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Proof that a certain group G contains a cyclic subgroup of order rs

  1. Sep 15, 2011 #1
    1. The problem statement, all variables and given/known data
    Let G be an abelian group and let H and K be finite cyclic subgroups with |H|=r and |K|=s.

    Show that if r and s are relatively prime, then G contains a cyclic subgroup of order rs


    2. Relevant equations

    Fundemental theorem of cyclic group which states that the order of any cyclic subgroup of some cyclic group G must be the divisor of the number of elements, n, in G.

    3. The attempt at a solution

    Well the big trouble I'm having here (i think) is that I can't apply the fundamental theorem of cyclic group since I don't know if G is cyclic. I just know that it's abelian. The problem doesn't even state if G is finite (problem 6-56 of fraleigh).

    I know that every cyclic group is abelian. But I also know that the converse is not true in general. Is there any way for this problem to deduce if G is cyclic? If I knew G was cyclic then this problem is (almost) trivial. Or am I barking up the wrong tree?
     
  2. jcsd
  3. Sep 15, 2011 #2

    micromass

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    You don't need to know that G is cyclic. In general, G won't be cyclic. You'll need to show that there is a cyclic subgroup of a certain order.

    You know that H is cyclic of order r, and K is cyclic of order s. What can you say about HK??
     
  4. Sep 15, 2011 #3
    To be honest, I really don't know. The author has not introduced group products as of yet. But after looking it up i can see that |HK| = |H||K|/|H and K|. I know that finite cycling groups of order o are isomorphic to the congruence group mod o. So |HK| = r*s/(r-s), given that r>s. (this is wrong and I realise that). Am I atleast on the right path? :)
     
  5. Sep 15, 2011 #4

    micromass

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    Yes, you are on the right path, but

    [tex]|H\cap K|\neq r-s[/tex]

    I claim that [itex]H\cap K=\{e\}[/itex]. Indeed, let g be in H and in K. What must the order of g be?? Can you come to a contradiction?
     
  6. Sep 15, 2011 #5
    If g is in both H and K then the order of g must be a divisor of r and at the same time a divisor of s. But since gcd(r,s) = 1 then g must be of order 1 and is the identity element.

    Is this correct? :)
     
  7. Sep 15, 2011 #6

    micromass

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    Yes!!! This is absolutely correct!!

    So that proves that |HK|={e}, right??

    The only thing you need to prove is that HK is cyclic. Take h a generator of H and k a generator of k. Maybe try to prove that hk is a generator of HK?
     
  8. Sep 15, 2011 #7
    Alright fantastic! Thank you so much for your help.

    I do have a couple of questions:

    1) (I realise that this may come up when i try to find a generator). I'm a bit worried that we haven't at all used the fact that G is abelian. Is it even necessary that G is abelian?

    2) Are there any alternative solutions to this problem? The reason I'm asking is that I had to use concepts that are not yet introduced (i.e. the order of the group product HK). Or is it common that I have to use concepts that will appear in future chapters in a course like this one? This is a first course in abstract algebra for me so the methods of solving problems are quite new to me.
     
  9. Sep 15, 2011 #8

    micromass

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    Well, I forgot that actually. You still need to check that HK is a subgroup of G. This will use abelianness of G.
    That HK is cyclic will also use that G is abelian.

    Yes, that worries me as well. Are you learning from a book?? Maybe I can check the book to see what they had in mind.

    It's usual that problems in abstract algebra have multiple solutions. And the more abstract algebra you know, the easier you will find it to solve something.
     
  10. Sep 15, 2011 #9
    I'm using the book "A First Course in Abstract Algebra by John B. Fraleigh (7th ed). The problem is the last of section 6.

    Yeah, If I had known to check the order of the subgroup product (or that such a thing could be calculated in that way) I probably wouldn't have gotten stuck.
     
  11. Sep 15, 2011 #10

    micromass

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    OK, here's a way to do it without using products of groups:

    Let h be a generator of H, let k be a generator of K. Show that hk has order rs. This shows that hk generates a group of order rs. This group is cyclic.
     
  12. Sep 15, 2011 #11
    I will ponder this while I try to sleep tonight!

    Thank you so much for your help micromass, I sincerely appreciate it!
     
  13. Sep 16, 2011 #12
    So if i assume that h gen H and k gen K. If i form the group HK generated by hk and form the subgroups of HK generated by h and k respectively theb their order must divide the order of HK since HK is cyclic. We then have by the fundamental theorem that the order of HK is rs since r and s are rel prime. Is it correct to reson like this? How can i know that hk generates a cyclic group?
     
  14. Sep 16, 2011 #13

    micromass

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    No, I think you misunderstood me. In the new method, we don't say anything about HK, so forget that.

    We just pick h in H and k in K, and we form the group generated by hk (this group will equal HK, but forget that). This group is cyclic because it is generated by one element (that is: hk). You only need to find the order of this group. That is, you must find the order of hk.
     
  15. Sep 16, 2011 #14
    Alright. So if we assume that r > s we know that HK contains h^s for the integer s. We can then form a cyclic subgroup generated by h^s and we can conclude that r divides |HK|. Am i on the right track?
     
    Last edited: Sep 16, 2011
  16. Sep 16, 2011 #15

    micromass

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    No, we don't do anything with HK here. We're working with the group generated by hk. Forget about HK.
     
  17. Sep 16, 2011 #16
    Im afraid I'm lost. How am I supposed to count the elements of the group generated by hk. Is there any general trick or hint that could be used here?
     
  18. Sep 16, 2011 #17

    micromass

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    Prove that [itex](hk)^{rs}=e[/itex] and that no smaller j>0 satisfies [itex](hk)^j=e[/itex].
     
  19. Sep 16, 2011 #18
    Eureka. Thank you so much! If I'll be back if i run into trouble but i think i should be able to handle it now :)
     
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