(O_1)}\\ &=\frac{\text{card}(A)}{\text{card}(O_1)}\\ &=\frac{r\text{card}(O_1)}{\text{card}(O_1)}\\ &=r\end{aligned}
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Transitive action, blocks, primitive action, maximal subgroups.
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: August 6th 2011, 07:39 PM
  2. [SOLVED] transitive action; conjugation
    Posted in the Advanced Algebra Forum
    Replies: 14
    Last Post: March 7th 2011, 06:13 PM
  3. Triply Transitive Group Action
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: December 5th 2010, 07:41 PM
  4. group action
    Posted in the Advanced Algebra Forum
    Replies: 5
    Last Post: August 26th 2010, 12:07 PM
  5. Group action
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: October 15th 2009, 12:58 PM

Search Tags


/mathhelpforum @mathhelpforum