Mixing
Profinite topology of a Group
$begingroup$
Let $G$ be a group. Consider now the set of all (left for instance) cosets in $G$ of subgroups of finite index. This set is a base for a topology in $G$.
I found somewhere that if $G$ is residually finite then $G$ is a compact space in this topology but I cannot see why. First of all, is it true? Second, any (maybe more group theoretical) hints?
abstract-algebra general-topology group-theory topological-groups profinite-groups
asked May 31 '14 at 23:35
W4cc0W4cc0
1,89621227
$endgroup$
add a comment |
$begingroup$
Let $G$ be a group. Consider now the set of all (left for instance) cosets in $G$ of subgroups of finite index. This set is a base for a topology in $G$.
I found somewhere that if $G$ is residually finite then $G$ is a compact space in this topology but I cannot see why. First of all, is it true? Second, any (maybe more group theoretical) hints?
abstract-algebra general-topology group-theory topological-groups profinite-groups
asked May 31 '14 at 23:35
W4cc0W4cc0
1,89621227
$endgroup$
$begingroup$
According to Wikipedia here, a group is residually finite iff its profinite topology is Hausdorff. According to Wikipedia here, a profinite group (one isomorphic to an inverse limit of finite groups) is automatically compact and Hausdorff. These facts might be relevant, I'm not sure.
$endgroup$
– blue
Jun 1 '14 at 0:44
add a comment |
$begingroup$
Let $G$ be a group. Consider now the set of all (left for instance) cosets in $G$ of subgroups of finite index. This set is a base for a topology in $G$.
I found somewhere that if $G$ is residually finite then $G$ is a compact space in this topology but I cannot see why. First of all, is it true? Second, any (maybe more group theoretical) hints?
abstract-algebra general-topology group-theory topological-groups profinite-groups
asked May 31 '14 at 23:35
W4cc0W4cc0
1,89621227
$endgroup$
Let $G$ be a group. Consider now the set of all (left for instance) cosets in $G$ of subgroups of finite index. This set is a base for a topology in $G$.
I found somewhere that if $G$ is residually finite then $G$ is a compact space in this topology but I cannot see why. First of all, is it true? Second, any (maybe more group theoretical) hints?
abstract-algebra general-topology group-theory topological-groups profinite-groups
abstract-algebra general-topology group-theory topological-groups profinite-groups
asked May 31 '14 at 23:35
W4cc0W4cc0
1,89621227
asked May 31 '14 at 23:35
W4cc0W4cc0
1,89621227
edited May 31 '14 at 23:47
W4cc0
asked May 31 '14 at 23:35
W4cc0W4cc0
1,89621227
asked May 31 '14 at 23:35
W4cc0W4cc0
1,89621227
asked May 31 '14 at 23:35
W4cc0W4cc0
1,89621227
1,89621227
$begingroup$
According to Wikipedia here, a group is residually finite iff its profinite topology is Hausdorff. According to Wikipedia here, a profinite group (one isomorphic to an inverse limit of finite groups) is automatically compact and Hausdorff. These facts might be relevant, I'm not sure.
$endgroup$
– blue
Jun 1 '14 at 0:44
add a comment |
$begingroup$
According to Wikipedia here, a group is residually finite iff its profinite topology is Hausdorff. According to Wikipedia here, a profinite group (one isomorphic to an inverse limit of finite groups) is automatically compact and Hausdorff. These facts might be relevant, I'm not sure.
$endgroup$
– blue
Jun 1 '14 at 0:44
$begingroup$
According to Wikipedia here, a group is residually finite iff its profinite topology is Hausdorff. According to Wikipedia here, a profinite group (one isomorphic to an inverse limit of finite groups) is automatically compact and Hausdorff. These facts might be relevant, I'm not sure.
$endgroup$
– blue
Jun 1 '14 at 0:44
$begingroup$
According to Wikipedia here, a group is residually finite iff its profinite topology is Hausdorff. According to Wikipedia here, a profinite group (one isomorphic to an inverse limit of finite groups) is automatically compact and Hausdorff. These facts might be relevant, I'm not sure.
$endgroup$
– blue
Jun 1 '14 at 0:44
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The statement is not true. The group of integers $mathbf Z$ is residually finite. On the other hand, consider the open sets $U_psubseteq mathbf Z$ defined by $U_p = pmathbf Z$, where $p$ runs over the primes. Clearly ${U_p} cup {5 mathbf{Z} +1} cup {5 mathbf{Z} -1}$ is an open covering of $mathbf Z$ (as any integer is either $pm 1$ or divisible by some prime). But this covering has no finite sub-covering: by the Chinese Remainder Theorem, there is no finite set of primes such that each integer which is $pm 2$ mod $5$ is divisible by one of those primes.
We can give a criterion. To say that $G$ is residually finite is to say that the canonical morphism $G to widehat{G}$ from $G$ to its profinite competion is injective. Thus, if $G$ is residually finite, $G$ embeds as a dense subgroup of the compact group $widehat{G}$. Since $widehat{G}$ is Hausdorff and compact, $G$ will be compact precisely when it is closed in $widehat{G}$ - which is to say, precisely when $G=widehat{G}$.
(Thanks to studiosus for pointing out my earlier mistake. My mistake lied in the silly statement that open subsets of profinite groups are closed - which is false. Open subgroups are closed, but open subsets need not be closed! For instance, the complement of a point in an infinite profinite group is obviously not closed.)
answered Jun 1 '14 at 0:57
Bruno JoyalBruno Joyal
42.5k693185
$endgroup$
3
$begingroup$
This is a completely wrong answer: Since $hat{G}$ is Hausdorff, its dense proper subset cannot be closed, hence, cannot be compact. The mistake in the proof is the implication: If $cup_i U_i$ contains $G$ then $cup_i U_i=hat{G}$. This implication is obviously false (just think about the set of nonzero real numbers as a subset in $R$.
$endgroup$
– Moishe Cohen
Jun 1 '14 at 2:05
$begingroup$
Compact subsets of Hausdorff spaces are closed. Thus, if $G$ were compact, it would be also closed, hence, would equal $hat{G}$, which is, in general, not the case.
$endgroup$
– Moishe Cohen
Jun 1 '14 at 2:13
1
$begingroup$
Dear Bruno: As a more relevant example with an open covering, consider the Cantor set $C$ and $Xsubset C$, the complement to one point. Then $X$ is open (and, of course, not closed!), but the open covering ${X}$ of $X$ clearly does not cover $C$. What confuses you is the fact that each TD space admits a basis consisting of clopen subsets.
$endgroup$
– Moishe Cohen
Jun 1 '14 at 2:29
1
$begingroup$
One can add that $G$ is compact in profinite topology if and only if $G$ is finite. The entire question, I think, stems from confusion between "compact" and "Hausdorff".
$endgroup$
– Moishe Cohen
Jun 1 '14 at 16:43
3
$begingroup$
@studiosus "$G$ is compact in the profinite topology iff $G$ is finite." is wrong, it should be "$G$ is compact iff it coincides with its profinite completion."
$endgroup$
– user 59363
Nov 21 '14 at 12:04
|
show 4 more comments
$begingroup$
I know it's several years late, but based on some of the comments on Bruno Joyal's answer, I thought it would be helpful to further clarify some details.
Suppose $G$ is a residually finite group. We consider $G$ as a topological group in the profinite topology. In the answer, it is stated that "$G$ embeds as a dense subgroup of $hat{G}$". It is worth clarifying that this use of the word "embeds" is indeed in the topological sense.
Proposition. Let $fcolon Gtohat{G}$ be the canonical embedding. Then $f$ is a homeomorphism between $G$ and $f(G)$ (with the subspace topology from $hat{G}$).
Proof. It suffices to fix a coset $C$ of some normal finite-index subgroup $H$ of $G$, and find an open set $Usubseteqhat{G}$ such that $f(G)cap U=f(C)$. Viewing elements of $hat{G}$ as sequences $(C_N)_{Ninmathcal{N}}in prod_{mathcal{N}}G/N$, where $mathcal{N}$ is the collection of normal finite-index subgroups of $G$, we can take $U={(C_N)in hat{G}:C_H=C}$.
So to simplify terminology, we view $G$ as a subgroup of $hat{G}$, and the profinite topology on $G$ as the subspace topology induced from $hat{G}$. The relevant result is:
Theorem. The following are equivalent.
$G$ is compact in the profinite topology.
$G$ is closed in $hat{G}$.
$G$ is open in $hat{G}$.
$G=hat{G}$.
The canonical embedding of $G$ into $hat{G}$ is a homeomorphism.
Proof.
(1) implies (2). If $G$ is compact in the profinite topology then (by the Proposition) $G$ is a compact set in $hat{G}$. Since $hat{G}$ is Hausdorff, $G$ is closed.
(2) implies (4). Trivial, since $G$ is dense in $hat{G}$.
(4) implies (5). Immediate from the Proposition.
(5) implies (1). Immediate from compactness of $hat{G}$.
(4) implies (3). Trivial.
(3) implies (2). Any open subgroup of a topological group is closed.
Some final remarks motivated by the comments on the accepted answer.
In general, $G$ is not necessarily open in $hat{G}$, and so the canonical embedding of $G$ into $hat{G}$ is not necessarily an open map (this was asked in a comment by W4cc0).
Since $mathbb{Z}neqhat{mathbb{Z}}$, it follows from the Theorem that $mathbb{Z}$ is not compact in the profinite topology (Bruno Joyal also gave an explicit demonstration). The same holds for any countably infinite residually finite group, since there are no countably infinite compact Hausdorff groups (see this link).
As pointed out by user 59363, it is not the case that $G$ is compact in the profinite topology if and only if $G$ is finite (which is claimed in a comment by Moishe Cohen). For example, $mathbb{Z}_p$ is a topologically finitely generated pro-$p$ group, and hence $widehat{mathbb{Z}_p}=mathbb{Z}_p$ (this follows from a result of Serre, which was generalized to any topologically finitely generated profinite group by Nikolov and Segal).
answered Jan 3 at 2:55
Gabe ConantGabe Conant
515
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add a comment |
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2 Answers
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$begingroup$
The statement is not true. The group of integers $mathbf Z$ is residually finite. On the other hand, consider the open sets $U_psubseteq mathbf Z$ defined by $U_p = pmathbf Z$, where $p$ runs over the primes. Clearly ${U_p} cup {5 mathbf{Z} +1} cup {5 mathbf{Z} -1}$ is an open covering of $mathbf Z$ (as any integer is either $pm 1$ or divisible by some prime). But this covering has no finite sub-covering: by the Chinese Remainder Theorem, there is no finite set of primes such that each integer which is $pm 2$ mod $5$ is divisible by one of those primes.
We can give a criterion. To say that $G$ is residually finite is to say that the canonical morphism $G to widehat{G}$ from $G$ to its profinite competion is injective. Thus, if $G$ is residually finite, $G$ embeds as a dense subgroup of the compact group $widehat{G}$. Since $widehat{G}$ is Hausdorff and compact, $G$ will be compact precisely when it is closed in $widehat{G}$ - which is to say, precisely when $G=widehat{G}$.
(Thanks to studiosus for pointing out my earlier mistake. My mistake lied in the silly statement that open subsets of profinite groups are closed - which is false. Open subgroups are closed, but open subsets need not be closed! For instance, the complement of a point in an infinite profinite group is obviously not closed.)
answered Jun 1 '14 at 0:57
Bruno JoyalBruno Joyal
42.5k693185
$endgroup$
3
$begingroup$
This is a completely wrong answer: Since $hat{G}$ is Hausdorff, its dense proper subset cannot be closed, hence, cannot be compact. The mistake in the proof is the implication: If $cup_i U_i$ contains $G$ then $cup_i U_i=hat{G}$. This implication is obviously false (just think about the set of nonzero real numbers as a subset in $R$.
$endgroup$
– Moishe Cohen
Jun 1 '14 at 2:05
$begingroup$
Compact subsets of Hausdorff spaces are closed. Thus, if $G$ were compact, it would be also closed, hence, would equal $hat{G}$, which is, in general, not the case.
$endgroup$
– Moishe Cohen
Jun 1 '14 at 2:13
1
$begingroup$
Dear Bruno: As a more relevant example with an open covering, consider the Cantor set $C$ and $Xsubset C$, the complement to one point. Then $X$ is open (and, of course, not closed!), but the open covering ${X}$ of $X$ clearly does not cover $C$. What confuses you is the fact that each TD space admits a basis consisting of clopen subsets.
$endgroup$
– Moishe Cohen
Jun 1 '14 at 2:29
1
$begingroup$
One can add that $G$ is compact in profinite topology if and only if $G$ is finite. The entire question, I think, stems from confusion between "compact" and "Hausdorff".
$endgroup$
– Moishe Cohen
Jun 1 '14 at 16:43
3
$begingroup$
@studiosus "$G$ is compact in the profinite topology iff $G$ is finite." is wrong, it should be "$G$ is compact iff it coincides with its profinite completion."
$endgroup$
– user 59363
Nov 21 '14 at 12:04
|
show 4 more comments
$begingroup$
The statement is not true. The group of integers $mathbf Z$ is residually finite. On the other hand, consider the open sets $U_psubseteq mathbf Z$ defined by $U_p = pmathbf Z$, where $p$ runs over the primes. Clearly ${U_p} cup {5 mathbf{Z} +1} cup {5 mathbf{Z} -1}$ is an open covering of $mathbf Z$ (as any integer is either $pm 1$ or divisible by some prime). But this covering has no finite sub-covering: by the Chinese Remainder Theorem, there is no finite set of primes such that each integer which is $pm 2$ mod $5$ is divisible by one of those primes.
We can give a criterion. To say that $G$ is residually finite is to say that the canonical morphism $G to widehat{G}$ from $G$ to its profinite competion is injective. Thus, if $G$ is residually finite, $G$ embeds as a dense subgroup of the compact group $widehat{G}$. Since $widehat{G}$ is Hausdorff and compact, $G$ will be compact precisely when it is closed in $widehat{G}$ - which is to say, precisely when $G=widehat{G}$.
(Thanks to studiosus for pointing out my earlier mistake. My mistake lied in the silly statement that open subsets of profinite groups are closed - which is false. Open subgroups are closed, but open subsets need not be closed! For instance, the complement of a point in an infinite profinite group is obviously not closed.)
answered Jun 1 '14 at 0:57
Bruno JoyalBruno Joyal
42.5k693185
$endgroup$
3
$begingroup$
This is a completely wrong answer: Since $hat{G}$ is Hausdorff, its dense proper subset cannot be closed, hence, cannot be compact. The mistake in the proof is the implication: If $cup_i U_i$ contains $G$ then $cup_i U_i=hat{G}$. This implication is obviously false (just think about the set of nonzero real numbers as a subset in $R$.
$endgroup$
– Moishe Cohen
Jun 1 '14 at 2:05
$begingroup$
Compact subsets of Hausdorff spaces are closed. Thus, if $G$ were compact, it would be also closed, hence, would equal $hat{G}$, which is, in general, not the case.
$endgroup$
– Moishe Cohen
Jun 1 '14 at 2:13
1
$begingroup$
Dear Bruno: As a more relevant example with an open covering, consider the Cantor set $C$ and $Xsubset C$, the complement to one point. Then $X$ is open (and, of course, not closed!), but the open covering ${X}$ of $X$ clearly does not cover $C$. What confuses you is the fact that each TD space admits a basis consisting of clopen subsets.
$endgroup$
– Moishe Cohen
Jun 1 '14 at 2:29
1
$begingroup$
One can add that $G$ is compact in profinite topology if and only if $G$ is finite. The entire question, I think, stems from confusion between "compact" and "Hausdorff".
$endgroup$
– Moishe Cohen
Jun 1 '14 at 16:43
3
$begingroup$
@studiosus "$G$ is compact in the profinite topology iff $G$ is finite." is wrong, it should be "$G$ is compact iff it coincides with its profinite completion."
$endgroup$
– user 59363
Nov 21 '14 at 12:04
|
show 4 more comments
$begingroup$
The statement is not true. The group of integers $mathbf Z$ is residually finite. On the other hand, consider the open sets $U_psubseteq mathbf Z$ defined by $U_p = pmathbf Z$, where $p$ runs over the primes. Clearly ${U_p} cup {5 mathbf{Z} +1} cup {5 mathbf{Z} -1}$ is an open covering of $mathbf Z$ (as any integer is either $pm 1$ or divisible by some prime). But this covering has no finite sub-covering: by the Chinese Remainder Theorem, there is no finite set of primes such that each integer which is $pm 2$ mod $5$ is divisible by one of those primes.
We can give a criterion. To say that $G$ is residually finite is to say that the canonical morphism $G to widehat{G}$ from $G$ to its profinite competion is injective. Thus, if $G$ is residually finite, $G$ embeds as a dense subgroup of the compact group $widehat{G}$. Since $widehat{G}$ is Hausdorff and compact, $G$ will be compact precisely when it is closed in $widehat{G}$ - which is to say, precisely when $G=widehat{G}$.
(Thanks to studiosus for pointing out my earlier mistake. My mistake lied in the silly statement that open subsets of profinite groups are closed - which is false. Open subgroups are closed, but open subsets need not be closed! For instance, the complement of a point in an infinite profinite group is obviously not closed.)
answered Jun 1 '14 at 0:57
Bruno JoyalBruno Joyal
42.5k693185
$endgroup$
The statement is not true. The group of integers $mathbf Z$ is residually finite. On the other hand, consider the open sets $U_psubseteq mathbf Z$ defined by $U_p = pmathbf Z$, where $p$ runs over the primes. Clearly ${U_p} cup {5 mathbf{Z} +1} cup {5 mathbf{Z} -1}$ is an open covering of $mathbf Z$ (as any integer is either $pm 1$ or divisible by some prime). But this covering has no finite sub-covering: by the Chinese Remainder Theorem, there is no finite set of primes such that each integer which is $pm 2$ mod $5$ is divisible by one of those primes.
We can give a criterion. To say that $G$ is residually finite is to say that the canonical morphism $G to widehat{G}$ from $G$ to its profinite competion is injective. Thus, if $G$ is residually finite, $G$ embeds as a dense subgroup of the compact group $widehat{G}$. Since $widehat{G}$ is Hausdorff and compact, $G$ will be compact precisely when it is closed in $widehat{G}$ - which is to say, precisely when $G=widehat{G}$.
(Thanks to studiosus for pointing out my earlier mistake. My mistake lied in the silly statement that open subsets of profinite groups are closed - which is false. Open subgroups are closed, but open subsets need not be closed! For instance, the complement of a point in an infinite profinite group is obviously not closed.)
answered Jun 1 '14 at 0:57
Bruno JoyalBruno Joyal
42.5k693185
edited Jun 1 '14 at 6:23
answered Jun 1 '14 at 0:57
Bruno JoyalBruno Joyal
42.5k693185
answered Jun 1 '14 at 0:57
Bruno JoyalBruno Joyal
42.5k693185
answered Jun 1 '14 at 0:57
Bruno JoyalBruno Joyal
42.5k693185
42.5k693185
3
$begingroup$
This is a completely wrong answer: Since $hat{G}$ is Hausdorff, its dense proper subset cannot be closed, hence, cannot be compact. The mistake in the proof is the implication: If $cup_i U_i$ contains $G$ then $cup_i U_i=hat{G}$. This implication is obviously false (just think about the set of nonzero real numbers as a subset in $R$.
$endgroup$
– Moishe Cohen
Jun 1 '14 at 2:05
$begingroup$
Compact subsets of Hausdorff spaces are closed. Thus, if $G$ were compact, it would be also closed, hence, would equal $hat{G}$, which is, in general, not the case.
$endgroup$
– Moishe Cohen
Jun 1 '14 at 2:13
1
$begingroup$
Dear Bruno: As a more relevant example with an open covering, consider the Cantor set $C$ and $Xsubset C$, the complement to one point. Then $X$ is open (and, of course, not closed!), but the open covering ${X}$ of $X$ clearly does not cover $C$. What confuses you is the fact that each TD space admits a basis consisting of clopen subsets.
$endgroup$
– Moishe Cohen
Jun 1 '14 at 2:29
1
$begingroup$
One can add that $G$ is compact in profinite topology if and only if $G$ is finite. The entire question, I think, stems from confusion between "compact" and "Hausdorff".
$endgroup$
– Moishe Cohen
Jun 1 '14 at 16:43
3
$begingroup$
@studiosus "$G$ is compact in the profinite topology iff $G$ is finite." is wrong, it should be "$G$ is compact iff it coincides with its profinite completion."
$endgroup$
– user 59363
Nov 21 '14 at 12:04
|
show 4 more comments
3
$begingroup$
This is a completely wrong answer: Since $hat{G}$ is Hausdorff, its dense proper subset cannot be closed, hence, cannot be compact. The mistake in the proof is the implication: If $cup_i U_i$ contains $G$ then $cup_i U_i=hat{G}$. This implication is obviously false (just think about the set of nonzero real numbers as a subset in $R$.
$endgroup$
– Moishe Cohen
Jun 1 '14 at 2:05
$begingroup$
Compact subsets of Hausdorff spaces are closed. Thus, if $G$ were compact, it would be also closed, hence, would equal $hat{G}$, which is, in general, not the case.
$endgroup$
– Moishe Cohen
Jun 1 '14 at 2:13
1
$begingroup$
Dear Bruno: As a more relevant example with an open covering, consider the Cantor set $C$ and $Xsubset C$, the complement to one point. Then $X$ is open (and, of course, not closed!), but the open covering ${X}$ of $X$ clearly does not cover $C$. What confuses you is the fact that each TD space admits a basis consisting of clopen subsets.
$endgroup$
– Moishe Cohen
Jun 1 '14 at 2:29
1
$begingroup$
One can add that $G$ is compact in profinite topology if and only if $G$ is finite. The entire question, I think, stems from confusion between "compact" and "Hausdorff".
$endgroup$
– Moishe Cohen
Jun 1 '14 at 16:43
3
$begingroup$
@studiosus "$G$ is compact in the profinite topology iff $G$ is finite." is wrong, it should be "$G$ is compact iff it coincides with its profinite completion."
$endgroup$
– user 59363
Nov 21 '14 at 12:04
3
3
$begingroup$
This is a completely wrong answer: Since $hat{G}$ is Hausdorff, its dense proper subset cannot be closed, hence, cannot be compact. The mistake in the proof is the implication: If $cup_i U_i$ contains $G$ then $cup_i U_i=hat{G}$. This implication is obviously false (just think about the set of nonzero real numbers as a subset in $R$.
$endgroup$
– Moishe Cohen
Jun 1 '14 at 2:05
$begingroup$
This is a completely wrong answer: Since $hat{G}$ is Hausdorff, its dense proper subset cannot be closed, hence, cannot be compact. The mistake in the proof is the implication: If $cup_i U_i$ contains $G$ then $cup_i U_i=hat{G}$. This implication is obviously false (just think about the set of nonzero real numbers as a subset in $R$.
$endgroup$
– Moishe Cohen
Jun 1 '14 at 2:05
$begingroup$
Compact subsets of Hausdorff spaces are closed. Thus, if $G$ were compact, it would be also closed, hence, would equal $hat{G}$, which is, in general, not the case.
$endgroup$
– Moishe Cohen
Jun 1 '14 at 2:13
$begingroup$
Compact subsets of Hausdorff spaces are closed. Thus, if $G$ were compact, it would be also closed, hence, would equal $hat{G}$, which is, in general, not the case.
$endgroup$
– Moishe Cohen
Jun 1 '14 at 2:13
1
1
$begingroup$
Dear Bruno: As a more relevant example with an open covering, consider the Cantor set $C$ and $Xsubset C$, the complement to one point. Then $X$ is open (and, of course, not closed!), but the open covering ${X}$ of $X$ clearly does not cover $C$. What confuses you is the fact that each TD space admits a basis consisting of clopen subsets.
$endgroup$
– Moishe Cohen
Jun 1 '14 at 2:29
$begingroup$
Dear Bruno: As a more relevant example with an open covering, consider the Cantor set $C$ and $Xsubset C$, the complement to one point. Then $X$ is open (and, of course, not closed!), but the open covering ${X}$ of $X$ clearly does not cover $C$. What confuses you is the fact that each TD space admits a basis consisting of clopen subsets.
$endgroup$
– Moishe Cohen
Jun 1 '14 at 2:29
1
1
$begingroup$
One can add that $G$ is compact in profinite topology if and only if $G$ is finite. The entire question, I think, stems from confusion between "compact" and "Hausdorff".
$endgroup$
– Moishe Cohen
Jun 1 '14 at 16:43
$begingroup$
One can add that $G$ is compact in profinite topology if and only if $G$ is finite. The entire question, I think, stems from confusion between "compact" and "Hausdorff".
$endgroup$
– Moishe Cohen
Jun 1 '14 at 16:43
3
3
$begingroup$
@studiosus "$G$ is compact in the profinite topology iff $G$ is finite." is wrong, it should be "$G$ is compact iff it coincides with its profinite completion."
$endgroup$
– user 59363
Nov 21 '14 at 12:04
$begingroup$
@studiosus "$G$ is compact in the profinite topology iff $G$ is finite." is wrong, it should be "$G$ is compact iff it coincides with its profinite completion."
$endgroup$
– user 59363
Nov 21 '14 at 12:04
|
show 4 more comments
$begingroup$
I know it's several years late, but based on some of the comments on Bruno Joyal's answer, I thought it would be helpful to further clarify some details.
Suppose $G$ is a residually finite group. We consider $G$ as a topological group in the profinite topology. In the answer, it is stated that "$G$ embeds as a dense subgroup of $hat{G}$". It is worth clarifying that this use of the word "embeds" is indeed in the topological sense.
Proposition. Let $fcolon Gtohat{G}$ be the canonical embedding. Then $f$ is a homeomorphism between $G$ and $f(G)$ (with the subspace topology from $hat{G}$).
Proof. It suffices to fix a coset $C$ of some normal finite-index subgroup $H$ of $G$, and find an open set $Usubseteqhat{G}$ such that $f(G)cap U=f(C)$. Viewing elements of $hat{G}$ as sequences $(C_N)_{Ninmathcal{N}}in prod_{mathcal{N}}G/N$, where $mathcal{N}$ is the collection of normal finite-index subgroups of $G$, we can take $U={(C_N)in hat{G}:C_H=C}$.
So to simplify terminology, we view $G$ as a subgroup of $hat{G}$, and the profinite topology on $G$ as the subspace topology induced from $hat{G}$. The relevant result is:
Theorem. The following are equivalent.
$G$ is compact in the profinite topology.
$G$ is closed in $hat{G}$.
$G$ is open in $hat{G}$.
$G=hat{G}$.
The canonical embedding of $G$ into $hat{G}$ is a homeomorphism.
Proof.
(1) implies (2). If $G$ is compact in the profinite topology then (by the Proposition) $G$ is a compact set in $hat{G}$. Since $hat{G}$ is Hausdorff, $G$ is closed.
(2) implies (4). Trivial, since $G$ is dense in $hat{G}$.
(4) implies (5). Immediate from the Proposition.
(5) implies (1). Immediate from compactness of $hat{G}$.
(4) implies (3). Trivial.
(3) implies (2). Any open subgroup of a topological group is closed.
Some final remarks motivated by the comments on the accepted answer.
In general, $G$ is not necessarily open in $hat{G}$, and so the canonical embedding of $G$ into $hat{G}$ is not necessarily an open map (this was asked in a comment by W4cc0).
Since $mathbb{Z}neqhat{mathbb{Z}}$, it follows from the Theorem that $mathbb{Z}$ is not compact in the profinite topology (Bruno Joyal also gave an explicit demonstration). The same holds for any countably infinite residually finite group, since there are no countably infinite compact Hausdorff groups (see this link).
As pointed out by user 59363, it is not the case that $G$ is compact in the profinite topology if and only if $G$ is finite (which is claimed in a comment by Moishe Cohen). For example, $mathbb{Z}_p$ is a topologically finitely generated pro-$p$ group, and hence $widehat{mathbb{Z}_p}=mathbb{Z}_p$ (this follows from a result of Serre, which was generalized to any topologically finitely generated profinite group by Nikolov and Segal).
answered Jan 3 at 2:55
Gabe ConantGabe Conant
515
$endgroup$
add a comment |
$begingroup$
I know it's several years late, but based on some of the comments on Bruno Joyal's answer, I thought it would be helpful to further clarify some details.
Suppose $G$ is a residually finite group. We consider $G$ as a topological group in the profinite topology. In the answer, it is stated that "$G$ embeds as a dense subgroup of $hat{G}$". It is worth clarifying that this use of the word "embeds" is indeed in the topological sense.
Proposition. Let $fcolon Gtohat{G}$ be the canonical embedding. Then $f$ is a homeomorphism between $G$ and $f(G)$ (with the subspace topology from $hat{G}$).
Proof. It suffices to fix a coset $C$ of some normal finite-index subgroup $H$ of $G$, and find an open set $Usubseteqhat{G}$ such that $f(G)cap U=f(C)$. Viewing elements of $hat{G}$ as sequences $(C_N)_{Ninmathcal{N}}in prod_{mathcal{N}}G/N$, where $mathcal{N}$ is the collection of normal finite-index subgroups of $G$, we can take $U={(C_N)in hat{G}:C_H=C}$.
So to simplify terminology, we view $G$ as a subgroup of $hat{G}$, and the profinite topology on $G$ as the subspace topology induced from $hat{G}$. The relevant result is:
Theorem. The following are equivalent.
$G$ is compact in the profinite topology.
$G$ is closed in $hat{G}$.
$G$ is open in $hat{G}$.
$G=hat{G}$.
The canonical embedding of $G$ into $hat{G}$ is a homeomorphism.
Proof.
(1) implies (2). If $G$ is compact in the profinite topology then (by the Proposition) $G$ is a compact set in $hat{G}$. Since $hat{G}$ is Hausdorff, $G$ is closed.
(2) implies (4). Trivial, since $G$ is dense in $hat{G}$.
(4) implies (5). Immediate from the Proposition.
(5) implies (1). Immediate from compactness of $hat{G}$.
(4) implies (3). Trivial.
(3) implies (2). Any open subgroup of a topological group is closed.
Some final remarks motivated by the comments on the accepted answer.
In general, $G$ is not necessarily open in $hat{G}$, and so the canonical embedding of $G$ into $hat{G}$ is not necessarily an open map (this was asked in a comment by W4cc0).
Since $mathbb{Z}neqhat{mathbb{Z}}$, it follows from the Theorem that $mathbb{Z}$ is not compact in the profinite topology (Bruno Joyal also gave an explicit demonstration). The same holds for any countably infinite residually finite group, since there are no countably infinite compact Hausdorff groups (see this link).
As pointed out by user 59363, it is not the case that $G$ is compact in the profinite topology if and only if $G$ is finite (which is claimed in a comment by Moishe Cohen). For example, $mathbb{Z}_p$ is a topologically finitely generated pro-$p$ group, and hence $widehat{mathbb{Z}_p}=mathbb{Z}_p$ (this follows from a result of Serre, which was generalized to any topologically finitely generated profinite group by Nikolov and Segal).
answered Jan 3 at 2:55
Gabe ConantGabe Conant
515
$endgroup$
add a comment |
$begingroup$
I know it's several years late, but based on some of the comments on Bruno Joyal's answer, I thought it would be helpful to further clarify some details.
Suppose $G$ is a residually finite group. We consider $G$ as a topological group in the profinite topology. In the answer, it is stated that "$G$ embeds as a dense subgroup of $hat{G}$". It is worth clarifying that this use of the word "embeds" is indeed in the topological sense.
Proposition. Let $fcolon Gtohat{G}$ be the canonical embedding. Then $f$ is a homeomorphism between $G$ and $f(G)$ (with the subspace topology from $hat{G}$).
Proof. It suffices to fix a coset $C$ of some normal finite-index subgroup $H$ of $G$, and find an open set $Usubseteqhat{G}$ such that $f(G)cap U=f(C)$. Viewing elements of $hat{G}$ as sequences $(C_N)_{Ninmathcal{N}}in prod_{mathcal{N}}G/N$, where $mathcal{N}$ is the collection of normal finite-index subgroups of $G$, we can take $U={(C_N)in hat{G}:C_H=C}$.
So to simplify terminology, we view $G$ as a subgroup of $hat{G}$, and the profinite topology on $G$ as the subspace topology induced from $hat{G}$. The relevant result is:
Theorem. The following are equivalent.
$G$ is compact in the profinite topology.
$G$ is closed in $hat{G}$.
$G$ is open in $hat{G}$.
$G=hat{G}$.
The canonical embedding of $G$ into $hat{G}$ is a homeomorphism.
Proof.
(1) implies (2). If $G$ is compact in the profinite topology then (by the Proposition) $G$ is a compact set in $hat{G}$. Since $hat{G}$ is Hausdorff, $G$ is closed.
(2) implies (4). Trivial, since $G$ is dense in $hat{G}$.
(4) implies (5). Immediate from the Proposition.
(5) implies (1). Immediate from compactness of $hat{G}$.
(4) implies (3). Trivial.
(3) implies (2). Any open subgroup of a topological group is closed.
Some final remarks motivated by the comments on the accepted answer.
In general, $G$ is not necessarily open in $hat{G}$, and so the canonical embedding of $G$ into $hat{G}$ is not necessarily an open map (this was asked in a comment by W4cc0).
Since $mathbb{Z}neqhat{mathbb{Z}}$, it follows from the Theorem that $mathbb{Z}$ is not compact in the profinite topology (Bruno Joyal also gave an explicit demonstration). The same holds for any countably infinite residually finite group, since there are no countably infinite compact Hausdorff groups (see this link).
As pointed out by user 59363, it is not the case that $G$ is compact in the profinite topology if and only if $G$ is finite (which is claimed in a comment by Moishe Cohen). For example, $mathbb{Z}_p$ is a topologically finitely generated pro-$p$ group, and hence $widehat{mathbb{Z}_p}=mathbb{Z}_p$ (this follows from a result of Serre, which was generalized to any topologically finitely generated profinite group by Nikolov and Segal).
answered Jan 3 at 2:55
Gabe ConantGabe Conant
515
$endgroup$
I know it's several years late, but based on some of the comments on Bruno Joyal's answer, I thought it would be helpful to further clarify some details.
Suppose $G$ is a residually finite group. We consider $G$ as a topological group in the profinite topology. In the answer, it is stated that "$G$ embeds as a dense subgroup of $hat{G}$". It is worth clarifying that this use of the word "embeds" is indeed in the topological sense.
Proposition. Let $fcolon Gtohat{G}$ be the canonical embedding. Then $f$ is a homeomorphism between $G$ and $f(G)$ (with the subspace topology from $hat{G}$).
Proof. It suffices to fix a coset $C$ of some normal finite-index subgroup $H$ of $G$, and find an open set $Usubseteqhat{G}$ such that $f(G)cap U=f(C)$. Viewing elements of $hat{G}$ as sequences $(C_N)_{Ninmathcal{N}}in prod_{mathcal{N}}G/N$, where $mathcal{N}$ is the collection of normal finite-index subgroups of $G$, we can take $U={(C_N)in hat{G}:C_H=C}$.
So to simplify terminology, we view $G$ as a subgroup of $hat{G}$, and the profinite topology on $G$ as the subspace topology induced from $hat{G}$. The relevant result is:
Theorem. The following are equivalent.
$G$ is compact in the profinite topology.
$G$ is closed in $hat{G}$.
$G$ is open in $hat{G}$.
$G=hat{G}$.
The canonical embedding of $G$ into $hat{G}$ is a homeomorphism.
Proof.
(1) implies (2). If $G$ is compact in the profinite topology then (by the Proposition) $G$ is a compact set in $hat{G}$. Since $hat{G}$ is Hausdorff, $G$ is closed.
(2) implies (4). Trivial, since $G$ is dense in $hat{G}$.
(4) implies (5). Immediate from the Proposition.
(5) implies (1). Immediate from compactness of $hat{G}$.
(4) implies (3). Trivial.
(3) implies (2). Any open subgroup of a topological group is closed.
Some final remarks motivated by the comments on the accepted answer.
In general, $G$ is not necessarily open in $hat{G}$, and so the canonical embedding of $G$ into $hat{G}$ is not necessarily an open map (this was asked in a comment by W4cc0).
Since $mathbb{Z}neqhat{mathbb{Z}}$, it follows from the Theorem that $mathbb{Z}$ is not compact in the profinite topology (Bruno Joyal also gave an explicit demonstration). The same holds for any countably infinite residually finite group, since there are no countably infinite compact Hausdorff groups (see this link).
As pointed out by user 59363, it is not the case that $G$ is compact in the profinite topology if and only if $G$ is finite (which is claimed in a comment by Moishe Cohen). For example, $mathbb{Z}_p$ is a topologically finitely generated pro-$p$ group, and hence $widehat{mathbb{Z}_p}=mathbb{Z}_p$ (this follows from a result of Serre, which was generalized to any topologically finitely generated profinite group by Nikolov and Segal).
answered Jan 3 at 2:55
Gabe ConantGabe Conant
515
edited Jan 4 at 16:35
answered Jan 3 at 2:55
Gabe ConantGabe Conant
515
answered Jan 3 at 2:55
Gabe ConantGabe Conant
515
answered Jan 3 at 2:55
Gabe ConantGabe Conant
515
515
add a comment |
add a comment |
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$begingroup$
According to Wikipedia here, a group is residually finite iff its profinite topology is Hausdorff. According to Wikipedia here, a profinite group (one isomorphic to an inverse limit of finite groups) is automatically compact and Hausdorff. These facts might be relevant, I'm not sure.
$endgroup$
– blue
Jun 1 '14 at 0:44