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I gave an example here of a topological proof that a pairproduct of two commutators in a free group is not itself always a commutator. In answer to the same question, Arturo Magidin indicated how to give a combinatorial proof. I think it's fair to say that the two proofs have completely different flavours, although you can judge for yourself which is 'simpler'.

I gave an example here of a topological proof that a pair of commutators in a free group is not itself always a commutator. In answer to the same question, Arturo Magidin indicated how to give a combinatorial proof. I think it's fair to say that the two proofs have completely different flavours, although you can judge for yourself which is 'simpler'.

I gave an example here of a topological proof that a product of two commutators in a free group is not itself always a commutator. In answer to the same question, Arturo Magidin indicated how to give a combinatorial proof. I think it's fair to say that the two proofs have completely different flavours, although you can judge for yourself which is 'simpler'.

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I gave an example herehere of a topological proof that a pair of commutators in a free group is not itself always a commutator. In answer to the same question, Arturo Magidin indicated how to give a combinatorial proof. I think it's fair to say that the two proofs have completely different flavours, although you can judge for yourself which is 'simpler'.

I gave an example here of a topological proof that a pair of commutators in a free group is not itself always a commutator. In answer to the same question, Arturo Magidin indicated how to give a combinatorial proof. I think it's fair to say that the two proofs have completely different flavours, although you can judge for yourself which is 'simpler'.

I gave an example here of a topological proof that a pair of commutators in a free group is not itself always a commutator. In answer to the same question, Arturo Magidin indicated how to give a combinatorial proof. I think it's fair to say that the two proofs have completely different flavours, although you can judge for yourself which is 'simpler'.

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I gave an example here of a topological proof that a pair of commutators in a free group is not itself always a commutator. In answer to the same question, Arturo Magidin indicated how to give a combinatorial proof. I think it's fair to say that the two proofs have completely different flavours, although you can judge for yourself which is 'simpler'.