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The posting of this question was suggested by Yemon Choi: see Discrete cyclic subgroup.. The question is not mine; it's just a rephrasing of Discrete cyclic subgroup.


EDIT 4. This post claims that the answer is No in general. I hope it's correct. END OF EDIT 4.


By page 110 of Weil's book L'intégration dans les groupes topologiques et ses applications, the answer is No in the abelian case.

I know almost nothing about locally compact groups. The question might be very easy for experts, and perhaps even for laymen. In the unlikely event the question is difficult, here is a particular case:

Let G be a non-compact connected Lie group. Does G admit a discrete infinite cyclic subgroup?

EDIT 1. I think that, by known results about lattices, the answer is No for semisimple Lie groups. Thanks for correcting me if I'm wrong, or (even better) for providing precise statements and references. Again, plenty of MathOverflowers know this stuff much better than I. I'm making it a Community Wiki. END OF EDIT 1.

EDIT 2. I scanned a few pages of Weil's L'intégration dans les groupes topologiques et ses applications and of Raghunathan's Discrete subgroups of Lie groups, and highlighted some statements. The highlighted statements from Raghunathan's book imply that the answer to the main question is No for semisimple Lie groups. (Of course, there might be more elementary arguments.)

[On page 100 of Raghunathan's book (one of the scanned pages) one reads "As will be seen later ... any lattice in a connected Lie group is finitely generated". Unfortunately, I haven't been able to find where, in the sequel of the book, this is proved. If somebody could indicate the appropriated page (and even scan it), it would be great!] END OF EDIT 2.

EDIT 3. Keivan Karai's answer convinced me that there are elementary arguments showing the negativity of the answer to the main question for semisimple Lie groups. END OF EDIT 3.

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you do not need a full lattice to show that there is a copy of $\mathbb Z$. However, assuming the existence of a lattice, the problem becomes easy. Suppose $\Gamma$ is a lattice in $G$: Apply Theorem 6.11 in Raghunathan to a find a torsion-free subgroup of finite index in $\Gamma$. This group is not trivial, since otherwise $\Gamma$ would be finite; Now, choose $\gamma \in \Gamma -\{ 1 \}$ to get a discrete copy $\langle g \rangle $. –  Keivan Karai Aug 2 '10 at 18:36
    
On a different note, the fact that a (non-uniform) lattice in a connected semi-simple Lie group is finitely generated is a deeper fact and considerably more difficult to prove. Raghunathan handles the case of rank one groups. In hight rank case (say for $SL_n(\mathbb R)$ for $n \ge 3$) the first proof is given by Kazhdan and uses property T. –  Keivan Karai Aug 2 '10 at 18:36
    
I agree that it was silly of me to invoke lattices in this connection, and that elementary arguments were sufficient. [I mentioned this in EDIT 3.] I don't have Raghunathan's book at hand, but I'll look at Thm. 6.11. About finite generation of lattices, you say that Raghunathan only "handles the case of rank one groups". But he writes on p. 100: "any lattice in a connected Lie group is finitely generated". He seems to consider any lattice of any connected Lie group. –  Pierre-Yves Gaillard Aug 2 '10 at 19:31
    
Several results in Chapter XIII of Raghunathan show that lattices with some extra properties (e.g. when the ambient group has a rank one factor) are finitely generated (or presented). In Remark 13.21, says that "Corollary 12.20 together with theorem 6.15 and the theorems of Kazhdan and Wang show that any lattice in a semi-simple Lie group is finitely generated." Corollary 13.20 is the fact I mentioned in the parentheses above, and the paper of Kazhdan is the one in which he introduces property T and shows that any lattice in a semi-simple Lie group of higher rank has property T, hence f.g. –  Keivan Karai Aug 3 '10 at 8:53
    
This is what he is alluding to in a remark 6.18. He says clearly that the two results by Wang and Kazhdan are needed in proving that every lattice is finitely generated. –  Keivan Karai Aug 3 '10 at 8:56
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3 Answers 3

The answer to the title question is No, that is

A locally compact group has a compact open subgroup or a discrete infinite cyclic subgroup.

[Keivan Karai helped me a lot, without being responsible for the possible mistakes in this post.]

Step 1. The case of a connected noncompact Lie group.

We claim that a connected noncompact Lie group $G$ has a discrete infinite cyclic subgroup.

Arguing by contradiction, assume $G$ is a counterexample of smallest dimension, and let $Z$ be its center.

Suppose $Ad(G)$ is compact. Then $Z$ is noncompact. If $\dim Z=0$, then $G$ is a noncompact connected covering of the connected compact group $Ad(G)$, which is impossible. Thus, $\dim Z\ge1$. Let $Z_1$ be the connected component of $Z$. If $Z_1$ were compact, $G/Z_1$ would be a counterexample of smaller dimension. If $Z_1$ were noncompact, it would contain a discrete infinite cyclic subgroup, again a contradiction.

Thus, $Ad(G)$ is noncompact. Replacing $G$ with $Ad(G)$, we can suppose $G\subset GL_n(\mathbb R)$.

Say that an endomorphism $x$ of a real finite dimension vector space $V$ satisfies condition (C) if it is semisimple with imaginary eigenvalues, or, equivalently, if the subgroup $exp(\mathbb Z x)$ of $GL(V)$ is not an infinite discrete subgroup.

Let $\mathfrak g$ be the Lie algebra of $G$, and $x$ be in $\mathfrak g$. If $x$ is nonzero and satisfies (C), the same holds for $ad(x)$, implying $Killing(x,x) < 0$. If it were so for all nonzero $x$ in $\mathfrak g$, then $\mathfrak g$, and thus $G$, would be compact.

Step 2. The general case.

Let $G$ be our locally compact group, $C$ its connected component, and recall the following facts (see [1], [2], and references therein):

(1) If $G$ is totally disconnected, then $G$ contains a compact open subgroup.

(2) $C$ is a normal subgroup of $G$, and $G/C$ is totally disconnected.

(3) If $G$ is connected, then every neighborhood of 1 contains a compact normal subgroup $K$ such that $G/K$ is a connected Lie group.

Assume $G$ has no compact open subgroups.

We must show that $G$ has a discrete infinite cyclic subgroup.

As $C$ is noncompact by (1) and (2), we can assume $G=C$, that is $G$ is connected. Then (3) implies that $G$ contains a compact normal subgroup $K$ such that $G/K$ is a connected noncompact Lie group, and we can assume $K=1$, that is $G$ is a connected noncompact Lie group, and the conclusion follows from Step 1.


[1] Willis, G. The structure of totally disconnected, locally compact groups. Math. Ann. 300 (1994), no. 2, 341-363.

http://gdz.sub.uni-goettingen.de/en/dms/load/img/?IDDOC=167209


[2] Willis, G. Totally disconnected, nilpotent, locally compact groups. Bull. Austral. Math. Soc. 55 (1997), no. 1, 143-146.

http://journals.cambridge.org/action/displayFulltext?type=1&fid=4856560&jid=BAZ&volumeId=55&issueId=01&aid=4856552


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You need to be more careful. What if $Ad(G)$ is compact? Even if $Ad(G)$ is not compact, it may not have any semi-simple elements. Consider just the case that $G$ is a linear nilpotent group, then all of the eigenvalues in question will be zero. –  Keivan Karai Aug 3 '10 at 16:26
    
Thanks a lot! I edited the post. I hope it is correct now; but even if it is correct, it doesn't answer the question. --- I agree that $Ad(G)$ may not have semisimple elements. But if $x\in M_n(\mathbb R)$ is not semisimple, then $exp(\mathbb Z x)$ is infinite and discrete. (Look at Jordan's blocks.) Don't hesitate to tell me if this is false! –  Pierre-Yves Gaillard Aug 3 '10 at 17:42
    
The above two comments refer to a previous version of the post. –  Pierre-Yves Gaillard Aug 4 '10 at 7:32
    
Nice; to me it seems correct. –  Keivan Karai Aug 4 '10 at 13:05
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Here is an argument for the semi-simple case: First, suppose that the center of $G$ is trivial. Then $G$ (via adjoint representation) can be considered as a subgroup of $GL(n)$. Consider a neighborhood $U$ of $0$ in the Lie algebra on which the exponential map is a diffeomorphism. If there is any $g \in \exp(U)$ with an eigenvalue which is off the unit circle in $\mathbb C$ then $\langle g \rangle$ is discrete. Otherwise the eigenvalue of every $X \in U$ will be in $\sqrt{-1} \mathbb{R}$, which then would be the case for all $X \in Lie(G)$. Now, it is easy to see that this can only happen for non-compact semi-simple groups, for instance because the Lie algebra must have a copy of $sl_2$. (Jacobson-Morozov). Now, the argument would work in general semi-simple case because, as long as $G/Z(G)$ is not compact, a pre-image of the generator of a non-discrete cyclic subgroup, generates such a subgroup and the universal cover of a semi-simple compact Lie group is also compact.

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I agree with the reduction to the case where our noncompact connected semisimple Lie group $G$ is in $GL_n(\mathbb R)$, but I'd express the rest of the argument as follows. [See next comment.] –  Pierre-Yves Gaillard Aug 2 '10 at 21:38
    
Say that an endomorphism $x$ of a real finite dimension vector space $V$ satisfies condition (C) if it is semisimple with imaginary eigenvalues, or, equivalently, if the subgroup $\exp(\mathbb Z x)$ of $GL(V)$ is not infinite discrete. Let $\mathfrak g$ be the Lie algebra of $G$, and $x$ be in $\mathfrak g$. If $x$ satisfies (C), so does $ad(x)$, implying $Killing(x,x)\le 0$. If it were so for all $x$ in $\mathfrak g$, then $\mathfrak g$ would be compact. –  Pierre-Yves Gaillard Aug 2 '10 at 21:38
    
Thanks. That's probably a better way of finishing the argument. –  Keivan Karai Aug 3 '10 at 6:29
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