Cantor Bernstein Property for Groups
Merit students
if, for any two groups A and B in the class, if A is embeddable in B, and
B is embeddable in A, then A and B are isomorphic.
Does the class of all finitely generated groups possess the Cantor Bernstein
Property?
Any ideas?
Allen Knutson
For the uninitiate: The Cantor-Bernstein theorem is the statement that bare
sets (not groups) have the Cantor-Bernstein property. It is easily proved
with the Axiom of Choice, but does not depend on it.
To your question, though: No. Take F_2 and F_3, the free groups on {a,b}
and {x,y,z} respectively. F_2 embeds in F_3 in an obvious way, a->x, b->y.
What's less obvious is a reverse embedding (which I will not specify, for
fear that this is a homework problem). I will give a hint: One can even embed
F_{countable infinity} in F_2!
F_2 and F_3 are clearly not isomorphic: mod out the commutator subgroups,
then all squares. The images are then (Z_2)^2 and (Z_2)^3.
Alternate question: is it true for finitely generated abelian? I say yes.
For countably generated abelian? No (Z_2 x Z x Z x... and Z_3 x Z x Z x ...).
Dr D F Holt
>me...@sun-4.cs.uct.ac.za (Merit students) writes:
>
>>A class of groups is said to have the Cantor Bernstein Property
>>if, for any two groups A and B in the class, if A is embeddable in B, and
>>B is embeddable in A, then A and B are isomorphic.
>
>>Does the class of all finitely generated groups possess the Cantor Bernstein
>>Property?
>
>and {x,y,z} respectively. F_2 embeds in F_3 in an obvious way, a->x, b->y.
>What's less obvious is a reverse embedding (which I will not specify, for
>fear that this is a homework problem). I will give a hint: One can even embed
>F_{countable infinity} in F_2!
>
>F_2 and F_3 are clearly not isomorphic: mod out the commutator subgroups,
>then all squares. The images are then (Z_2)^2 and (Z_2)^3.
>
An example in which the groups embed in each other with finite index is:
G1 = < x,y | x^11=1, y^-1 x y = x^4 > and
G2 = < x,y | x^11=1, y^-1 x y = x^5 >.
Then G1 embeds in G2 with index 3 by x->x, y->y^3
and G2 embeds in G1 with index 2 by x->x y->y^2,
but they are not isomorphic (exercise).
>Alternate question: is it true for finitely generated abelian? I say yes.
>For countably generated abelian? No (Z_2 x Z x Z x... and Z_3 x Z x Z x ...).
I don't find that last example convincing. How do you embed Z_2 in
Z_3 x Z x Z x .... ?
Charles Yeomans
I have an idea or two. Try answering the question first for some
simpler subclasses of f.g. groups. For f.g. abelian groups, finding
an answer should be straightforward. What about finite groups in
general? The answer is pretty easy there. Now consider free groups.
Verification is trivial when one of the groups is (isomorphic to) the
free group on one generator. Can you embed the free group on three
generators into the free group on two generators? If you can answer this I
think you might be able to answer the question in general.
Shouldn't there be a "-" between "Cantor" and "Bernstein"?
Charles Yeomans
yeomans@austin@onu.edu
cyeo...@ms.uky.edu
P.S. Please post your answer.
Allen Knutson
>In article <allenk.687726273@irk> all...@irk.ugcs.caltech.edu (Allen Knutson) writes:
Blah blah blah about the Cantor-Bernstein property for groups, and is it true
>>For countably generated abelian? No (Z_2 x Z x Z x... and Z_3 x Z x Z x ...).
>I don't find that last example convincing. How do you embed Z_2 in
>Z_3 x Z x Z x .... ?
Oops!
You're right, of course. This is rather an example of SURjections both ways.
For INjections both ways, replace all those things by their character groups,
so Z_2 and Z_3 are replaced by Z_2 and Z_3 (no change), but Z is replaced
by U(1), the group of rotations of a circle. Sorry! Allen K