Modern Abstract
Posted in Uncategorized on 09/13/2008 06:52 pm by adminModern Abstract
Modern/Abstract Algebra: Let a and b be elements of a group G. Show that if ab has finite order n, then ba?
Modern/Abstract Algebra:
Let a and b be elements of a group G. Show that if ab has finite order n, then ba also has order n.
Please help me!!!
First we prove an identity -- for any positive integer n, (ab)^(n+1) = a (ba)^n b. We proceed by induction: when n=1, we have (ab)^(1+1) = abab = a (ba)^1 b, so this holds when n=1. Suppose it holds for some n. Then (ab)^(n+2) = ab(ab)^(n+1) = aba (ba)^n b = a(ba)(ba)^n b = a (ba)^(n+1) b, so it holds for n+1. By induction, it holds for all n.
Now, suppose ab has finite order n. Then (ab)^(n+1) = ab(ab)^n = abe = ab. However, by the formula above, (ab)^(n+1) = a (ba)^n b. Thus, ab = a (ba)^n b, so e = a⁻¹abb⁻¹ = a⁻¹a (ba)^n bb⁻¹ = (ba)^n, which means |ba| ≤ n. But if |ba|
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