Does o(ab)=lcm(o(a),o(b)) for Group G Elements a & b?

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In a group G, the equation o(ab) = lcm(o(a), o(b)) holds true when elements a and b commute. However, when a and b do not commute, this relationship does not hold. A counterexample can be found in the symmetric group S3, where specific elements a and b have orders o(a) = 2 and o(b) = 3, resulting in o(ab) = 2. Thus, the statement is valid only under the condition of commutativity. Understanding these conditions is crucial for analyzing group element orders.
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Suppose G is a group, a,b are two elements of G, does o(ab)=lcm(o(a),o(b))?
o(ab) denotes the order of ab, lcm(o(a),o(b)) denotes the Least Common Multiple of o(a) & o(b).
 
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It is true when a and b commute, and can easily be proven in that case.
When a and b do not commute, the statement is false. For a counterexample, you could try to find some elements in S3 (e.g. there are elements a and b such that o(a) = 2, o(b) = 3, o(a, b) = 2).
 
I am studying the mathematical formalism behind non-commutative geometry approach to quantum gravity. I was reading about Hopf algebras and their Drinfeld twist with a specific example of the Moyal-Weyl twist defined as F=exp(-iλ/2θ^(μν)∂_μ⊗∂_ν) where λ is a constant parametar and θ antisymmetric constant tensor. {∂_μ} is the basis of the tangent vector space over the underlying spacetime Now, from my understanding the enveloping algebra which appears in the definition of the Hopf algebra...

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