Results 1 to 7 of 7

Math Help - abelian group

  1. #1
    Member
    Joined
    May 2009
    Posts
    86

    abelian group

    Suppose that G is a finite Abelian group that has exactly one subgroup for each divisor of |G|. Show that G is cyclic.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Swlabr's Avatar
    Joined
    May 2009
    Posts
    1,176
    Use the fundamental theorem for finite Abelian groups twice. Once to show that it is sufficient to prove the result for p-groups, and the second time to investigate said p-groups.

    Note also that if p divides the order of a group then that group contains an element of order p.

    You would have been better posting this in the abstract algebra forum, btw...
    Last edited by Swlabr; May 25th 2009 at 06:59 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    Quote Originally Posted by poorna View Post

    Suppose that G is a finite Abelian group that has exactly one subgroup for each divisor of |G|. Show that G is cyclic.
    let n=\max \{o(g): \ g \in G \} and o(a)=n. if n=|G|, then we're done. otherwise, choose b \notin <a>. let o(b)=m. if m \mid n, then o(b)=o(a^{n/m}) and thus by the uniqueness property of G

    we must have b \in <a^{n/m}>, which is not possible. so m \nmid n and hence \text{lcm}(n,m) > n. but, since G is abelian, there exists an element of G of order ....................... contradiction!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member TheAbstractionist's Avatar
    Joined
    Apr 2009
    Posts
    328
    Thanks
    1
    Another proof?

    Let |G|=n. If G is not cyclic, then G would be the internal direct product of distinct cyclic subgroups C_{n_1}C_{n_2}\cdots C_{n_r} where n_i\mid n_{i-1} and n=n_1n_2\cdots n_r. Since n_2\mid n_1, it follows that C_{n_1} would have a subgroup of order n_2 and so G would have two subgroups of order n_2. Contradiction!
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    Quote Originally Posted by TheAbstractionist View Post
    Another proof?

    Let |G|=n. If G is not cyclic, then G would be the internal direct product of distinct cyclic subgroups C_{n_1}C_{n_2}\cdots C_{n_r} where n_i\mid n_{i-1} and n=n_1n_2\cdots n_r. Since n_2\mid n_1, it follows that C_{n_1} would have a subgroup of order n_2 and so G would have two subgroups of order n_2. Contradiction!
    yes, but i was actually avoiding using the fundamental theorem of finite(ly generated) abelian groups. now, i was thinking, what if we try to make poorna's question more interesting:

    this time suppose G is a finitely generated abelian group such that for any positive integer n there exists at most one subgroup H of G with [G:H]=n. what can we say about G?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor Swlabr's Avatar
    Joined
    May 2009
    Posts
    1,176
    Quote Originally Posted by NonCommAlg View Post
    this time suppose G is a finitely generated abelian group such that for any positive integer n there exists at most one subgroup H of G with [G:H]=n. what can we say about G?
    Then G = I(G) \times C_m, m \geq 0 \in \mathbb{N} where I(G) is the subgroup of all non-torsion elements (along with the identity). I(G) is a subgroup because G is abelian. We get that G = I(G) \times C_m by inducting on the number of generators of the group, n. Clearly the result holds for a 1-generated group.

    Assume the result holds for all groups generated by less than or equal to n elements. Let G be a subgroup generated by n+1 elements. Then, I(G) \unlhd G is of finite index as G/I(G) contains only torsion elements. Thus, G/I(G) is generated by less than or equal to n elements (this can be seen by viewing G as an n+1-dimensional vector space). Is it of the form we wish? Let [G/I(G):K_1] = [G/I(G):K_2] \leq \infty. Then as we are in an Abelian group these subgroup as normal and so we can apply the correspondence theorem to get that they are normal subgroups of G:  K_1, K_2 \lhd G. Then,  [G:K_1] = [G:I(G)].[I(G):K_1] = [G:I(G)].[I(G):K_2] = [G:K_2], and so they are the same subgroups. Thus, G/I(G) is of the correct form. As it consists only torsion elements it is cyclic.

    Viewing our group as a vector space we can glue these two bits together - their intersection is trivial and G/I(G) + I(G) = G and so we have a direct sum, and so a direct product. Thus, G = I(G) \times C_m.

    I think we cannot do anything with I(G). Clearly it is just C_{\infty} \times ... \times C_{\infty}, but C_{\infty} \times ... \times C_{\infty} conforms to the conditions stipulated.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    Quote Originally Posted by Swlabr View Post
    Then G = I(G) \times C_m, m \geq 0 \in \mathbb{N} where I(G) is the subgroup of all non-torsion elements (along with the identity). I(G) is a subgroup because G is abelian. We get that G = I(G) \times C_m by inducting on the number of generators of the group, n. Clearly the result holds for a 1-generated group.

    Assume the result holds for all groups generated by less than or equal to n elements. Let G be a subgroup generated by n+1 elements. Then, I(G) \unlhd G is of finite index as G/I(G) contains only torsion elements. Thus, G/I(G) is generated by less than or equal to n elements (this can be seen by viewing G as an n+1-dimensional vector space). Is it of the form we wish? Let [G/I(G):K_1] = [G/I(G):K_2] \leq \infty. Then as we are in an Abelian group these subgroup as normal and so we can apply the correspondence theorem to get that they are normal subgroups of G:  K_1, K_2 \lhd G. Then,  [G:K_1] = [G:I(G)].[I(G):K_1] = [G:I(G)].[I(G):K_2] = [G:K_2], and so they are the same subgroups. Thus, G/I(G) is of the correct form. As it consists only torsion elements it is cyclic.

    Viewing our group as a vector space we can glue these two bits together - their intersection is trivial and G/I(G) + I(G) = G and so we have a direct sum, and so a direct product. Thus, G = I(G) \times C_m.

    I think we cannot do anything with I(G). Clearly it is just C_{\infty} \times ... \times C_{\infty}, but C_{\infty} \times ... \times C_{\infty} conforms to the conditions stipulated.
    yes, we have G=T(G) \times \mathbb{Z}^m, where T(G) is a a product of finite cyclic groups. then G/\mathbb{Z}^m \cong T(G) has the same property as G because [G/\mathbb{Z}^m : H/\mathbb{Z}^m]=[G:H]. thus T(G) has to be

    cyclic because it's finite. so G=C_n \times \mathbb{Z}^m. now the question is: does any G of this form have the property that if [G:H_1]=[G:H_2] < \infty, for some subgroups H_1,H_2 of G, then H_1=H_2?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Group, abelian
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: October 14th 2010, 05:05 AM
  2. abelian group
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: May 16th 2010, 04:41 PM
  3. Is the subgroup of an abelian group always abelian?
    Posted in the Advanced Algebra Forum
    Replies: 6
    Last Post: December 6th 2009, 11:38 PM
  4. Abelian Group
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: March 29th 2009, 05:38 AM
  5. Non- Abelian Group
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: August 25th 2008, 02:00 PM

Search Tags


/mathhelpforum @mathhelpforum