Re: group problem

Li Yi wrote:

Let G be an abelian group and suppose that G has elements of orders
m and n, respectively. Prove that G has an element whose order is
the least common mutiple of m and n.

The following proof is presented in 'problems from herstein'.
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Proof. Let o(a) = m and o(b) = n. If (m, n) = 1, then let c = ab.
Clearly, c^[m,n] = c^mn = a^mn b^mn = e. If o(c) = r, we have a^r =
b^(-r). Thus, e = a^mr = b^(-mr). But then n|mr. Since (m, n) = 1,
n|r. Similarly, m|r. This implies that [m, n] = mn|r and thus mn = r.

If (m, n) = d > 1, then o(a^d) = m/d and o(b^d) = n/d and (m/d, n/d) =
1. Thus (ab)^d has order mn/d^2 by previous case and thus ab has order
[m, n] = mn/d.
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I wonder the last step. Why does ab have order mn/d if (ab)^d has the
order of mn/d^2?

If o(a^2)=5, o(a) is not necessary to be 10. It can be 5, too.

Hmm, I am not an expert in group theory, but what you say sound right to me. You take b = a^(-1) with o(a) = o(b) = m = n = d, and o(ab) = 1 < d if d > 2.

If n | m, then you can take a^d b. This or m | n is trivially the case when n and m are prime powers, and in the general case one can take a^d1 b^d1 with d1*d2 = d, where d1 and d2 depend on the prime decompositions of n and m. But that you probably knew already. More interesting is to find an argument without the word "prime" in it.
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