Re: group problem

Li Yi wrote: 
So how to prove it?

Take c = a^d1 * b^d2. Then c^[n,m] = 1, i.e. o(c) | [n,m], no matter what d1 and d2 are. Now we choose d1 and d2 in order to get o(c) = [n,m]. For that purpose, we look at prime factor decomposition. First, we look at the case n and m are prime powers. Then (1) n | m and we can take d1 = d and d2 = 1 or (2) m | n and n > m and we can take d1 = 1 and d2 = d. In the general case, we choose d1 * d2 = d as well, and do the same as for the prime power case, but now for every prime separately. So we take

d1 = prod_{p prime | 0 < v_p(n) <= v_p(m)} p^{v_p(n)}

and

d2 = prod_{p prime | v_p(n) > v_p(m)} p^{v_p(m)}

where v_p(n) is how often n is divisible by prime p. Now o(a^d1) = n/d1 and o(b^d2) = m/d2, and (n/d1, m/d2) = 1. Now apply the case (n, m) = 1 of the book.

But as I said, it is more interesting to have a proof without the word "prime" in it (or to make it conceivable that no such proof exists).
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