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Homework Help: 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

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    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

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    That sounds better.
     
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