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Order of elements in finite abelian groups

  • Thread starter jacobrhcp
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prove that if G is a finite and abelian group and m is the least common multiple of the order of it's element, that there is an element of order m.

My idea:

if ai are the elements of G, the order of a1*a2 is lcm(a1,a2) and the result follows directly when applied to all ai... but why is this correct and why is this only for abelian groups?
 
Last edited:

morphism

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The order of a1*a2 is not lcm(o(a1),o(a2)), e.g. take a nonidentity element and its inverse: the order of their product is 1, but the lcm of their orders is >1.

And yes, abelian is necessary here. (Try to find an example of a finite nonabelian group in which this is not true.)
 
169
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I did that, that was the next question in the book =P,
D3 is a finite nonabelian group, in which the elements have order 1,2, or 3. The least common multiple of these is 6 and the result is not true.
 

morphism

Science Advisor
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Yup, that works. For the original problem try looking at the structure theorem for finite abelian groups.
 

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