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Abstract Algebra: Abelian group order

  1. May 12, 2014 #1
    1. The problem statement, all variables and given/known data

    Let G be an abelian group and let x, y be elements in G. Suppose that x and y are of finite order. Show that xy is of finite order and that, in fact, o(xy) divides o(x)o(y). Assume in addition that (o(x),(o(y)) = 1. Prove that o(xy) = o(x)o(y).


    3. The attempt at a solution

    I was able to prove the first part, that xy is of finite order and that o(xy) divides o(x)o(y). I'm having trouble with the second part, proving that o(xy) = o(x)o(y) if the greatest common factor of o(x) and o(y) is 1.
     
  2. jcsd
  3. May 12, 2014 #2

    jbunniii

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    If ##(xy)^k = 1##, then ##x^k## and ##y^k## are inverses of each other. Can this happen if ##x^k## and ##y^k## are not the identity element, and ##((o(x),o(y)) = 1##?
     
    Last edited: May 12, 2014
  4. May 12, 2014 #3
    So since (xy)^(o(xy)) = e, x^(o(xy)) and y^(o(xy)) are inverses of each other. We also know x^(o(xy)) = e because it's the same as x^(o(x))^(o(y), and the same goes for y^(o(xy)) = e. I feel like I'm really close but I'm still not grasping why it has to be the same.
     
  5. May 12, 2014 #4

    jbunniii

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    The question is to find out whether ##(xy)^k = e## is possible where ##k## is a proper divisor of ##o(x)o(y)##. We want to show that this is impossible if ##(o(x),o(y)) = 1##.

    So suppose ##(xy)^k = e##. Since the group is abelian, this means ##x^k y^k = e##, and so ##x^k## and ##y^k## are inverses of each other. I claim that this implies that ##x^k = y^k = e##. Therefore, assuming this claim is true, ##k## must be a multiple of both ##o(x)## and ##o(y)## and therefore a multiple of ##o(x)o(y)## since ##(o(x),o(y)) = 1##.

    So the key is to prove the claim. This brings us back to the question I asked in my first post. Can ##x^k## and ##y^k## be inverses of each other if they are not the identity?

    Hint: what can you say about the intersection of ##\langle x\rangle## and ##\langle y \rangle##, the subgroups generated by ##x## and ##y## respectively?
     
    Last edited: May 12, 2014
  6. May 18, 2014 #5
    Ok, I've spent a lot of time struggling with this one. I don't know why it's so hard for me! I understand completely how if ##x^k = y^k = e##, then the desired result that ##o(xy) = o(x)o(y)## is true. So, to prove your claim. I think this works, please let me know if it does not:

    ##x^k## and ##y^k## are inverses of each other. An element must have the same order as its inverse, so ##o(x^k) = o(y^k)##. But ##o(x^k) = o(x)/(k,o(x))##, and ##o(y^k) = o(y)/(k,o(y))##, so ##o(x)/(k,o(x)) = o(y)/(k,o(y)) = p##. If ##p > 1##, both ##o(x)## and ##o(y)## are divisible by ##p##. But since ##(o(x),o(y)) = 1##, ##p = 1##, so both sides of the equation must equal ##1##, which means ##x^k = y^k = e##.

    Does this proof make sense? And even if it does, is there a more elegant way to prove this? Thank you so much for your help!
     
  7. May 18, 2014 #6

    jbunniii

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    I think your proof is OK, but the argument I had in mind may be simpler because it doesn't require the formula involving the gcd: ##x^k y^k = e##, so ##x^k## and ##y^k## are inverses of each other. Now ##\langle x\rangle## must contain both ##x^k## and its inverse ##y^k##, and similarly, ##\langle y \rangle## must contain both ##x^k## and ##y^k##. This means that ##x^k## and ##y^k## are both elements of ##\langle x \rangle \cap \langle y \rangle##. Since ##\langle x \rangle \cap \langle y \rangle## is a subgroup of both ##\langle x \rangle## and ##\langle y \rangle##, its order must divide ##o(x)## and ##o(y)##. But ##o(x)## and ##o(y)## are relatively prime, so this forces ##|\langle x \rangle \cap \langle y \rangle| = 1## and therefore ##\langle x \rangle \cap \langle y \rangle = \{e\}##. We conclude that ##x^k = y^k = e##.
     
    Last edited: May 18, 2014
  8. May 19, 2014 #7

    micromass

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    Follow-up question: Is the same true in non-abelian groups

    Generalization of the question: Prove that if in a group ##G## the elements ##x## and ##y## commute, then
    [tex]o(xy)~\text{divides}~\textrm{lcm}(o(x),o(y)) = \frac{o(xy)}{\textrm{gcd}(o(x),o(y))}[/tex]
    and
    [tex]\frac{o(xy)}{\textrm{gcd}(o(x),o(y))^2}~\text{divides}~o(xy)[/tex]

    Find a counterexample where ##o(xy) \neq \textrm{lcm}(o(x),o(y))## even if they ##x## and ##y## commute.
     
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