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Proving H is a subgroup of G, given

  1. Nov 27, 2011 #1
    1. The problem statement, all variables and given/known data

    Let G be an abelian group, k a fixed positive integer, and H = {a is an element of G; |a| divides k}. Prove that H is a subgroup of G.

    2. Relevant equations

    Definition of groups, subgroups, and general knowledge of division algorithm.

    3. The attempt at a solution

    I know that to prove H is a subgroup of G, I need to:
    1) prove that it is closed (so if a and b are in H, ab is in H)
    2) prove that if a is in H, then a-1 is in H as well.

    I'm somewhat confused on how to proceed. To prove the first part, I suppose that ab is an element of H. Then we can write abn = k for some integer n. Because G is abelian, we can use associativity and commutativity to write (a)(bn) = k and (b)(an) = k. This clearly shows that a divides k and b divides k.

    The problem is that I'm not sure if I went the wrong direction - meaning I supposed ab was in H, and proved a and b are in H.

    I'm also at a loss on how to approach the second part. It's driving me mad because this is an "easy" problem and it's the one that's stumping me on this problem set.
     
  2. jcsd
  3. Nov 27, 2011 #2

    micromass

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    You also need to prove

    3) e is in H

    Yes, you went in the wrong direction. You need to start from a,b in H. So you start from knowing that ak=e and bk=e. And you need to prove (ab)k=e.

    Here you need to prove that if a in H, then a-1 is in H. So if ak=e, then (a-1)k=e.
     
  4. Nov 27, 2011 #3
    I'm not sure I follow; the problem states that |a| divides k, not that k is the order of any element in the subgroup. So how do we know that ak = e for any a in H?
     
  5. Nov 27, 2011 #4
    Since the order of any element a divides k, let the order of some element a be n. then, n times some integer, say, j, equals k. So, ak=anj=(an)j....see how this equals e?
     
  6. Nov 27, 2011 #5
    I'm a little ashamed to say this, but I was reading |a| as absolute value of a, not order of a. This makes perfect sense then, thank you both for your help!
     
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