Suppose that a and b are group elements that commute. If |a| is
finite and |b| infinite, prove that |ab| has infinite order.
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Answer:
Suppose O(a) and O(b) are finite and also O(ab) and O(ba) are finite. Then lcm(|a|,|b|)=lcm(|b|,|a|). (Is that correct?)
Suppose O(a) and O(b) is finite but O(ab) is infinite, how to prove that O(ab)=O(ba)?
If not give elements a,b in a group G, such that O(ab) is infinite but O(ba) is finite.
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asked
Sep 29 '14 at 20:38
Rising Star
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Sep 11 '18 at 8:39
Jendrik Stelzner
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5 Answers
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(1) Conjugate elements have the same order, i.e. o(a)=o(g−1ag)
(2) ab=b−1(ba)b
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answered
Sep 29 '14 at 20:42
Timbuc
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Just to be clear, the order of ab isn't necessarily the LCM of the orders of a and b. For example, consider the cycles a=(1,2) and b=(1,2,3) in S3: the LCM of their orders is 6, but ab=(1,3) (or (2,3) if right-to-left) has order 2.
However order is invariant under isomorphisms. More precisely, if ϕ:G→H is a group isomorphism and g∈G, the order |⟨g⟩| of g in G is the same as |⟨ϕ(g)⟩|, the order of ϕ(g) in H.
In particular, conjugation by b is an automorphism of G: define ϕ:G→G by ϕ(g)=b−1gb. Then ab=ϕ(ba) so the two have the same order (whether finite or infinite).