Results 1 to 9 of 9

Math Help - Abstract Algebra: subgroups and order proof

  1. #1
    Junior Member
    Joined
    Feb 2010
    Posts
    42

    Abstract Algebra: subgroups and order proof

    Hi sorry I did not know what to title this question!

    Here is the problem:

    Suppose that G has exactly three subgroups. Prove that G is finite and the order of G is a prime squared.

    ATTEMPT:

    G is either finite or infinite. Assume it is infinite. For g in G consider <g> ={g^n | n an integer} and <g^2> = {g^2n | n an integer} these both are subgroups but <g> =/= <g^2> so G has more than 3 subgroups so G cannot be infinite.

    Now we must show that its order is a prime squared but i dont know what to do?

    Any hints?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Feb 2010
    Posts
    42
    If i can show G is cyclic this will be easy since for Each divisor of the order of a cyclic group there is a subgroup of that order
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by mulaosmanovicben View Post
    Hi sorry I did not know what to title this question!

    Here is the problem:

    Suppose that G has exactly three subgroups. Prove that G is finite and the order of G is a prime squared.

    ATTEMPT:

    G is either finite or infinite. Assume it is infinite. For g in G consider <g> ={g^n | n an integer} and <g^2> = {g^2n | n an integer} these both are subgroups but <g> =/= <g^2> so G has more than 3 subgroups so G cannot be infinite.

    Now we must show that its order is a prime squared but i dont know what to do?

    Any hints?

    Suppose p,q\mid |G| , with p,q two different primes, and then use Cauchy's theorem to show G has at least 4 subgroups ...

    After the above you already know G is a finite p-group, and now a simple counting argument finishes the task.

    Tonio
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Feb 2010
    Posts
    42
    so why can't G have order p^3 or greater?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member roninpro's Avatar
    Joined
    Nov 2009
    Posts
    485
    Quote Originally Posted by mulaosmanovicben View Post
    so why can't G have order p^3 or greater?
    If your group has order p^3, then you are guaranteed to have subgroups of order 1, p, p^2, p^3.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Feb 2010
    Posts
    42
    why? that sounds like converse of lagranges theorem which is not always true.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by mulaosmanovicben View Post
    why? that sounds like converse of lagranges theorem which is not always true.
    No, he is citing one of the Sylow Theorem's which says that a group has a subgroup of every prime multiplicity dividing it.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by mulaosmanovicben View Post
    why? that sounds like converse of lagranges theorem which is not always true.

    With finite p-groups it is true...in fact, much more can be said: if |G|=p^n , with p a prime, then G has

    a normal subgroup of order p^k\,\,\forall\, 0\leq k\leq n .

    Tonio
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Newbie Halmos Rules's Avatar
    Joined
    Feb 2009
    Posts
    14
    I think you can avoid Cauchy and Sylow and just use some basic results concerning cyclic groups. Here's a hint :
    If {e}, H, G are the three distinct subgroups, what can you say about the subgroups of H? From this deduce the order of H and continue on from there...
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Abstract Algebra II: Normal Subgroups
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: September 7th 2010, 06:35 PM
  2. abstract algebra proof
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: January 27th 2010, 02:47 PM
  3. Abstract Algebra- order of a group
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: July 1st 2009, 06:23 PM
  4. Abstract algebra proof
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: September 30th 2008, 12:22 PM
  5. order of x (Abstract Algebra)
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: September 14th 2008, 09:47 AM

Search Tags


/mathhelpforum @mathhelpforum