The rigidity of the countable product of free groupsTopologizing a free product G*H of discrete groups? ...
The rigidity of the countable product of free groups
Topologizing a free product G*H of discrete groups? Second homology group of free nilpotent p-groupThe finest countably generated free topological group so that $x^{m_{n}}_nrightarrow 1$?Salvaging Howson's theorem for free profinite groupsIs this concrete set generically Haar-null?Is each closed subgroups of $mathbb R^omega$ isomorphic to a Tychonoff product of locally compact Abelian groups?Closed free subgroups of the automorphism group of the countable atomless boolean algebraClassification of compact connected abelian groupsDoes each $omega$-narrow topological group have countable discrete cellularity?Conjugating generators in free groups
$begingroup$
For a natural number $n$ let $F_n$ be the free group with $n$ generators.
The group $F_n$ is endowed with the discrete topology.
Given an increasing sequence $vec p=(p_k)_{kinomega}$ of prime numbers, consider the Polish group $F_{vec p}=prod_{kinomega}F_{p_k}$.
Problem. Let $vec p,vec q$ be two increasing sequences of prime numbers such that the Polish groups $F_{vec p}$ and $F_{vec q}$ are topologically isomorphic. Is $vec p=vec q$?
gr.group-theory geometric-group-theory topological-groups
edited 9 hours ago
YCor
30.1k4 gold badges89 silver badges144 bronze badges
asked 9 hours ago
Taras BanakhTaras Banakh
18.1k1 gold badge38 silver badges99 bronze badges
$endgroup$
add a comment |
$begingroup$
For a natural number $n$ let $F_n$ be the free group with $n$ generators.
The group $F_n$ is endowed with the discrete topology.
Given an increasing sequence $vec p=(p_k)_{kinomega}$ of prime numbers, consider the Polish group $F_{vec p}=prod_{kinomega}F_{p_k}$.
Problem. Let $vec p,vec q$ be two increasing sequences of prime numbers such that the Polish groups $F_{vec p}$ and $F_{vec q}$ are topologically isomorphic. Is $vec p=vec q$?
gr.group-theory geometric-group-theory topological-groups
edited 9 hours ago
YCor
30.1k4 gold badges89 silver badges144 bronze badges
asked 9 hours ago
Taras BanakhTaras Banakh
18.1k1 gold badge38 silver badges99 bronze badges
$endgroup$
add a comment |
$begingroup$
For a natural number $n$ let $F_n$ be the free group with $n$ generators.
The group $F_n$ is endowed with the discrete topology.
Given an increasing sequence $vec p=(p_k)_{kinomega}$ of prime numbers, consider the Polish group $F_{vec p}=prod_{kinomega}F_{p_k}$.
Problem. Let $vec p,vec q$ be two increasing sequences of prime numbers such that the Polish groups $F_{vec p}$ and $F_{vec q}$ are topologically isomorphic. Is $vec p=vec q$?
gr.group-theory geometric-group-theory topological-groups
edited 9 hours ago
YCor
30.1k4 gold badges89 silver badges144 bronze badges
asked 9 hours ago
Taras BanakhTaras Banakh
18.1k1 gold badge38 silver badges99 bronze badges
$endgroup$
For a natural number $n$ let $F_n$ be the free group with $n$ generators.
The group $F_n$ is endowed with the discrete topology.
Given an increasing sequence $vec p=(p_k)_{kinomega}$ of prime numbers, consider the Polish group $F_{vec p}=prod_{kinomega}F_{p_k}$.
Problem. Let $vec p,vec q$ be two increasing sequences of prime numbers such that the Polish groups $F_{vec p}$ and $F_{vec q}$ are topologically isomorphic. Is $vec p=vec q$?
gr.group-theory geometric-group-theory topological-groups
gr.group-theory geometric-group-theory topological-groups
edited 9 hours ago
YCor
30.1k4 gold badges89 silver badges144 bronze badges
asked 9 hours ago
Taras BanakhTaras Banakh
18.1k1 gold badge38 silver badges99 bronze badges
edited 9 hours ago
YCor
30.1k4 gold badges89 silver badges144 bronze badges
asked 9 hours ago
Taras BanakhTaras Banakh
18.1k1 gold badge38 silver badges99 bronze badges
edited 9 hours ago
YCor
30.1k4 gold badges89 silver badges144 bronze badges
edited 9 hours ago
YCor
30.1k4 gold badges89 silver badges144 bronze badges
edited 9 hours ago
YCor
30.1k4 gold badges89 silver badges144 bronze badges
30.1k4 gold badges89 silver badges144 bronze badges
asked 9 hours ago
Taras BanakhTaras Banakh
18.1k1 gold badge38 silver badges99 bronze badges
asked 9 hours ago
Taras BanakhTaras Banakh
18.1k1 gold badge38 silver badges99 bronze badges
asked 9 hours ago
Taras BanakhTaras Banakh
18.1k1 gold badge38 silver badges99 bronze badges
18.1k1 gold badge38 silver badges99 bronze badges
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Yes, and even in the group category. More generally, I claim that if $prod_{iin I}G_i$ and $prod_{jin J}H_j$ are isomorphic groups, for two sets $I,J$ and two families $(G_i)$, $(H_j)$ of groups that are center-free and directly indecomposable, then there is a bijection $f:Ito J$ such that $G_i$ and $H_{f(i)}$ are isomorphic for all $i$.
(Recall that a group is directly indecomposable if it's nontrivial and not direct product of two nontrivial subgroups.)
To show this, it is enough to recognize the subgroups $G_i$ in the product $G=prod_i G_i$, purely relying on the structure of the group $G$.
Indeed, $G_i$ is directly indecomposable, and $G$ is direct product of $G_i$ and its centralizer (using that $G_i$ has trivial center).
I claim that this characterizes the subgroups $G_i$.
Claim if $H$ is a directly indecomposable subgroup of $G$ such that $G$ is direct product of $H$ and its centralizer, then $H=G_i$ for some unique $i$.
Trivial lemma: Let $H$ be a subgroup of $G=prod G_i$ ($G_i$ arbitrary groups), and $H_i$ its projection on $G_i$. Let $K_i$ be the centralizer of $H_i$. Then the centralizer of $H$ is $prod_i K_i$. $Box$
Now to prove the claim, let $H$ be a subgroup with these properties. Then $Hsubsetprod H_i$, which has trivial intersection with the centralizer $K=prod K_i$ of $H$. Since by assumption $G=Htimes K$, we deduce that $H=prod H_i$. Since $H$ is directly indecomposable, $H_i$ is nontrivial for a single $i$. So $Hsubset H_i$, and $G_i=H_itimes K_i$. Since $G_i$ is directly indecomposable, we deduce that $H=H_i=G_i$.
answered 8 hours ago
YCorYCor
30.1k4 gold badges89 silver badges144 bronze badges
$endgroup$
$begingroup$
Thank you for the quick answer, which answers some other problem related to product of surfaces, which was asked at the 34th Summer Comference in Johannesburg.
$endgroup$
– Taras Banakh
8 hours ago
add a comment |
Your Answer
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "504"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f335723%2fthe-rigidity-of-the-countable-product-of-free-groups%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Yes, and even in the group category. More generally, I claim that if $prod_{iin I}G_i$ and $prod_{jin J}H_j$ are isomorphic groups, for two sets $I,J$ and two families $(G_i)$, $(H_j)$ of groups that are center-free and directly indecomposable, then there is a bijection $f:Ito J$ such that $G_i$ and $H_{f(i)}$ are isomorphic for all $i$.
(Recall that a group is directly indecomposable if it's nontrivial and not direct product of two nontrivial subgroups.)
To show this, it is enough to recognize the subgroups $G_i$ in the product $G=prod_i G_i$, purely relying on the structure of the group $G$.
Indeed, $G_i$ is directly indecomposable, and $G$ is direct product of $G_i$ and its centralizer (using that $G_i$ has trivial center).
I claim that this characterizes the subgroups $G_i$.
Claim if $H$ is a directly indecomposable subgroup of $G$ such that $G$ is direct product of $H$ and its centralizer, then $H=G_i$ for some unique $i$.
Trivial lemma: Let $H$ be a subgroup of $G=prod G_i$ ($G_i$ arbitrary groups), and $H_i$ its projection on $G_i$. Let $K_i$ be the centralizer of $H_i$. Then the centralizer of $H$ is $prod_i K_i$. $Box$
Now to prove the claim, let $H$ be a subgroup with these properties. Then $Hsubsetprod H_i$, which has trivial intersection with the centralizer $K=prod K_i$ of $H$. Since by assumption $G=Htimes K$, we deduce that $H=prod H_i$. Since $H$ is directly indecomposable, $H_i$ is nontrivial for a single $i$. So $Hsubset H_i$, and $G_i=H_itimes K_i$. Since $G_i$ is directly indecomposable, we deduce that $H=H_i=G_i$.
answered 8 hours ago
YCorYCor
30.1k4 gold badges89 silver badges144 bronze badges
$endgroup$
$begingroup$
Thank you for the quick answer, which answers some other problem related to product of surfaces, which was asked at the 34th Summer Comference in Johannesburg.
$endgroup$
– Taras Banakh
8 hours ago
add a comment |
$begingroup$
Yes, and even in the group category. More generally, I claim that if $prod_{iin I}G_i$ and $prod_{jin J}H_j$ are isomorphic groups, for two sets $I,J$ and two families $(G_i)$, $(H_j)$ of groups that are center-free and directly indecomposable, then there is a bijection $f:Ito J$ such that $G_i$ and $H_{f(i)}$ are isomorphic for all $i$.
(Recall that a group is directly indecomposable if it's nontrivial and not direct product of two nontrivial subgroups.)
To show this, it is enough to recognize the subgroups $G_i$ in the product $G=prod_i G_i$, purely relying on the structure of the group $G$.
Indeed, $G_i$ is directly indecomposable, and $G$ is direct product of $G_i$ and its centralizer (using that $G_i$ has trivial center).
I claim that this characterizes the subgroups $G_i$.
Claim if $H$ is a directly indecomposable subgroup of $G$ such that $G$ is direct product of $H$ and its centralizer, then $H=G_i$ for some unique $i$.
Trivial lemma: Let $H$ be a subgroup of $G=prod G_i$ ($G_i$ arbitrary groups), and $H_i$ its projection on $G_i$. Let $K_i$ be the centralizer of $H_i$. Then the centralizer of $H$ is $prod_i K_i$. $Box$
Now to prove the claim, let $H$ be a subgroup with these properties. Then $Hsubsetprod H_i$, which has trivial intersection with the centralizer $K=prod K_i$ of $H$. Since by assumption $G=Htimes K$, we deduce that $H=prod H_i$. Since $H$ is directly indecomposable, $H_i$ is nontrivial for a single $i$. So $Hsubset H_i$, and $G_i=H_itimes K_i$. Since $G_i$ is directly indecomposable, we deduce that $H=H_i=G_i$.
answered 8 hours ago
YCorYCor
30.1k4 gold badges89 silver badges144 bronze badges
$endgroup$
$begingroup$
Thank you for the quick answer, which answers some other problem related to product of surfaces, which was asked at the 34th Summer Comference in Johannesburg.
$endgroup$
– Taras Banakh
8 hours ago
add a comment |
$begingroup$
Yes, and even in the group category. More generally, I claim that if $prod_{iin I}G_i$ and $prod_{jin J}H_j$ are isomorphic groups, for two sets $I,J$ and two families $(G_i)$, $(H_j)$ of groups that are center-free and directly indecomposable, then there is a bijection $f:Ito J$ such that $G_i$ and $H_{f(i)}$ are isomorphic for all $i$.
(Recall that a group is directly indecomposable if it's nontrivial and not direct product of two nontrivial subgroups.)
To show this, it is enough to recognize the subgroups $G_i$ in the product $G=prod_i G_i$, purely relying on the structure of the group $G$.
Indeed, $G_i$ is directly indecomposable, and $G$ is direct product of $G_i$ and its centralizer (using that $G_i$ has trivial center).
I claim that this characterizes the subgroups $G_i$.
Claim if $H$ is a directly indecomposable subgroup of $G$ such that $G$ is direct product of $H$ and its centralizer, then $H=G_i$ for some unique $i$.
Trivial lemma: Let $H$ be a subgroup of $G=prod G_i$ ($G_i$ arbitrary groups), and $H_i$ its projection on $G_i$. Let $K_i$ be the centralizer of $H_i$. Then the centralizer of $H$ is $prod_i K_i$. $Box$
Now to prove the claim, let $H$ be a subgroup with these properties. Then $Hsubsetprod H_i$, which has trivial intersection with the centralizer $K=prod K_i$ of $H$. Since by assumption $G=Htimes K$, we deduce that $H=prod H_i$. Since $H$ is directly indecomposable, $H_i$ is nontrivial for a single $i$. So $Hsubset H_i$, and $G_i=H_itimes K_i$. Since $G_i$ is directly indecomposable, we deduce that $H=H_i=G_i$.
answered 8 hours ago
YCorYCor
30.1k4 gold badges89 silver badges144 bronze badges
$endgroup$
Yes, and even in the group category. More generally, I claim that if $prod_{iin I}G_i$ and $prod_{jin J}H_j$ are isomorphic groups, for two sets $I,J$ and two families $(G_i)$, $(H_j)$ of groups that are center-free and directly indecomposable, then there is a bijection $f:Ito J$ such that $G_i$ and $H_{f(i)}$ are isomorphic for all $i$.
(Recall that a group is directly indecomposable if it's nontrivial and not direct product of two nontrivial subgroups.)
To show this, it is enough to recognize the subgroups $G_i$ in the product $G=prod_i G_i$, purely relying on the structure of the group $G$.
Indeed, $G_i$ is directly indecomposable, and $G$ is direct product of $G_i$ and its centralizer (using that $G_i$ has trivial center).
I claim that this characterizes the subgroups $G_i$.
Claim if $H$ is a directly indecomposable subgroup of $G$ such that $G$ is direct product of $H$ and its centralizer, then $H=G_i$ for some unique $i$.
Trivial lemma: Let $H$ be a subgroup of $G=prod G_i$ ($G_i$ arbitrary groups), and $H_i$ its projection on $G_i$. Let $K_i$ be the centralizer of $H_i$. Then the centralizer of $H$ is $prod_i K_i$. $Box$
Now to prove the claim, let $H$ be a subgroup with these properties. Then $Hsubsetprod H_i$, which has trivial intersection with the centralizer $K=prod K_i$ of $H$. Since by assumption $G=Htimes K$, we deduce that $H=prod H_i$. Since $H$ is directly indecomposable, $H_i$ is nontrivial for a single $i$. So $Hsubset H_i$, and $G_i=H_itimes K_i$. Since $G_i$ is directly indecomposable, we deduce that $H=H_i=G_i$.
answered 8 hours ago
YCorYCor
30.1k4 gold badges89 silver badges144 bronze badges
answered 8 hours ago
YCorYCor
30.1k4 gold badges89 silver badges144 bronze badges
answered 8 hours ago
YCorYCor
30.1k4 gold badges89 silver badges144 bronze badges
answered 8 hours ago
YCorYCor
30.1k4 gold badges89 silver badges144 bronze badges
30.1k4 gold badges89 silver badges144 bronze badges
$begingroup$
Thank you for the quick answer, which answers some other problem related to product of surfaces, which was asked at the 34th Summer Comference in Johannesburg.
$endgroup$
– Taras Banakh
8 hours ago
add a comment |
$begingroup$
Thank you for the quick answer, which answers some other problem related to product of surfaces, which was asked at the 34th Summer Comference in Johannesburg.
$endgroup$
– Taras Banakh
8 hours ago
$begingroup$
Thank you for the quick answer, which answers some other problem related to product of surfaces, which was asked at the 34th Summer Comference in Johannesburg.
$endgroup$
– Taras Banakh
8 hours ago
$begingroup$
Thank you for the quick answer, which answers some other problem related to product of surfaces, which was asked at the 34th Summer Comference in Johannesburg.
$endgroup$
– Taras Banakh
8 hours ago
add a comment |
Thanks for contributing an answer to MathOverflow!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f335723%2fthe-rigidity-of-the-countable-product-of-free-groups%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
