Word norm in compact subsets of finitely generated groups
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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
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up vote
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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
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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 |
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D,FGp6x
Can you please provide the definition of "word norm"?
– uniquesolution
9 hours ago
I edited the question
– user404944
9 hours ago