Word norm in compact subsets of finitely generated groups
up vote
1
down vote
favorite
Let $G$ be a finitely generated topological group, not necessarily discrete. Fix a finite generating set $S$ and denote by $|x|$ the word norm of $x in G$ with respect to this generating set, i.e., the minimal $n in mathbb{N}$ such that there exist $s_1, ldots, s_n in S$ with $x = s_1 cdots s_n$. Let $K$ be a compact subset of $G$. Is it true that ${ |k| : k in K } subseteq mathbb{N}$ is bounded?
abstract-algebra group-theory topological-groups geometric-group-theory
asked 10 hours ago
user404944
735211
add a comment |
up vote
1
down vote
favorite
Let $G$ be a finitely generated topological group, not necessarily discrete. Fix a finite generating set $S$ and denote by $|x|$ the word norm of $x in G$ with respect to this generating set, i.e., the minimal $n in mathbb{N}$ such that there exist $s_1, ldots, s_n in S$ with $x = s_1 cdots s_n$. Let $K$ be a compact subset of $G$. Is it true that ${ |k| : k in K } subseteq mathbb{N}$ is bounded?
abstract-algebra group-theory topological-groups geometric-group-theory
asked 10 hours ago
user404944
735211
Can you please provide the definition of "word norm"?
– uniquesolution
9 hours ago
I edited the question
– user404944
9 hours ago
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $G$ be a finitely generated topological group, not necessarily discrete. Fix a finite generating set $S$ and denote by $|x|$ the word norm of $x in G$ with respect to this generating set, i.e., the minimal $n in mathbb{N}$ such that there exist $s_1, ldots, s_n in S$ with $x = s_1 cdots s_n$. Let $K$ be a compact subset of $G$. Is it true that ${ |k| : k in K } subseteq mathbb{N}$ is bounded?
abstract-algebra group-theory topological-groups geometric-group-theory
asked 10 hours ago
user404944
735211
Let $G$ be a finitely generated topological group, not necessarily discrete. Fix a finite generating set $S$ and denote by $|x|$ the word norm of $x in G$ with respect to this generating set, i.e., the minimal $n in mathbb{N}$ such that there exist $s_1, ldots, s_n in S$ with $x = s_1 cdots s_n$. Let $K$ be a compact subset of $G$. Is it true that ${ |k| : k in K } subseteq mathbb{N}$ is bounded?
abstract-algebra group-theory topological-groups geometric-group-theory
abstract-algebra group-theory topological-groups geometric-group-theory
asked 10 hours ago
user404944
735211
asked 10 hours ago
user404944
735211
edited 9 hours ago
asked 10 hours ago
user404944
735211
asked 10 hours ago
user404944
735211
asked 10 hours ago
user404944
735211
735211
Can you please provide the definition of "word norm"?
– uniquesolution
9 hours ago
I edited the question
– user404944
9 hours ago
add a comment |
Can you please provide the definition of "word norm"?
– uniquesolution
9 hours ago
I edited the question
– user404944
9 hours ago
Can you please provide the definition of "word norm"?
– uniquesolution
9 hours ago
Can you please provide the definition of "word norm"?
– uniquesolution
9 hours ago
I edited the question
– user404944
9 hours ago
I edited the question
– user404944
9 hours ago
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
If I understood the question correctly, no.
Pick $G=mathbb{Z}$, $S={ 1 }$, but endow $G$ with the $p$-adic topology (for example $p=2$).
Then $K={p^k}_{k geq 0} cup {0}$ is a converging sequence and its endpoint, so is a compact subset, and clearly the lengths are not bounded on $K$.
answered 8 hours ago
user120527
1,631211
Great, that's what I imagined. Just one more question, is $mathbb{Z}$ locally compact with the $p$-adic topology? I know the definition but I am not that familiar with it...
– user404944
8 hours ago
It is not locally compact; I think a countable group cannot be locally compact unless the topology is discrete, it has to do with perfect sets being uncountable...
– user120527
8 hours ago
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
If I understood the question correctly, no.
Pick $G=mathbb{Z}$, $S={ 1 }$, but endow $G$ with the $p$-adic topology (for example $p=2$).
Then $K={p^k}_{k geq 0} cup {0}$ is a converging sequence and its endpoint, so is a compact subset, and clearly the lengths are not bounded on $K$.
answered 8 hours ago
user120527
1,631211
Great, that's what I imagined. Just one more question, is $mathbb{Z}$ locally compact with the $p$-adic topology? I know the definition but I am not that familiar with it...
– user404944
8 hours ago
It is not locally compact; I think a countable group cannot be locally compact unless the topology is discrete, it has to do with perfect sets being uncountable...
– user120527
8 hours ago
add a comment |
up vote
1
down vote
accepted
If I understood the question correctly, no.
Pick $G=mathbb{Z}$, $S={ 1 }$, but endow $G$ with the $p$-adic topology (for example $p=2$).
Then $K={p^k}_{k geq 0} cup {0}$ is a converging sequence and its endpoint, so is a compact subset, and clearly the lengths are not bounded on $K$.
answered 8 hours ago
user120527
1,631211
Great, that's what I imagined. Just one more question, is $mathbb{Z}$ locally compact with the $p$-adic topology? I know the definition but I am not that familiar with it...
– user404944
8 hours ago
It is not locally compact; I think a countable group cannot be locally compact unless the topology is discrete, it has to do with perfect sets being uncountable...
– user120527
8 hours ago
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
If I understood the question correctly, no.
Pick $G=mathbb{Z}$, $S={ 1 }$, but endow $G$ with the $p$-adic topology (for example $p=2$).
Then $K={p^k}_{k geq 0} cup {0}$ is a converging sequence and its endpoint, so is a compact subset, and clearly the lengths are not bounded on $K$.
answered 8 hours ago
user120527
1,631211
If I understood the question correctly, no.
Pick $G=mathbb{Z}$, $S={ 1 }$, but endow $G$ with the $p$-adic topology (for example $p=2$).
Then $K={p^k}_{k geq 0} cup {0}$ is a converging sequence and its endpoint, so is a compact subset, and clearly the lengths are not bounded on $K$.
answered 8 hours ago
user120527
1,631211
answered 8 hours ago
user120527
1,631211
answered 8 hours ago
user120527
1,631211
answered 8 hours ago
user120527
1,631211
1,631211
Great, that's what I imagined. Just one more question, is $mathbb{Z}$ locally compact with the $p$-adic topology? I know the definition but I am not that familiar with it...
– user404944
8 hours ago
It is not locally compact; I think a countable group cannot be locally compact unless the topology is discrete, it has to do with perfect sets being uncountable...
– user120527
8 hours ago
add a comment |
Great, that's what I imagined. Just one more question, is $mathbb{Z}$ locally compact with the $p$-adic topology? I know the definition but I am not that familiar with it...
– user404944
8 hours ago
It is not locally compact; I think a countable group cannot be locally compact unless the topology is discrete, it has to do with perfect sets being uncountable...
– user120527
8 hours ago
Great, that's what I imagined. Just one more question, is $mathbb{Z}$ locally compact with the $p$-adic topology? I know the definition but I am not that familiar with it...
– user404944
8 hours ago
Great, that's what I imagined. Just one more question, is $mathbb{Z}$ locally compact with the $p$-adic topology? I know the definition but I am not that familiar with it...
– user404944
8 hours ago
It is not locally compact; I think a countable group cannot be locally compact unless the topology is discrete, it has to do with perfect sets being uncountable...
– user120527
8 hours ago
It is not locally compact; I think a countable group cannot be locally compact unless the topology is discrete, it has to do with perfect sets being uncountable...
– user120527
8 hours ago
add a comment |
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
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2995339%2fword-norm-in-compact-subsets-of-finitely-generated-groups%23new-answer', 'question_page');
}
);
Post as a guest
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
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
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
Can you please provide the definition of "word norm"?
– uniquesolution
9 hours ago
I edited the question
– user404944
9 hours ago