Product of Two Commutators as a Square in a Free Group
Canadian mathematical bulletin, Tome 33 (1990) no. 2, pp. 190-196
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We show that, if [s,t][u, v] = x2 in a free group, x need not be a commutator. We arrive at our example by use of a result of D. Piollet which characterizes solutions of such equations using an algebraic interpretation of the mapping class group of the corresponding surface.
Comerford, Jonell A.; Lee, Y. Product of Two Commutators as a Square in a Free Group. Canadian mathematical bulletin, Tome 33 (1990) no. 2, pp. 190-196. doi: 10.4153/CMB-1990-032-4
@article{10_4153_CMB_1990_032_4,
author = {Comerford, Jonell A. and Lee, Y.},
title = {Product of {Two} {Commutators} as a {Square} in a {Free} {Group}},
journal = {Canadian mathematical bulletin},
pages = {190--196},
year = {1990},
volume = {33},
number = {2},
doi = {10.4153/CMB-1990-032-4},
url = {http://geodesic.mathdoc.fr/articles/10.4153/CMB-1990-032-4/}
}
TY - JOUR AU - Comerford, Jonell A. AU - Lee, Y. TI - Product of Two Commutators as a Square in a Free Group JO - Canadian mathematical bulletin PY - 1990 SP - 190 EP - 196 VL - 33 IS - 2 UR - http://geodesic.mathdoc.fr/articles/10.4153/CMB-1990-032-4/ DO - 10.4153/CMB-1990-032-4 ID - 10_4153_CMB_1990_032_4 ER -
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