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Thread: abelian group

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

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

    You would have been better posting this in the abstract algebra forum, btw...
    Last edited by Swlabr; May 25th 2009 at 06:59 AM.
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  3. #3
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    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 $\displaystyle n=\max \{o(g): \ g \in G \}$ and $\displaystyle o(a)=n.$ if $\displaystyle n=|G|,$ then we're done. otherwise, choose $\displaystyle b \notin <a>.$ let $\displaystyle o(b)=m.$ if $\displaystyle m \mid n,$ then $\displaystyle o(b)=o(a^{n/m})$ and thus by the uniqueness property of G

    we must have $\displaystyle b \in <a^{n/m}>,$ which is not possible. so $\displaystyle m \nmid n$ and hence $\displaystyle \text{lcm}(n,m) > n.$ but, since G is abelian, there exists an element of G of order ....................... contradiction!
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  4. #4
    Senior Member TheAbstractionist's Avatar
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    Another proof?

    Let $\displaystyle |G|=n.$ If $\displaystyle G$ is not cyclic, then $\displaystyle G$ would be the internal direct product of distinct cyclic subgroups $\displaystyle C_{n_1}C_{n_2}\cdots C_{n_r}$ where $\displaystyle n_i\mid n_{i-1}$ and $\displaystyle n=n_1n_2\cdots n_r.$ Since $\displaystyle n_2\mid n_1,$ it follows that $\displaystyle C_{n_1}$ would have a subgroup of order $\displaystyle n_2$ and so $\displaystyle G$ would have two subgroups of order $\displaystyle n_2.$ Contradiction!
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  5. #5
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    Quote Originally Posted by TheAbstractionist View Post
    Another proof?

    Let $\displaystyle |G|=n.$ If $\displaystyle G$ is not cyclic, then $\displaystyle G$ would be the internal direct product of distinct cyclic subgroups $\displaystyle C_{n_1}C_{n_2}\cdots C_{n_r}$ where $\displaystyle n_i\mid n_{i-1}$ and $\displaystyle n=n_1n_2\cdots n_r.$ Since $\displaystyle n_2\mid n_1,$ it follows that $\displaystyle C_{n_1}$ would have a subgroup of order $\displaystyle n_2$ and so $\displaystyle G$ would have two subgroups of order $\displaystyle n_2.$ 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 $\displaystyle G$ is a finitely generated abelian group such that for any positive integer $\displaystyle n$ there exists at most one subgroup $\displaystyle H$ of $\displaystyle G$ with $\displaystyle [G:H]=n.$ what can we say about $\displaystyle G$?
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  6. #6
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by NonCommAlg View Post
    this time suppose $\displaystyle G$ is a finitely generated abelian group such that for any positive integer $\displaystyle n$ there exists at most one subgroup $\displaystyle H$ of $\displaystyle G$ with $\displaystyle [G:H]=n.$ what can we say about $\displaystyle G$?
    Then $\displaystyle G = I(G) \times C_m, m \geq 0 \in \mathbb{N}$ where $\displaystyle I(G)$ is the subgroup of all non-torsion elements (along with the identity). $\displaystyle I(G)$ is a subgroup because $\displaystyle G$ is abelian. We get that $\displaystyle G = I(G) \times C_m$ by inducting on the number of generators of the group, $\displaystyle n$. Clearly the result holds for a 1-generated group.

    Assume the result holds for all groups generated by less than or equal to $\displaystyle n$ elements. Let $\displaystyle G$ be a subgroup generated by $\displaystyle n+1$ elements. Then, $\displaystyle I(G) \unlhd G$ is of finite index as $\displaystyle G/I(G)$ contains only torsion elements. Thus, $\displaystyle G/I(G)$ is generated by less than or equal to $\displaystyle n$ elements (this can be seen by viewing $\displaystyle G$ as an $\displaystyle n+1$-dimensional vector space). Is it of the form we wish? Let $\displaystyle [G/I(G):K_1] = [G/I(G):K_2] \leq \infty$. 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 $\displaystyle G$: $\displaystyle K_1, K_2 \lhd G$. Then, $\displaystyle [G:K_1] = [G:I(G)].[I(G):K_1] = [G:I(G)].[I(G):K_2] = [G:K_2]$, and so they are the same subgroups. Thus, $\displaystyle G/I(G)$ 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 $\displaystyle G/I(G) + I(G) = G$ and so we have a direct sum, and so a direct product. Thus, $\displaystyle G = I(G) \times C_m$.

    I think we cannot do anything with $\displaystyle I(G)$. Clearly it is just $\displaystyle C_{\infty} \times ... \times C_{\infty}$, but $\displaystyle C_{\infty} \times ... \times C_{\infty}$ conforms to the conditions stipulated.
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  7. #7
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    Quote Originally Posted by Swlabr View Post
    Then $\displaystyle G = I(G) \times C_m, m \geq 0 \in \mathbb{N}$ where $\displaystyle I(G)$ is the subgroup of all non-torsion elements (along with the identity). $\displaystyle I(G)$ is a subgroup because $\displaystyle G$ is abelian. We get that $\displaystyle G = I(G) \times C_m$ by inducting on the number of generators of the group, $\displaystyle n$. Clearly the result holds for a 1-generated group.

    Assume the result holds for all groups generated by less than or equal to $\displaystyle n$ elements. Let $\displaystyle G$ be a subgroup generated by $\displaystyle n+1$ elements. Then, $\displaystyle I(G) \unlhd G$ is of finite index as $\displaystyle G/I(G)$ contains only torsion elements. Thus, $\displaystyle G/I(G)$ is generated by less than or equal to $\displaystyle n$ elements (this can be seen by viewing $\displaystyle G$ as an $\displaystyle n+1$-dimensional vector space). Is it of the form we wish? Let $\displaystyle [G/I(G):K_1] = [G/I(G):K_2] \leq \infty$. 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 $\displaystyle G$: $\displaystyle K_1, K_2 \lhd G$. Then, $\displaystyle [G:K_1] = [G:I(G)].[I(G):K_1] = [G:I(G)].[I(G):K_2] = [G:K_2]$, and so they are the same subgroups. Thus, $\displaystyle G/I(G)$ 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 $\displaystyle G/I(G) + I(G) = G$ and so we have a direct sum, and so a direct product. Thus, $\displaystyle G = I(G) \times C_m$.

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

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