how a free group of rank greater than or equal to 3 contains every free group of countable ranks as a sugroup?
$\begingroup$
$\endgroup$
2
-
18$\begingroup$ math.stackexchange would have been more appropriate for this question. $\endgroup$– Alain ValetteCommented Sep 8, 2011 at 19:06
-
$\begingroup$ Given that so many people agree with Alain, why has no one else voted to close this question? $\endgroup$– HJRWCommented Sep 13, 2011 at 14:03
Add a comment
|
4 Answers
$\begingroup$
$\endgroup$
4
Actually a free group of rank 2 contains a free group of countable rank as a subgroup, namely the commutator subgroup. The easiest way to see this is to view the free group as the fundamental group of a wedge of 2 circles. The covering space associated to the commutator subgroup is the Cayley graph of ZxZ, or if you like, the grid formed by the integer lattice in R^2. Clearly the fundamental group of this covering space is free of countable rank since there are countably infinitely many edges off a spanning tree. You can take the x-axis and all vertical lines as the spanning tree.
answered Sep 8, 2011 at 16:53
-
$\begingroup$ Great answer! Is there any trick to finding the covering space associated to subgroups? I can verify what you said is indeed true, but how to come up with it? $\endgroup$ Commented Feb 10, 2020 at 7:51
-
$\begingroup$ @HritRoy You get a graph with vertices the cosets of the subgroup H and edges Hg->Hgx for each free generator x. $\endgroup$ Commented Feb 10, 2020 at 12:31
-
$\begingroup$ and it's fundamental group is always H? Is it true even if G is not free? $\endgroup$ Commented Feb 12, 2020 at 9:18
-
$\begingroup$ @HritRoy, yes it is H. For general groups you need 2-cells for the relations $\endgroup$ Commented Feb 12, 2020 at 12:19
$\begingroup$
$\endgroup$
In case an explicit description of such a subgroup would help, let $F_2 = \langle a,b|\rangle$ be the free group on 2 generators. Then the subgroup generated by the elements $\{a^nb^n\}_{n\in\mathbb{N}}$ is free since it is a subgroup of a free group. Next, it is easy to check that the generator $a^kb^k$ is not an element of the free group generated by the $a^\ell b^\ell$ for $\ell\neq k$ so one concludes that this subgroup has infinite rank.
$\begingroup$
$\endgroup$
8
Even better, no normal subgroup of infinite index of a group of cohomological dimension at most two is finitely presented (for free groups, it must be free, so countable rank).
EDIT Bieri's theorem does state:
If $G$ is a group of cohomological dimension of at most two, while $N$ is a normal subgroup of G of infinite index, then either $N$ is free, or $N$ is not finitely presentable. Sigh. I was thinking of $G$ a free or a surface group, in which case my statement is correct (it is a theorem of Jaco (easy mod a not so easy theorem of Whitehead) that a subgroup of a surface group is free if and only if it is of infinite index...
answered Sep 8, 2011 at 18:01
-
$\begingroup$ This isn't quite right: $F\times\mathbb{Z}$ is a counterexample! I think Bieri proved that no non-free normal subgroup of infinite index of a group of cohomological dimension two is finitely presented. $\endgroup$– HJRWCommented Sep 8, 2011 at 19:58
-
1$\begingroup$ $\mathbb{Z} \times \mathbb{Z}$ is even better. ;) $\endgroup$ Commented Sep 8, 2011 at 20:00
-
-
$\begingroup$ In any event, it is trivial to prove that a finitely generated subgroup of a free group that is normal must have finite index. The covering space of a wedge of circles associated to a finitely generated subgroup of infinite index has a finite core (the Stallings graph) with a bunch of infinite trees attached to it. The covering space associated to a normal subgroup is a Cayley graph of the quotient group and looks the same at each vertex. Clearly, these two situations cannot hold at the same time. $\endgroup$ Commented Sep 8, 2011 at 21:58
-
1$\begingroup$ Benjamin - I never actually look at the proofs in Lyndon and Schupp. $\endgroup$– HJRWCommented Sep 13, 2011 at 14:02
$\begingroup$
$\endgroup$
There is a very nice theorem which states:
Thm: If $F$ is a free group on $n$ generators and $H$ is a normal subgroup of index $j$ then if both $n$ and $j$ are finite $H$ is a free group on $j(n-1)+1$ generators. If $n$ is infinite and $j$ is finite then $H$ is a free group on infinitely many generators. Finally, if $j$ is infinite, then $H$ may be finitely or infinitely generated; however, if $H$ contains a normal subgroup $N$ of $F$, $N\neq 1$, then $H$ is a free group on infinitely many generators.
This is theorem 2.10 in Magnus, Karrass and Solitar "Combinatorial Group Theory" (and has a name...Schreier's formula?)
Once you have this theorem you can prove, for example, that the commutator subgroup is a free group on infinitely many generators (it contains a normal subgroup and is itself characteristic so you can apply the last line of the theorem). But doing it the way Benjamin Steinberg did it is much neater. I just thought it would be useful to mention this formula!
