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Math Help - Algebra, Problems For Fun! (1)

  1. #1
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    Algebra, Problems For Fun! (1)

    i decided to occasionally post a problem for fun in here. they are going to be as elementary as possible (first course in algebra) but certainly not straightforward. here's the first one:

    .................................................. .................................................. .................................................. .................................................. ......................................

    Let G be an abelian group and 1 \neq x \in G. Suppose that every subgroup H \neq \{1\} of G contains x. Prove that every two subgroups of G are comparable, i.e. if H_1, H_2 are two subgroups

    of G, then either H_1 \subseteq H_2 or H_2 \subseteq H_1.
    Last edited by NonCommAlg; December 24th 2008 at 07:34 PM.
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  2. #2
    Senior Member TheAbstractionist's Avatar
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    Quote Originally Posted by NonCommAlg View Post
    Let G be an abelian group and 1 \neq x \in G. Suppose that every subgroup H \neq \{1\} of G contains x. Prove that every two subgroups of G are comparable, i.e. if H_1, H_2 are two subgroups

    of G, then either H_1 \subseteq H_2 or H_2 \subseteq H_1.
    Hi NonCommAlg.

    We have either x^2=1 or else x\in\left<x^2\right>\ \implies\ x=x^{2m} for some m\ \implies\ x^{2m-1}=1. In either case, x has finite order, say p.

    Indeed p must be prime. For suppose p=ab for some 1<a,b<k. Then x^a\ne1\ \implies\ x\in\left<x^a\right>\ \implies\ x=x^{na} for some n\ \implies\ x^b=x^{nab}=1 contrary to the minimality of p.

    Now let g be any nontrivial element of G. Then x\in\left<g\right>\ \implies\ x=g^m for some m\ \implies\ g^{mp}=x^p=1. Hence g has finite order too.

    Since \left<x\right>\subseteq\left<g\right>, p divides the order of g by Lagrange’s theorem, so the order of g is of the form p^rk for some positive integers r,k with k coprime with p. If k>1, then k would be divisible by a prime q\ne p and then \left<g\right> would have a Sylow q-subgroup which cannot possibly contain x, an element of order p. Hence k=1 and so every nontrivial element g of G has order a power of p.

    Note that if K is any finite subgroup of G of order p, then, since x\in K, we must have K=\left<x\right>. Hence \left<x\right> is the unique finite subgroup of G of order p.

    So let g have order p^r. Assume that r>1. Then \left<g\right> has a totally ordered chain of subgroups \left<g\right>\supseteq\left<g^p\right>\supseteq\l  eft<g^{p^2}\right>\supseteq\cdots\supseteq\left<g^  {p^r}\right>=\{1\} containing every one of its subgroups. (Indeed this is a composition series in Jordan–Hölder theory.) (Note that \left<x\right>=\left<g^{p^{r-1}}\right>.) Of course if r=1, then \left<g\right>=\left<x\right> and so all subgroups of \left<g\right> (namely \left<x\right> and {1}) are comparable as well. Hence every finite cylic subgroup of G has subgroups that are all comparable.

    The next thing is to show that every finite subgroup A of G is cyclic. It would then follow by the above that all subgroups of any finite subgroup of G are comparable. Since G is Abelian, so is A. If A is not cyclic, then, by the theory of finite Abelian groups, A would be the internal direct product of cyclic subgroups A_1,A_2,\ldots,A_s. This would imply that A_i\cap A_j=\{1\} if i\ne j, meaning that x can’t be in all the A_i’s. As this is not what we want, we must take it that A is cyclic.

    (NB: For more on direct products and the classification of finite Abelian groups, see Chapters 13 and 14 of John F. Humphreys, A Course in Group Theory, Oxford University Press, 1996, which contains an excellent and readable account of the subject. )

    Now we are ready to prove the main result. Let H_1,H_2 be subgroups of G and suppose that H_2\not\subseteq H_1, i.e. \exists\,h_0\in H_2 with h_0\notin H_1. Let h\in H_1. As G is Abelian, it is easily checked that B=\left\{yz:y\in\left<h\right>,z\in\left<h_0\right  >\right\} is a subgroup of G containing both \left<h\right> and \left<h_0\right>. Since it is also finite, all its subgroups are comparable by the results established above; in particular, \left<h\right> and \left<h_0\right> are comparable. Since h_0\notin \left<h\right>, we must have \left<h\right>\subseteq\left<h_0\right>. Thus h is in \left<h_0\right> for any h\in H_1. Hence H_1\subseteq\left<h_0\right>\subseteq H_2 proving that H_1 and H_2 are comparable.
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    Quote Originally Posted by TheAbstractionist View Post
    Hi NonCommAlg.

    We have either x^2=1 or else x\in\left<x^2\right>\ \implies\ x=x^{2m} for some m\ \implies\ x^{2m-1}=1. In either case, x has finite order, say p.

    Indeed p must be prime. For suppose p=ab for some 1<a,b<k. Then x^a\ne1\ \implies\ x\in\left<x^a\right>\ \implies\ x=x^{na} for some n\ \implies\ x^b=x^{nab}=1 contrary to the minimality of p.

    Now let g be any nontrivial element of G. Then x\in\left<g\right>\ \implies\ x=g^m for some m\ \implies\ g^{mp}=x^p=1. Hence g has finite order too.

    Since \left<x\right>\subseteq\left<g\right>, p divides the order of g by Lagrange’s theorem, so the order of g is of the form p^rk for some positive integers r,k with k coprime with p. If k>1, then k would be divisible by a prime q\ne p and then \left<g\right> would have a Sylow q-subgroup which cannot possibly contain x, an element of order p. Hence k=1 and so every nontrivial element g of G has order a power of p.

    Note that if K is any finite subgroup of G of order p, then, since x\in K, we must have K=\left<x\right>. Hence \left<x\right> is the unique finite subgroup of G of order p.

    So let g have order p^r. Assume that r>1. Then \left<g\right> has a totally ordered chain of subgroups \left<g\right>\supseteq\left<g^p\right>\supseteq\l  eft<g^{p^2}\right>\supseteq\cdots\supseteq\left<g^  {p^r}\right>=\{1\} containing every one of its subgroups. (Indeed this is a composition series in Jordan–Hölder theory.) (Note that \left<x\right>=\left<g^{p^{r-1}}\right>.) Of course if r=1, then \left<g\right>=\left<x\right> and so all subgroups of \left<g\right> (namely \left<x\right> and {1}) are comparable as well. Hence every finite cylic subgroup of G has subgroups that are all comparable.

    The next thing is to show that every finite subgroup A of G is cyclic. It would then follow by the above that all subgroups of any finite subgroup of G are comparable. Since G is Abelian, so is A. If A is not cyclic, then, by the theory of finite Abelian groups, A would be the internal direct product of cyclic subgroups A_1,A_2,\ldots,A_s. This would imply that A_i\cap A_j=\{1\} if i\ne j, meaning that x can’t be in all the A_i’s. As this is not what we want, we must take it that A is cyclic.

    (NB: For more on direct products and the classification of finite Abelian groups, see Chapters 13 and 14 of John F. Humphreys, A Course in Group Theory, Oxford University Press, 1996, which contains an excellent and readable account of the subject. )

    Now we are ready to prove the main result. Let H_1,H_2 be subgroups of G and suppose that H_2\not\subseteq H_1, i.e. \exists\,h_0\in H_2 with h_0\notin H_1. Let h\in H_1. As G is Abelian, it is easily checked that B=\left\{yz:y\in\left<h\right>,z\in\left<h_0\right  >\right\} is a subgroup of G containing both \left<h\right> and \left<h_0\right>. Since it is also finite, all its subgroups are comparable by the results established above; in particular, \left<h\right> and \left<h_0\right> are comparable. Since h_0\notin \left<h\right>, we must have \left<h\right>\subseteq\left<h_0\right>. Thus h is in \left<h_0\right> for any h\in H_1. Hence H_1\subseteq\left<h_0\right>\subseteq H_2 proving that H_1 and H_2 are comparable.
    i posted this problem long time ago and i can't remember my solution anymore! haha ... i guess it was something like yours.
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    Senior Member TheAbstractionist's Avatar
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    Quote Originally Posted by NonCommAlg View Post
    i posted this problem long time ago and i can't remember my solution anymore! haha ... i guess it was something like yours.
    Hi NonCommAlg.

    That was a very interesting problem. Thanks for posting it!

    It’s amazing how a simple and innocuous assumption can lead to all sorts of interesting consequences and results. In this case, just assuming that the intersection of all nontrivial subgroups of G is nontrivial forces G to be a p-group. I suppose this is the sort of problem the Hungarian mathematician Pál Erdős would have loved.
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    Quote Originally Posted by TheAbstractionist View Post
    Hi NonCommAlg.

    That was a very interesting problem. Thanks for posting it!

    It’s amazing how a simple and innocuous assumption can lead to all sorts of interesting consequences and results. In this case, just assuming that the intersection of all nontrivial subgroups of G is nontrivial forces G to be a p-group. I suppose this is the sort of problem the Hungarian mathematician Pál Erdős would have loved.
    yeah, the crazy Hungarian! haha ... now, is there any (non-trivial) abelian group in which every two subgroups are comparable but the intersection of all non-trivial subgroups is trivial?
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