Return to Answer

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

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

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

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

I posted what I thought was a solution. Keivan Karai pointed out a mistake. Then I posted a partial solution. And now I'm again posting what I hope is a complete solution. [I apologize to those who are not interested in this story.]

I'll try to prove that the answer to the title question is No.

Let $G$ be a 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.

I 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.

Let $Z$ be the center of $G$.

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. If dim $Z\ge1$, let $Z_0$ be the connected component of $Z$. If $Z_0$ is compact, we can mod out by it, and we are in the previous case. If $Z_0$ is noncompact, we're done.

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

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.

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 $G\subset GL_n(\mathbb R)$.

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.