Results 1 to 11 of 11

Math Help - A Few Questions Involving Subgroups

  1. #1
    Member
    Joined
    Nov 2008
    Posts
    152

    A Few Questions Involving Subgroups

    1) Let G = <x> be a cyclic group of order n. Show that x^m is a generator of G if and only if (m,n) = 1. Thus the number of generators of a cyclic group of order n is the number of integers m in the set (0,1,...,n-1) such that (m,n) = 1. This is called Euler's function and plays a prominent role in number theory.

    2) Let mZ and nZ be subgroups of (Z, +). What condition on m and n is equivalent to mZ is a subset of nZ? What condition on m and n is equivalent to mZ union nZ being a subgroup of (Z,+)?
    (I don't really even get what this is asking...)

    3) Prove that the intersection of two subgroups of a group G is itself a subgroup of G.

    4) Give an example of a group G and a subset H of G such that H is closed under multiplication but H is not a subgroup of G.

    5) a) Show that it is impossible for a group G to be the union of two proper subgroups.
    b) Give an example of a group that is the union of three proper subgroups.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Dec 2008
    Posts
    130
    Quote Originally Posted by Janu42 View Post
    1) Let G = <x> be a cyclic group of order n. Show that x^m is a generator of G if and only if (m,n) = 1. Thus the number of generators of a cyclic group of order n is the number of integers m in the set (0,1,...,n-1) such that (m,n) = 1. This is called Euler's function and plays a prominent role in number theory.

    2) Let mZ and nZ be subgroups of (Z, +). What condition on m and n is equivalent to mZ is a subset of nZ? What condition on m and n is equivalent to mZ union nZ being a subgroup of (Z,+)?
    (I don't really even get what this is asking...)

    3) Prove that the intersection of two subgroups of a group G is itself a subgroup of G.

    4) Give an example of a group G and a subset H of G such that H is closed under multiplication but H is not a subgroup of G.

    5) a) Show that it is impossible for a group G to be the union of two proper subgroups.
    b) Give an example of a group that is the union of three proper subgroups.
    3) here
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by Janu42 View Post
    1) Let G = <x> be a cyclic group of order n. Show that x^m is a generator of G if and only if (m,n) = 1. Thus the number of generators of a cyclic group of order n is the number of integers m in the set (0,1,...,n-1) such that (m,n) = 1. This is called Euler's function and plays a prominent role in number theory.
    To show \left< x^m \right> = G it is sufficient to show the order of x^m is n. Say that k is the order of x^m then it means x^{km} = e. But x has order n therefore n|km \implies n|k \implies n=k since (n,m)=1. Conversely, say that \left< x^m \right> = G and say that (n,m)=d>1. We will arrive at a contradiction. Note that (x^m)^{n/d} = (x^n)^{m/d} = e. But then it means the order of x^m is less than n since n/d<n - this is a contradiction.

    2) Let mZ and nZ be subgroups of (Z, +). What condition on m and n is equivalent to mZ is a subset of nZ? What condition on m and n is equivalent to mZ union nZ being a subgroup of (Z,+)?
    (I don't really even get what this is asking...)
    This problem is asking if m\mathbb{Z}\subseteq n\mathbb{Z} what can you conclude about the relationship between n\text{ and }m.

    4) Give an example of a group G and a subset H of G such that H is closed under multiplication but H is not a subgroup of G.
    Let G = \mathbb{R} and let H = \{ 2^r | r>0\}.

    5) a) Show that it is impossible for a group G to be the union of two proper subgroups.
    Hint: If a union of two subgroups is a group then one subgroup is contained in another one.

    b) Give an example of a group that is the union of three proper subgroups.
    Let G=\mathbb{Z}_2\times \mathbb{Z}_2.
    Can you think of it now?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Nov 2008
    Posts
    152
    For 5(a), why does one have to be contained in the other? Like, if G = {1,2,3,4,5,6}, couldn't one subgroup be {1,2,3} and the other {4,5,6}?

    For (b), would it be (0,0), (0,1), and (1,0) as the proper subgroups?

    Also for 4, this is closed under multiplication obviously. But it is not a subgroup because it is not closed under inverses?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Nov 2008
    Posts
    152
    And 2 more:

    1) Let G be a group and let H be a nonempty subset of G such that whenever x,y are in H we have xy^-1 in H. Prove that H is a subgroup of G.

    2) Let G be a group and let a be some fixed element of G. Let H be a subgroup of G and let aHa^-1 be the subset of G consisting of all elements that are of the form aha^-1, with h in H. Show that aHa^-1 is a subgroup of G. It is called the conjugate subgroup of H by a.
    Last edited by Janu42; February 12th 2009 at 08:19 PM.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Dec 2008
    Posts
    130
    Quote Originally Posted by Janu42 View Post

    Also for 4, this is closed under multiplication obviously. But it is not a subgroup because it is not closed under inverses?
    there is no identity element

    Also, please make a separate thread for your next 2 questions.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Nov 2008
    Posts
    152
    Oh, for 5, does something have to do with the fact that the identity is in every subgroup?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by Janu42 View Post
    Oh, for 5, does something have to do with the fact that the identity is in every subgroup?
    By definition a subgroup needs to have an identity element.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Member
    Joined
    Nov 2008
    Posts
    152
    And for 5(b), I don't know if what I said is right. Is it the union of (0,0), (0,1), and (1,0)?
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by Janu42 View Post
    And for 5(b), I don't know if what I said is right. Is it the union of (0,0), (0,1), and (1,0)?
    Let H_1 = \{ (0,0),(1,0)\}, H_2 = \{ (0,0),(0,1)\},H_3 = \{ (0,0),(1,1)\}.
    Then it means H_1\cup H_2\cup H_3 = \mathbb{Z}_2\times \mathbb{Z}_2.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Member
    Joined
    Nov 2008
    Posts
    152
    Ahh, right. OK. I had the right idea but overlooked the fact that the identity needs to be in every subgroup. Thanks.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Two questions involving integration
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: April 12th 2011, 06:12 PM
  2. Homomorphisms and Normal Subgroups Questions
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: March 29th 2009, 10:44 AM
  3. Subgroups involving disjoint cycle notation.
    Posted in the Advanced Algebra Forum
    Replies: 8
    Last Post: January 1st 2009, 08:18 AM
  4. Questions involving continuity
    Posted in the Calculus Forum
    Replies: 3
    Last Post: September 18th 2008, 05:07 PM
  5. A few questions involving limits
    Posted in the Calculus Forum
    Replies: 2
    Last Post: January 24th 2008, 02:27 AM

Search Tags


/mathhelpforum @mathhelpforum