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

    of $\displaystyle G,$ then either $\displaystyle H_1 \subseteq H_2$ or $\displaystyle H_2 \subseteq H_1.$
    Last edited by NonCommAlg; Dec 24th 2008 at 06:34 PM.
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  2. #2
    Senior Member TheAbstractionist's Avatar
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    Quote Originally Posted by NonCommAlg View Post
    Let $\displaystyle G$ be an abelian group and $\displaystyle 1 \neq x \in G.$ Suppose that every subgroup $\displaystyle H \neq \{1\}$ of $\displaystyle G$ contains $\displaystyle x.$ Prove that every two subgroups of $\displaystyle G$ are comparable, i.e. if $\displaystyle H_1, H_2$ are two subgroups

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

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

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

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

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

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

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

    The next thing is to show that every finite subgroup $\displaystyle A$ of $\displaystyle G$ is cyclic. It would then follow by the above that all subgroups of any finite subgroup of $\displaystyle G$ are comparable. Since $\displaystyle G$ is Abelian, so is $\displaystyle A.$ If $\displaystyle A$ is not cyclic, then, by the theory of finite Abelian groups, $\displaystyle A$ would be the internal direct product of cyclic subgroups $\displaystyle A_1,A_2,\ldots,A_s.$ This would imply that $\displaystyle A_i\cap A_j=\{1\}$ if $\displaystyle i\ne j,$ meaning that $\displaystyle x$ can’t be in all the $\displaystyle A_i$’s. As this is not what we want, we must take it that $\displaystyle A$ 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 $\displaystyle H_1,H_2$ be subgroups of $\displaystyle G$ and suppose that $\displaystyle H_2\not\subseteq H_1,$ i.e. $\displaystyle \exists\,h_0\in H_2$ with $\displaystyle h_0\notin H_1.$ Let $\displaystyle h\in H_1.$ As $\displaystyle G$ is Abelian, it is easily checked that $\displaystyle B=\left\{yz:y\in\left<h\right>,z\in\left<h_0\right >\right\}$ is a subgroup of $\displaystyle G$ containing both $\displaystyle \left<h\right>$ and $\displaystyle \left<h_0\right>.$ Since it is also finite, all its subgroups are comparable by the results established above; in particular, $\displaystyle \left<h\right>$ and $\displaystyle \left<h_0\right>$ are comparable. Since $\displaystyle h_0\notin \left<h\right>,$ we must have $\displaystyle \left<h\right>\subseteq\left<h_0\right>.$ Thus $\displaystyle h$ is in $\displaystyle \left<h_0\right>$ for any $\displaystyle h\in H_1.$ Hence $\displaystyle H_1\subseteq\left<h_0\right>\subseteq H_2$ proving that $\displaystyle H_1$ and $\displaystyle H_2$ are comparable.
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    Quote Originally Posted by TheAbstractionist View Post
    Hi NonCommAlg.

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

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

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

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

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

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

    The next thing is to show that every finite subgroup $\displaystyle A$ of $\displaystyle G$ is cyclic. It would then follow by the above that all subgroups of any finite subgroup of $\displaystyle G$ are comparable. Since $\displaystyle G$ is Abelian, so is $\displaystyle A.$ If $\displaystyle A$ is not cyclic, then, by the theory of finite Abelian groups, $\displaystyle A$ would be the internal direct product of cyclic subgroups $\displaystyle A_1,A_2,\ldots,A_s.$ This would imply that $\displaystyle A_i\cap A_j=\{1\}$ if $\displaystyle i\ne j,$ meaning that $\displaystyle x$ can’t be in all the $\displaystyle A_i$’s. As this is not what we want, we must take it that $\displaystyle A$ 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 $\displaystyle H_1,H_2$ be subgroups of $\displaystyle G$ and suppose that $\displaystyle H_2\not\subseteq H_1,$ i.e. $\displaystyle \exists\,h_0\in H_2$ with $\displaystyle h_0\notin H_1.$ Let $\displaystyle h\in H_1.$ As $\displaystyle G$ is Abelian, it is easily checked that $\displaystyle B=\left\{yz:y\in\left<h\right>,z\in\left<h_0\right >\right\}$ is a subgroup of $\displaystyle G$ containing both $\displaystyle \left<h\right>$ and $\displaystyle \left<h_0\right>.$ Since it is also finite, all its subgroups are comparable by the results established above; in particular, $\displaystyle \left<h\right>$ and $\displaystyle \left<h_0\right>$ are comparable. Since $\displaystyle h_0\notin \left<h\right>,$ we must have $\displaystyle \left<h\right>\subseteq\left<h_0\right>.$ Thus $\displaystyle h$ is in $\displaystyle \left<h_0\right>$ for any $\displaystyle h\in H_1.$ Hence $\displaystyle H_1\subseteq\left<h_0\right>\subseteq H_2$ proving that $\displaystyle H_1$ and $\displaystyle H_2$ 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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  4. #4
    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 $\displaystyle G$ is nontrivial forces $\displaystyle G$ to be a $\displaystyle p$-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 $\displaystyle G$ is nontrivial forces $\displaystyle G$ to be a $\displaystyle p$-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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