Definition of infinite cyclic groupWhat happens to the 0 element in a Finite Group?Polyhedra with symmetries order threeProve the isomorphism of cyclic groups $C_mncong C_mtimes C_n$ via categorical considerationsConfused on group notationFour groups of order 20 that are not isomorphicProve: Every cyclic group with order > 2 has at least 2 distinct generatorsCardinality of set of groupsFinite matrix groups as subgroups of $S_n$.Group Theory-IsomorphismsWhat is the definition of the least common multiple of elements of a group rather than their orders?
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Definition of infinite cyclic group
What happens to the 0 element in a Finite Group?Polyhedra with symmetries order threeProve the isomorphism of cyclic groups $C_mncong C_mtimes C_n$ via categorical considerationsConfused on group notationFour groups of order 20 that are not isomorphicProve: Every cyclic group with order > 2 has at least 2 distinct generatorsCardinality of set of groupsFinite matrix groups as subgroups of $S_n$.Group Theory-IsomorphismsWhat is the definition of the least common multiple of elements of a group rather than their orders?
$begingroup$
I'm having some conceptual issues with the infinite cyclic group $C_infty$. Finite groups $C_n$ have a clear representation as integers $0,1,cdots,n-1$ under addition $(operatornamemod n)$, or as the rotation group of the $n$-gon for $ngeq 3$. The rotation group of a circle, which is what I interpreted $C_infty$ to be, has uncountable order since any real angle $[0,2pi)$ is valid. This would make it isomorphic to $[0,2pi)$ under addition $(operatornamemod 2pi)$. But online it says $(mathbbZ,+)$ is also isomorphic, which doesn't make sense to me because it has order $aleph_0$. Also, the first group has two inverses $0$ and $pi$, while this group only has $0$.
I'm guessing my interpretation is wrong. The textbook never defines what $C_infty$. What exactly is it?
group-theory
asked 8 hours ago
Ovinus RealOvinus Real
365
New contributor
Ovinus Real is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I'm having some conceptual issues with the infinite cyclic group $C_infty$. Finite groups $C_n$ have a clear representation as integers $0,1,cdots,n-1$ under addition $(operatornamemod n)$, or as the rotation group of the $n$-gon for $ngeq 3$. The rotation group of a circle, which is what I interpreted $C_infty$ to be, has uncountable order since any real angle $[0,2pi)$ is valid. This would make it isomorphic to $[0,2pi)$ under addition $(operatornamemod 2pi)$. But online it says $(mathbbZ,+)$ is also isomorphic, which doesn't make sense to me because it has order $aleph_0$. Also, the first group has two inverses $0$ and $pi$, while this group only has $0$.
I'm guessing my interpretation is wrong. The textbook never defines what $C_infty$. What exactly is it?
group-theory
asked 8 hours ago
Ovinus RealOvinus Real
365
New contributor
Ovinus Real is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
"The rotation group of a circle, which is what I interpreted $C_infty$ to be". Well, there we are.
$endgroup$
– Lord Shark the Unknown
8 hours ago
$begingroup$
Is there a good geometric interpretation then? The best one I can think of is rotating a circle by an integer number of radians.
$endgroup$
– Ovinus Real
8 hours ago
$begingroup$
The order of the rotation group of a circle is $2^aleph_0$. Whether or not this is $aleph_1$ is not decidable from the axioms of set theory.
$endgroup$
– DanielWainfleet
8 hours ago
$begingroup$
Should have just said uncountable, sorry
$endgroup$
– Ovinus Real
8 hours ago
add a comment |
$begingroup$
I'm having some conceptual issues with the infinite cyclic group $C_infty$. Finite groups $C_n$ have a clear representation as integers $0,1,cdots,n-1$ under addition $(operatornamemod n)$, or as the rotation group of the $n$-gon for $ngeq 3$. The rotation group of a circle, which is what I interpreted $C_infty$ to be, has uncountable order since any real angle $[0,2pi)$ is valid. This would make it isomorphic to $[0,2pi)$ under addition $(operatornamemod 2pi)$. But online it says $(mathbbZ,+)$ is also isomorphic, which doesn't make sense to me because it has order $aleph_0$. Also, the first group has two inverses $0$ and $pi$, while this group only has $0$.
I'm guessing my interpretation is wrong. The textbook never defines what $C_infty$. What exactly is it?
group-theory
asked 8 hours ago
Ovinus RealOvinus Real
365
New contributor
Ovinus Real is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I'm having some conceptual issues with the infinite cyclic group $C_infty$. Finite groups $C_n$ have a clear representation as integers $0,1,cdots,n-1$ under addition $(operatornamemod n)$, or as the rotation group of the $n$-gon for $ngeq 3$. The rotation group of a circle, which is what I interpreted $C_infty$ to be, has uncountable order since any real angle $[0,2pi)$ is valid. This would make it isomorphic to $[0,2pi)$ under addition $(operatornamemod 2pi)$. But online it says $(mathbbZ,+)$ is also isomorphic, which doesn't make sense to me because it has order $aleph_0$. Also, the first group has two inverses $0$ and $pi$, while this group only has $0$.
I'm guessing my interpretation is wrong. The textbook never defines what $C_infty$. What exactly is it?
group-theory
group-theory
asked 8 hours ago
Ovinus RealOvinus Real
365
New contributor
Ovinus Real is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 8 hours ago
Ovinus RealOvinus Real
365
New contributor
Ovinus Real is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 8 hours ago
Ovinus Real
asked 8 hours ago
Ovinus RealOvinus Real
365
New contributor
Ovinus Real is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 8 hours ago
Ovinus RealOvinus Real
365
asked 8 hours ago
Ovinus RealOvinus Real
365
365
New contributor
Ovinus Real is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Ovinus Real is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
"The rotation group of a circle, which is what I interpreted $C_infty$ to be". Well, there we are.
$endgroup$
– Lord Shark the Unknown
8 hours ago
$begingroup$
Is there a good geometric interpretation then? The best one I can think of is rotating a circle by an integer number of radians.
$endgroup$
– Ovinus Real
8 hours ago
$begingroup$
The order of the rotation group of a circle is $2^aleph_0$. Whether or not this is $aleph_1$ is not decidable from the axioms of set theory.
$endgroup$
– DanielWainfleet
8 hours ago
$begingroup$
Should have just said uncountable, sorry
$endgroup$
– Ovinus Real
8 hours ago
add a comment |
$begingroup$
"The rotation group of a circle, which is what I interpreted $C_infty$ to be". Well, there we are.
$endgroup$
– Lord Shark the Unknown
8 hours ago
$begingroup$
Is there a good geometric interpretation then? The best one I can think of is rotating a circle by an integer number of radians.
$endgroup$
– Ovinus Real
8 hours ago
$begingroup$
The order of the rotation group of a circle is $2^aleph_0$. Whether or not this is $aleph_1$ is not decidable from the axioms of set theory.
$endgroup$
– DanielWainfleet
8 hours ago
$begingroup$
Should have just said uncountable, sorry
$endgroup$
– Ovinus Real
8 hours ago
$begingroup$
"The rotation group of a circle, which is what I interpreted $C_infty$ to be". Well, there we are.
$endgroup$
– Lord Shark the Unknown
8 hours ago
$begingroup$
"The rotation group of a circle, which is what I interpreted $C_infty$ to be". Well, there we are.
$endgroup$
– Lord Shark the Unknown
8 hours ago
$begingroup$
Is there a good geometric interpretation then? The best one I can think of is rotating a circle by an integer number of radians.
$endgroup$
– Ovinus Real
8 hours ago
$begingroup$
Is there a good geometric interpretation then? The best one I can think of is rotating a circle by an integer number of radians.
$endgroup$
– Ovinus Real
8 hours ago
$begingroup$
The order of the rotation group of a circle is $2^aleph_0$. Whether or not this is $aleph_1$ is not decidable from the axioms of set theory.
$endgroup$
– DanielWainfleet
8 hours ago
$begingroup$
The order of the rotation group of a circle is $2^aleph_0$. Whether or not this is $aleph_1$ is not decidable from the axioms of set theory.
$endgroup$
– DanielWainfleet
8 hours ago
$begingroup$
Should have just said uncountable, sorry
$endgroup$
– Ovinus Real
8 hours ago
$begingroup$
Should have just said uncountable, sorry
$endgroup$
– Ovinus Real
8 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The (up to isomorphism) infinite cyclic group is just $mathbbZ$ under addition.
You can visualize it as the group of integer shifts of the integers.
You can also visualize it as the rotations of the circle through integer numbers of radians, but that's not pretty geometrically since the orbit of any point on the circle is dense.
The group of all rotations of the circle is not cyclic.
answered 8 hours ago
Ethan BolkerEthan Bolker
49k556125
$endgroup$
add a comment |
$begingroup$
A group is by definition cyclic if
$$G = langle g rangle = g^n mid n in mathbbZ$$
for some $g in G$.
Being infinite, $g$ has infinite order and $(g^n)_n in mathbbZ$ is a list of different elements in $G$.
Hence, we have a natural isomorphism
$$(mathbbZ,+) to G: n mapsto g^n$$
and thus a cyclic group is just $mathbbZ$ with addition.
answered 7 hours ago
EpsilonDeltaEpsilonDelta
39819
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
The (up to isomorphism) infinite cyclic group is just $mathbbZ$ under addition.
You can visualize it as the group of integer shifts of the integers.
You can also visualize it as the rotations of the circle through integer numbers of radians, but that's not pretty geometrically since the orbit of any point on the circle is dense.
The group of all rotations of the circle is not cyclic.
answered 8 hours ago
Ethan BolkerEthan Bolker
49k556125
$endgroup$
add a comment |
$begingroup$
The (up to isomorphism) infinite cyclic group is just $mathbbZ$ under addition.
You can visualize it as the group of integer shifts of the integers.
You can also visualize it as the rotations of the circle through integer numbers of radians, but that's not pretty geometrically since the orbit of any point on the circle is dense.
The group of all rotations of the circle is not cyclic.
answered 8 hours ago
Ethan BolkerEthan Bolker
49k556125
$endgroup$
add a comment |
$begingroup$
The (up to isomorphism) infinite cyclic group is just $mathbbZ$ under addition.
You can visualize it as the group of integer shifts of the integers.
You can also visualize it as the rotations of the circle through integer numbers of radians, but that's not pretty geometrically since the orbit of any point on the circle is dense.
The group of all rotations of the circle is not cyclic.
answered 8 hours ago
Ethan BolkerEthan Bolker
49k556125
$endgroup$
The (up to isomorphism) infinite cyclic group is just $mathbbZ$ under addition.
You can visualize it as the group of integer shifts of the integers.
You can also visualize it as the rotations of the circle through integer numbers of radians, but that's not pretty geometrically since the orbit of any point on the circle is dense.
The group of all rotations of the circle is not cyclic.
answered 8 hours ago
Ethan BolkerEthan Bolker
49k556125
answered 8 hours ago
Ethan BolkerEthan Bolker
49k556125
answered 8 hours ago
Ethan BolkerEthan Bolker
49k556125
answered 8 hours ago
Ethan BolkerEthan Bolker
49k556125
49k556125
add a comment |
add a comment |
$begingroup$
A group is by definition cyclic if
$$G = langle g rangle = g^n mid n in mathbbZ$$
for some $g in G$.
Being infinite, $g$ has infinite order and $(g^n)_n in mathbbZ$ is a list of different elements in $G$.
Hence, we have a natural isomorphism
$$(mathbbZ,+) to G: n mapsto g^n$$
and thus a cyclic group is just $mathbbZ$ with addition.
answered 7 hours ago
EpsilonDeltaEpsilonDelta
39819
$endgroup$
add a comment |
$begingroup$
A group is by definition cyclic if
$$G = langle g rangle = g^n mid n in mathbbZ$$
for some $g in G$.
Being infinite, $g$ has infinite order and $(g^n)_n in mathbbZ$ is a list of different elements in $G$.
Hence, we have a natural isomorphism
$$(mathbbZ,+) to G: n mapsto g^n$$
and thus a cyclic group is just $mathbbZ$ with addition.
answered 7 hours ago
EpsilonDeltaEpsilonDelta
39819
$endgroup$
add a comment |
$begingroup$
A group is by definition cyclic if
$$G = langle g rangle = g^n mid n in mathbbZ$$
for some $g in G$.
Being infinite, $g$ has infinite order and $(g^n)_n in mathbbZ$ is a list of different elements in $G$.
Hence, we have a natural isomorphism
$$(mathbbZ,+) to G: n mapsto g^n$$
and thus a cyclic group is just $mathbbZ$ with addition.
answered 7 hours ago
EpsilonDeltaEpsilonDelta
39819
$endgroup$
A group is by definition cyclic if
$$G = langle g rangle = g^n mid n in mathbbZ$$
for some $g in G$.
Being infinite, $g$ has infinite order and $(g^n)_n in mathbbZ$ is a list of different elements in $G$.
Hence, we have a natural isomorphism
$$(mathbbZ,+) to G: n mapsto g^n$$
and thus a cyclic group is just $mathbbZ$ with addition.
answered 7 hours ago
EpsilonDeltaEpsilonDelta
39819
answered 7 hours ago
EpsilonDeltaEpsilonDelta
39819
answered 7 hours ago
EpsilonDeltaEpsilonDelta
39819
answered 7 hours ago
EpsilonDeltaEpsilonDelta
39819
39819
add a comment |
add a comment |
Ovinus Real is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
"The rotation group of a circle, which is what I interpreted $C_infty$ to be". Well, there we are.
$endgroup$
– Lord Shark the Unknown
8 hours ago
$begingroup$
Is there a good geometric interpretation then? The best one I can think of is rotating a circle by an integer number of radians.
$endgroup$
– Ovinus Real
8 hours ago
$begingroup$
The order of the rotation group of a circle is $2^aleph_0$. Whether or not this is $aleph_1$ is not decidable from the axioms of set theory.
$endgroup$
– DanielWainfleet
8 hours ago
$begingroup$
Should have just said uncountable, sorry
$endgroup$
– Ovinus Real
8 hours ago