Thread: 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:

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Let be an abelian group and Suppose that every subgroup of contains Prove that every two subgroups of are comparable, i.e. if are two subgroups

of then either or

Originally Posted by NonCommAlg

Let be an abelian group and Suppose that every subgroup of contains Prove that every two subgroups of are comparable, i.e. if are two subgroups

of then either or

Hi NonCommAlg.

We have either or else for some In either case, has finite order, say .

Indeed must be prime. For suppose for some Then for some contrary to the minimality of

Now let be any nontrivial element of Then for some Hence has finite order too.

Since divides the order of by Lagrange’s theorem, so the order of is of the form for some positive integers with coprime with If then would be divisible by a prime and then would have a Sylow -subgroup which cannot possibly contain , an element of order Hence and so every nontrivial element of has order a power of

Note that if is any finite subgroup of of order then, since we must have Hence is the unique finite subgroup of of order

So let have order Assume that Then has a totally ordered chain of subgroups containing every one of its subgroups. (Indeed this is a composition series in Jordan–Hölder theory.) (Note that Of course if then and so all subgroups of (namely and {1}) are comparable as well. Hence every finite cylic subgroup of has subgroups that are all comparable.

The next thing is to show that every finite subgroup of is cyclic. It would then follow by the above that all subgroups of any finite subgroup of are comparable. Since is Abelian, so is If is not cyclic, then, by the theory of finite Abelian groups, would be the internal direct product of cyclic subgroups This would imply that if meaning that can’t be in all the ’s. As this is not what we want, we must take it that 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 be subgroups of and suppose that i.e. with Let As is Abelian, it is easily checked that is a subgroup of containing both and Since it is also finite, all its subgroups are comparable by the results established above; in particular, and are comparable. Since we must have Thus is in for any Hence proving that and are comparable.

Originally Posted by TheAbstractionist

Hi NonCommAlg.

We have either or else for some In either case, has finite order, say .

Indeed must be prime. For suppose for some Then for some contrary to the minimality of

Now let be any nontrivial element of Then for some Hence has finite order too.

Since divides the order of by Lagrange’s theorem, so the order of is of the form for some positive integers with coprime with If then would be divisible by a prime and then would have a Sylow -subgroup which cannot possibly contain , an element of order Hence and so every nontrivial element of has order a power of

Note that if is any finite subgroup of of order then, since we must have Hence is the unique finite subgroup of of order

So let have order Assume that Then has a totally ordered chain of subgroups containing every one of its subgroups. (Indeed this is a composition series in Jordan–Hölder theory.) (Note that Of course if then and so all subgroups of (namely and {1}) are comparable as well. Hence every finite cylic subgroup of has subgroups that are all comparable.

The next thing is to show that every finite subgroup of is cyclic. It would then follow by the above that all subgroups of any finite subgroup of are comparable. Since is Abelian, so is If is not cyclic, then, by the theory of finite Abelian groups, would be the internal direct product of cyclic subgroups This would imply that if meaning that can’t be in all the ’s. As this is not what we want, we must take it that 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 be subgroups of and suppose that i.e. with Let As is Abelian, it is easily checked that is a subgroup of containing both and Since it is also finite, all its subgroups are comparable by the results established above; in particular, and are comparable. Since we must have Thus is in for any Hence proving that and 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.

Originally Posted by NonCommAlg

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 is nontrivial forces to be a -group. I suppose this is the sort of problem the Hungarian mathematician Pál Erdős would have loved.

Originally Posted by TheAbstractionist

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 is nontrivial forces to be a -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?