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In a Wikipedia article on the Commutator Subgroup, the following is stated:

"The commutator subgroup can also be defined as the set of elements $g$ of the group which have an expression as a product $g = g_1 g_2 \cdots g_k$ that can be rearranged to give the identity."

I was wondering if there was a reference for this (that contains a proof) or if someone could provide a proof since I thought that this was a neat property.

I spent some time searching and did not come up with anything. The only part of the proof that I was able to do myself was the trivial fact that every element of the commutator subgroup has this property, but there may be other elements that have this property as well.

Michael Hardy
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user357980
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    You can rearrange the product on the right by introducing commutators. By the time you've finished the part that isn't a commutator reduces to the identity and you have expressed $g$ as a product of commutators, as required. – Mark Bennet Aug 26 '17 at 21:05
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    Do you have an algorithm for going about that? – user357980 Aug 27 '17 at 04:24

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Let $G$ be your group. Any product $g_1\ldots g_n$ that can be rearranged to give $1$ must map to $1$ under any homomorphism from $G$ to an abelian group. If $[G, G]$ is the commutator subgroup, the projection $p : G \to G/[G, G]$ is a homomorphism from $G$ to an abelian group and $\ker(p) = [G, G]$.

Rob Arthan
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