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- Thread starter charlamov
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So if you can't see a proof for the infinite case as I haven't, try constructing a counterexample.

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thanks, i finally have found that it is not true generally

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did you find the counterexample?

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mathwonk

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i guess the usual example is of free (non abelian) groups on different sets of generators.

as i recall, a figure eight knot has a fundamental group which is free on two generators. Then you construct a covering space which is a wedge of more than two loops.

The covering space has fundamental group free on more generators, but injects into the fundamental group of the base.

Just recalling a lecture by Eilenberg from 30-40 years ago. I'll check it out.

here's a reference in hatcher's free algebraic topology book, pages 57-61.

http://www.math.cornell.edu/~hatcher/AT/ATpage.html

as i recall, a figure eight knot has a fundamental group which is free on two generators. Then you construct a covering space which is a wedge of more than two loops.

The covering space has fundamental group free on more generators, but injects into the fundamental group of the base.

Just recalling a lecture by Eilenberg from 30-40 years ago. I'll check it out.

here's a reference in hatcher's free algebraic topology book, pages 57-61.

http://www.math.cornell.edu/~hatcher/AT/ATpage.html

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i guess the usual example is of free (non abelian) groups on different sets of generators.

as i recall, a figure eight knot has a fundamental group which is free on two generators. Then you construct a covering space which is a wedge of more than two loops.

The covering space has fundamental group free on more generators, but injects into the fundamental group of the base.

Just recalling a lecture by Eilenberg from 30-40 years ago. I'll check it out.

here's a reference in hatcher's free algebraic topology book, pages 57-61.

http://www.math.cornell.edu/~hatcher/AT/ATpage.html

Ah, the above reminded me: the free group [itex]\,F_2\,[/itex] on two generators contains as a subgroup the free group on any number of generators up to and including the free group on infinite countable generators (for example, the group's commutator subgroup [itex]\,(F_2)'=[F_2:F_2]\cong F_\infty\,[/itex]) , so we have injections [tex]F_2\to F_\infty\,\,,\,\,F_\infty\to F_2[/tex] but the two groups are clearly non-isomorphic.

DonAntonio

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