1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Problem concerning cyclic groups.

  1. Jun 12, 2012 #1
    The question states:
    "Let G be a group and let Gn={gn|g ε G}. Under what hypothesis about G can we show that Gn is a subgroup of G?


    The set Gn is taking each element of G and raising it to a fixed number. I started my investigation by examining what happens if I take n=3 and considering the groups Z4 and the Klein 4-group. I noticed that Gn did not create a subgroup with the integers modulo 4, but it did create one with the Klein 4-group. Thus, I believe that in order for Gn be a subgroup, we need the condition that G must NOT be cyclic.

    That being said, I have a proof, which seems to work, but nowhere did I use the fact that G is not cyclic. I have an outline below.

    Thrm: If G is a group and is not cyclic, then Gn={gn|g ε G} is a subgroup.

    Proof (SKETCH): Let G be a group and suppose G is not cyclic. We wish to show Gn={gn|g ε G} is a subgroup.
    *Associativity*
    ... inherited from G
    *Identity*
    ... The identity element e from G is in Gn, and e raised to some fixed number n is still e.
    *Inverse*
    ... a and a-1 are in G... an ε Gn .. a-1n= a-n ε Gn ... ana-n=a-nan=e


    The above is just an outline. But, even with all the details, I never use the fact that G is not cyclic. I must be doing something wrong. Please help! Thanks
     
  2. jcsd
  3. Jun 12, 2012 #2
    ^ Forgot to show closure. Let a,b ε Gn. Then a=g1n and b=g2n for some g1,g2εG. Then,
    ab = g1ng2n=(g1g2)n. Hence ab ε Gn
     
  4. Jun 12, 2012 #3

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    I think there's some confusion here. If you mean Z4 to be the integers mod 4 that's not a group under multiplication. You want the group action to be multiplication. A cyclic group of order 4 would be {e,a,a^2,a^3} with the relation a^4=e.
     
    Last edited: Jun 12, 2012
  5. Jun 12, 2012 #4
    I was completely wrong when I went about investigating Z4. Ignore that completely....


    I was also completely wrong about that G having to be NONCYCLIC. It can be cyclic. G must be ABELIAN. And, I did end up using this fact ... Right here ..
    ab = g1ng2n=(g1g2)n

    If G wasn't commutative, then I wouldn't be able to show that Gn is closed.
     
  6. Jun 12, 2012 #5

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    That sounds better.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Problem concerning cyclic groups.
Loading...