Results 1 to 4 of 4

Math Help - Abstract Algebra Help!!!

  1. #1
    Junior Member
    Joined
    Oct 2008
    From
    Green Bay
    Posts
    30

    Abstract Algebra Help!!!

    Hey im preparing for my second exam and I need major help with these problems. I would greatly appreciate any help, hints or answers, that could help me understand how to do these problems. So that I can do well thanks!!!

    G denotes a group

    1. Let |G|=p*g*r with different primes p, g, r, and let H,K<=G with |H|=p*g, |K|=g*r. Show:|H (intersection) K|=g.

    2. Let H be the normal subgroup G, a in G and |H|=10. The element aH of the group G/H has order 3; what are the possibilities for |a|? Explain.

    3. Let G be a group and G' be the subgroup generated by S={x^-1*y^-1*x*y| x,y in G}. Then:

    a. Prove that G' is normal in G.
    b. Prove that G/G' is abelian.
    c. If G/N is abelian, prove that G'<=N.
    d. Prove that if H<=G and G'<=H then H is a subgroup of G.

    4. Suppose that G is an abelian group of order 120 and that G has exactly 3 elements of order 2. Determine the isomorphism class of G. Explain.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by GreenandGold View Post
    1. Let |G|=p*g*r with different primes p, g, r, and let H,K<=G with |H|=p*g, |K|=g*r. Show:|H (intersection) K|=g.
    If H,K are subgroups of a finite group G then |HK|\cdot |H\cap K| = |H||K|.

    Now since H\cap K is a subgroup of H and K it means its order divides |H| = pg and gr. Therefore, |H\cap K| = 1\text{ or }g. We cannot have |H\cap K| = 1 for that would imply pgr=|G| \geq |HK| = |H||K| = pg^2r. A contradiction.

    Therefore, the only possible case is that |H\cap K| = g.

    2. Let H be the normal subgroup G, a in G and |H|=10. The element aH of the group G/H has order 3; what are the possibilities for |a|? Explain.
    If |aH| = 3 in the group G/H it means (aH)^3 = H and therefore a^3 \in H. But |H| = 10. Therefore, (a^3)^{10} = a^{30} = 1. Since a\not = 1 it means the possible values for |a| are 2,3,5,6,10,15,30.
    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 GreenandGold View Post
    3. Let G be a group and G' be the subgroup generated by S={x^-1*y^-1*x*y| x,y in G}. Then:

    a. Prove that G' is normal in G.
    b. Prove that G/G' is abelian.
    c. If G/N is abelian, prove that G'<=N.
    d. Prove that if H<=G and G'<=H then H is a subgroup of G.
    For x,y\in G define [x,y] = xyx^{-1}y^{-1} then show that g[x,y]g^{-1} = [gxg^{-1},gyg^{-1}].
    Therefore, we see that G' is a normal subgroup.

    Since this group is normal we can form the factor group G/G'. Let aG',bG' \in G/G'. To show (aG')(bG') = (bG')(aG') it is necessary and sufficient to prove that (ab)(ba)^{-1} \in G' \implies aba^{-1}b^{-1} \in G'. This is true by definition of G'.

    If G/N is abelian it means (aN)(bN) = (bN)(aN) and so aba^{-1}b^{-1} \in N. This is true for all a,b\in G. Therefore, G'\subseteq N.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by GreenandGold View Post
    4. Suppose that G is an abelian group of order 120 and that G has exactly 3 elements of order 2. Determine the isomorphism class of G. Explain.
    Let P be the Sylow 2-subgroup of order 8.

    Consider, G/P. This group has order 15 and therefore G/P \simeq \mathbb{Z}_{15} because that is the only group up to isomorphism.

    Since |P| = 8 and it is abelian implies that P must be isomorphic to one of the following groups: \mathbb{Z}_8, \mathbb{Z}_2^3, \mathbb{Z}_4\times \mathbb{Z}_2.
    Elements of order 2 must be contained in P. Therefore, by hypothesis, we are told that P has exactly 3 elements of order 2.
    The first two groups (on the list) do not satisfy this requirement.
    However, \mathbb{Z}_4\times \mathbb{Z}_2 has that requirement with elements (0,1),(2,1),(2,0).

    Therefore, G/P \simeq G/(\mathbb{Z}_4 \times \mathbb{Z}_2) \simeq \mathbb{Z}_3 \times \mathbb{Z}_5.

    Now we "cross multiply" and we get, G\simeq \mathbb{Z}_2\times \mathbb{Z}_3 \times \mathbb{Z}_4 \times \mathbb{Z}_5.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: December 6th 2010, 03:03 PM
  2. Replies: 0
    Last Post: April 23rd 2010, 11:37 PM
  3. Abstract Algebra
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: October 22nd 2008, 05:21 PM
  4. abstract algebra
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: October 4th 2008, 06:00 PM
  5. Abstract Algebra help
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: April 12th 2008, 01:59 AM

Search Tags


/mathhelpforum @mathhelpforum