Closed subgroups of abelian groupsEmbedded Lie subgroups are closed.The injectivity of torus in the category of abelian Lie groupsCenter of compact lie group closed?Closedness of connected semisimple Lie subgroups of semisimple groupsIntersection of a family of closed Lie subgroupsExamples about maximal Abelian subgroup is not a maximal torus in compact connected Lie group $G$.A question on abelian Lie groups and maximal compact subgroupReal and complex nilpotent Lie groupsClosed Subgroups of Lie Groups are closed Lie Subgroups?Lattice and abelian Lie groups
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Closed subgroups of abelian groups
Embedded Lie subgroups are closed.The injectivity of torus in the category of abelian Lie groupsCenter of compact lie group closed?Closedness of connected semisimple Lie subgroups of semisimple groupsIntersection of a family of closed Lie subgroupsExamples about maximal Abelian subgroup is not a maximal torus in compact connected Lie group $G$.A question on abelian Lie groups and maximal compact subgroupReal and complex nilpotent Lie groupsClosed Subgroups of Lie Groups are closed Lie Subgroups?Lattice and abelian Lie groups
$begingroup$
What is an example of an abelian Lie group $G$ and a closed subgroup $H$ such that $Gnotcong G/H times H$?
Would the circle $S^1$ in $mathbb R^2$ be an example? what is $mathbb R^2/S^1$?
general-topology differential-geometry lie-groups lie-algebras
edited 4 hours ago
Clayton
19.6k33288
asked 4 hours ago
Amrat AAmrat A
340111
$endgroup$
add a comment |
$begingroup$
What is an example of an abelian Lie group $G$ and a closed subgroup $H$ such that $Gnotcong G/H times H$?
Would the circle $S^1$ in $mathbb R^2$ be an example? what is $mathbb R^2/S^1$?
general-topology differential-geometry lie-groups lie-algebras
edited 4 hours ago
Clayton
19.6k33288
asked 4 hours ago
Amrat AAmrat A
340111
$endgroup$
1
$begingroup$
Isn't $G=mathbbR$ and $H=mathbbZ$ an example of the non-iso you want? All you need to show is that $mathbbR$ is not iso to $S^1 times mathbbZ$. That's easy.
$endgroup$
– Randall
2 hours ago
add a comment |
$begingroup$
What is an example of an abelian Lie group $G$ and a closed subgroup $H$ such that $Gnotcong G/H times H$?
Would the circle $S^1$ in $mathbb R^2$ be an example? what is $mathbb R^2/S^1$?
general-topology differential-geometry lie-groups lie-algebras
edited 4 hours ago
Clayton
19.6k33288
asked 4 hours ago
Amrat AAmrat A
340111
$endgroup$
What is an example of an abelian Lie group $G$ and a closed subgroup $H$ such that $Gnotcong G/H times H$?
Would the circle $S^1$ in $mathbb R^2$ be an example? what is $mathbb R^2/S^1$?
general-topology differential-geometry lie-groups lie-algebras
general-topology differential-geometry lie-groups lie-algebras
edited 4 hours ago
Clayton
19.6k33288
asked 4 hours ago
Amrat AAmrat A
340111
edited 4 hours ago
Clayton
19.6k33288
asked 4 hours ago
Amrat AAmrat A
340111
edited 4 hours ago
Clayton
19.6k33288
edited 4 hours ago
Clayton
19.6k33288
edited 4 hours ago
Clayton
19.6k33288
19.6k33288
asked 4 hours ago
Amrat AAmrat A
340111
asked 4 hours ago
Amrat AAmrat A
340111
asked 4 hours ago
Amrat AAmrat A
340111
340111
1
$begingroup$
Isn't $G=mathbbR$ and $H=mathbbZ$ an example of the non-iso you want? All you need to show is that $mathbbR$ is not iso to $S^1 times mathbbZ$. That's easy.
$endgroup$
– Randall
2 hours ago
add a comment |
1
$begingroup$
Isn't $G=mathbbR$ and $H=mathbbZ$ an example of the non-iso you want? All you need to show is that $mathbbR$ is not iso to $S^1 times mathbbZ$. That's easy.
$endgroup$
– Randall
2 hours ago
1
1
$begingroup$
Isn't $G=mathbbR$ and $H=mathbbZ$ an example of the non-iso you want? All you need to show is that $mathbbR$ is not iso to $S^1 times mathbbZ$. That's easy.
$endgroup$
– Randall
2 hours ago
$begingroup$
Isn't $G=mathbbR$ and $H=mathbbZ$ an example of the non-iso you want? All you need to show is that $mathbbR$ is not iso to $S^1 times mathbbZ$. That's easy.
$endgroup$
– Randall
2 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
EDIT: To be clear I was doing the case when $H$ was assumed connected. The disconnected case is handled below by Randall.
Every connected real abelian Lie group $G$ is isomorphic to $mathbbR^mtimes (S^1)^n$ for some $n$. In fact, given $G$ you can read off $n$ and $m$ as $n=mathrmrank(pi_1(G))$ and $m=dim G-n$.
Now, if you have a short exact sequence of abelian Lie groups
$$0to Hto Gto G/Hto 0$$
Then evidentily $dim G=dim H+dim G/H$. Moreover, since this is fibration, the groups are connected, and have vanishing second homotopy groups you also get a short exact sequence
$$0to pi_1(H)topi_1(G)topi_1(G/H)to 0$$
So, $mathrmrank(pi_1(G))=mathrmrank(pi_1(H))+mathrmrank(pi_1(G/H))$. Combining these two gives that $Gcong Htimes G/H$ as desired
EDIT: Here are more details. To show that $Gcong Htimes (G/H)$ it suffices to show that
$$mathrmrk(pi_1(G))=mathrmrk(pi_1(Htimes (G/H))=mathrmrk(pi_1(G))+mathrmrk(pi_1(G/H))$$
and
$$mathrmdim(G)-mathrmrk(pi_1(G))=dim(Gtimes (G/H))-mathrmrk(pi_1(Htimes (G/H))$$
The first equality holds by remark about the long exact sequence on homotopy groups from the fibration. The second is given as follows:
$$beginaligneddim(G)-mathrmrk(pi_1(G)) &= dim(H)+dim(G/H)-(mathrmrk(pi_1(H))+mathrmrk(pi_1(G/H))\ &= dim(Gtimes G/H))-mathrmrank(pi_1(Gtimes (G/H)))endaligned$$
(Below is for the non-abelian situation)
Here's a simple interesting example.
Take $mathrmGL_2(mathbbC)$ with its center $Z:=lambda I_2:lambdainmathbbC^times$. Then, $mathrmGL_2(mathbbC)/Zcong mathrmPGL_2(mathbbC)$. To see that $mathrmGL_2(mathbbC)notcong ZtimesmathrmPGL_2(mathbbC)$ note that the derived (i.e. commutative) subgroup of the former is $mathrmSL_2(mathbbC)$ whereas the latter is $mathrmPGL_2(mathbbC)$. Of course, these groups aren't isomorphic as the former is simply connected and the latter is not.
answered 3 hours ago
Alex YoucisAlex Youcis
36k775115
$endgroup$
$begingroup$
Thank you very much Alex. So the bundle $Gto G/H$ is always trivial!
$endgroup$
– Amrat A
3 hours ago
$begingroup$
@AmratA No problem. Did you see the updated affirmative answer to the abelian situation?
$endgroup$
– Alex Youcis
3 hours ago
$begingroup$
Oh yes, I just did. Thanks again!
$endgroup$
– Amrat A
3 hours ago
1
$begingroup$
@AmratA This is not true. Be careful, I didn't even necessarily claim that the fibration is trivial in my proof. I just proved that abstractly $Gcong Htimes (G/H)$, not that the sequence splits.
$endgroup$
– Alex Youcis
3 hours ago
1
$begingroup$
@AmratA Updated.
$endgroup$
– Alex Youcis
3 hours ago
|
show 5 more comments
$begingroup$
Take $G = mathbbR$ and $H=mathbbZ$. The quotient $G/H$ is the circle $S^1$. The question is now to compare $mathbbR$ to $S^1 times mathbbZ$. Now, whether you interpret $ncong$ as "not topologically iso" or "not group iso" doesn't matter, as this is a counterexample to both at once. Topologically they are distinct as $mathbbR$ is connected but $S^1 times mathbbZ$ is not (it's a stack of circles). Algebraically they're also distinct by looking at elements of order $2$ ($mathbbR$ has none, $S^1 times mathbbZ$ has at least one).
answered 1 hour ago
RandallRandall
10.7k11431
$endgroup$
2
$begingroup$
As I said in response to your comment, I generally only think about connected groups, so made that assumption (perhaps unfairly for the OP). The disconnected case as you've mentioned is quite obviously no. I edited my post to reflect this.
$endgroup$
– Alex Youcis
1 hour ago
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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votes
active
oldest
votes
$begingroup$
EDIT: To be clear I was doing the case when $H$ was assumed connected. The disconnected case is handled below by Randall.
Every connected real abelian Lie group $G$ is isomorphic to $mathbbR^mtimes (S^1)^n$ for some $n$. In fact, given $G$ you can read off $n$ and $m$ as $n=mathrmrank(pi_1(G))$ and $m=dim G-n$.
Now, if you have a short exact sequence of abelian Lie groups
$$0to Hto Gto G/Hto 0$$
Then evidentily $dim G=dim H+dim G/H$. Moreover, since this is fibration, the groups are connected, and have vanishing second homotopy groups you also get a short exact sequence
$$0to pi_1(H)topi_1(G)topi_1(G/H)to 0$$
So, $mathrmrank(pi_1(G))=mathrmrank(pi_1(H))+mathrmrank(pi_1(G/H))$. Combining these two gives that $Gcong Htimes G/H$ as desired
EDIT: Here are more details. To show that $Gcong Htimes (G/H)$ it suffices to show that
$$mathrmrk(pi_1(G))=mathrmrk(pi_1(Htimes (G/H))=mathrmrk(pi_1(G))+mathrmrk(pi_1(G/H))$$
and
$$mathrmdim(G)-mathrmrk(pi_1(G))=dim(Gtimes (G/H))-mathrmrk(pi_1(Htimes (G/H))$$
The first equality holds by remark about the long exact sequence on homotopy groups from the fibration. The second is given as follows:
$$beginaligneddim(G)-mathrmrk(pi_1(G)) &= dim(H)+dim(G/H)-(mathrmrk(pi_1(H))+mathrmrk(pi_1(G/H))\ &= dim(Gtimes G/H))-mathrmrank(pi_1(Gtimes (G/H)))endaligned$$
(Below is for the non-abelian situation)
Here's a simple interesting example.
Take $mathrmGL_2(mathbbC)$ with its center $Z:=lambda I_2:lambdainmathbbC^times$. Then, $mathrmGL_2(mathbbC)/Zcong mathrmPGL_2(mathbbC)$. To see that $mathrmGL_2(mathbbC)notcong ZtimesmathrmPGL_2(mathbbC)$ note that the derived (i.e. commutative) subgroup of the former is $mathrmSL_2(mathbbC)$ whereas the latter is $mathrmPGL_2(mathbbC)$. Of course, these groups aren't isomorphic as the former is simply connected and the latter is not.
answered 3 hours ago
Alex YoucisAlex Youcis
36k775115
$endgroup$
$begingroup$
Thank you very much Alex. So the bundle $Gto G/H$ is always trivial!
$endgroup$
– Amrat A
3 hours ago
$begingroup$
@AmratA No problem. Did you see the updated affirmative answer to the abelian situation?
$endgroup$
– Alex Youcis
3 hours ago
$begingroup$
Oh yes, I just did. Thanks again!
$endgroup$
– Amrat A
3 hours ago
1
$begingroup$
@AmratA This is not true. Be careful, I didn't even necessarily claim that the fibration is trivial in my proof. I just proved that abstractly $Gcong Htimes (G/H)$, not that the sequence splits.
$endgroup$
– Alex Youcis
3 hours ago
1
$begingroup$
@AmratA Updated.
$endgroup$
– Alex Youcis
3 hours ago
|
show 5 more comments
$begingroup$
EDIT: To be clear I was doing the case when $H$ was assumed connected. The disconnected case is handled below by Randall.
Every connected real abelian Lie group $G$ is isomorphic to $mathbbR^mtimes (S^1)^n$ for some $n$. In fact, given $G$ you can read off $n$ and $m$ as $n=mathrmrank(pi_1(G))$ and $m=dim G-n$.
Now, if you have a short exact sequence of abelian Lie groups
$$0to Hto Gto G/Hto 0$$
Then evidentily $dim G=dim H+dim G/H$. Moreover, since this is fibration, the groups are connected, and have vanishing second homotopy groups you also get a short exact sequence
$$0to pi_1(H)topi_1(G)topi_1(G/H)to 0$$
So, $mathrmrank(pi_1(G))=mathrmrank(pi_1(H))+mathrmrank(pi_1(G/H))$. Combining these two gives that $Gcong Htimes G/H$ as desired
EDIT: Here are more details. To show that $Gcong Htimes (G/H)$ it suffices to show that
$$mathrmrk(pi_1(G))=mathrmrk(pi_1(Htimes (G/H))=mathrmrk(pi_1(G))+mathrmrk(pi_1(G/H))$$
and
$$mathrmdim(G)-mathrmrk(pi_1(G))=dim(Gtimes (G/H))-mathrmrk(pi_1(Htimes (G/H))$$
The first equality holds by remark about the long exact sequence on homotopy groups from the fibration. The second is given as follows:
$$beginaligneddim(G)-mathrmrk(pi_1(G)) &= dim(H)+dim(G/H)-(mathrmrk(pi_1(H))+mathrmrk(pi_1(G/H))\ &= dim(Gtimes G/H))-mathrmrank(pi_1(Gtimes (G/H)))endaligned$$
(Below is for the non-abelian situation)
Here's a simple interesting example.
Take $mathrmGL_2(mathbbC)$ with its center $Z:=lambda I_2:lambdainmathbbC^times$. Then, $mathrmGL_2(mathbbC)/Zcong mathrmPGL_2(mathbbC)$. To see that $mathrmGL_2(mathbbC)notcong ZtimesmathrmPGL_2(mathbbC)$ note that the derived (i.e. commutative) subgroup of the former is $mathrmSL_2(mathbbC)$ whereas the latter is $mathrmPGL_2(mathbbC)$. Of course, these groups aren't isomorphic as the former is simply connected and the latter is not.
answered 3 hours ago
Alex YoucisAlex Youcis
36k775115
$endgroup$
$begingroup$
Thank you very much Alex. So the bundle $Gto G/H$ is always trivial!
$endgroup$
– Amrat A
3 hours ago
$begingroup$
@AmratA No problem. Did you see the updated affirmative answer to the abelian situation?
$endgroup$
– Alex Youcis
3 hours ago
$begingroup$
Oh yes, I just did. Thanks again!
$endgroup$
– Amrat A
3 hours ago
1
$begingroup$
@AmratA This is not true. Be careful, I didn't even necessarily claim that the fibration is trivial in my proof. I just proved that abstractly $Gcong Htimes (G/H)$, not that the sequence splits.
$endgroup$
– Alex Youcis
3 hours ago
1
$begingroup$
@AmratA Updated.
$endgroup$
– Alex Youcis
3 hours ago
|
show 5 more comments
$begingroup$
EDIT: To be clear I was doing the case when $H$ was assumed connected. The disconnected case is handled below by Randall.
Every connected real abelian Lie group $G$ is isomorphic to $mathbbR^mtimes (S^1)^n$ for some $n$. In fact, given $G$ you can read off $n$ and $m$ as $n=mathrmrank(pi_1(G))$ and $m=dim G-n$.
Now, if you have a short exact sequence of abelian Lie groups
$$0to Hto Gto G/Hto 0$$
Then evidentily $dim G=dim H+dim G/H$. Moreover, since this is fibration, the groups are connected, and have vanishing second homotopy groups you also get a short exact sequence
$$0to pi_1(H)topi_1(G)topi_1(G/H)to 0$$
So, $mathrmrank(pi_1(G))=mathrmrank(pi_1(H))+mathrmrank(pi_1(G/H))$. Combining these two gives that $Gcong Htimes G/H$ as desired
EDIT: Here are more details. To show that $Gcong Htimes (G/H)$ it suffices to show that
$$mathrmrk(pi_1(G))=mathrmrk(pi_1(Htimes (G/H))=mathrmrk(pi_1(G))+mathrmrk(pi_1(G/H))$$
and
$$mathrmdim(G)-mathrmrk(pi_1(G))=dim(Gtimes (G/H))-mathrmrk(pi_1(Htimes (G/H))$$
The first equality holds by remark about the long exact sequence on homotopy groups from the fibration. The second is given as follows:
$$beginaligneddim(G)-mathrmrk(pi_1(G)) &= dim(H)+dim(G/H)-(mathrmrk(pi_1(H))+mathrmrk(pi_1(G/H))\ &= dim(Gtimes G/H))-mathrmrank(pi_1(Gtimes (G/H)))endaligned$$
(Below is for the non-abelian situation)
Here's a simple interesting example.
Take $mathrmGL_2(mathbbC)$ with its center $Z:=lambda I_2:lambdainmathbbC^times$. Then, $mathrmGL_2(mathbbC)/Zcong mathrmPGL_2(mathbbC)$. To see that $mathrmGL_2(mathbbC)notcong ZtimesmathrmPGL_2(mathbbC)$ note that the derived (i.e. commutative) subgroup of the former is $mathrmSL_2(mathbbC)$ whereas the latter is $mathrmPGL_2(mathbbC)$. Of course, these groups aren't isomorphic as the former is simply connected and the latter is not.
answered 3 hours ago
Alex YoucisAlex Youcis
36k775115
$endgroup$
EDIT: To be clear I was doing the case when $H$ was assumed connected. The disconnected case is handled below by Randall.
Every connected real abelian Lie group $G$ is isomorphic to $mathbbR^mtimes (S^1)^n$ for some $n$. In fact, given $G$ you can read off $n$ and $m$ as $n=mathrmrank(pi_1(G))$ and $m=dim G-n$.
Now, if you have a short exact sequence of abelian Lie groups
$$0to Hto Gto G/Hto 0$$
Then evidentily $dim G=dim H+dim G/H$. Moreover, since this is fibration, the groups are connected, and have vanishing second homotopy groups you also get a short exact sequence
$$0to pi_1(H)topi_1(G)topi_1(G/H)to 0$$
So, $mathrmrank(pi_1(G))=mathrmrank(pi_1(H))+mathrmrank(pi_1(G/H))$. Combining these two gives that $Gcong Htimes G/H$ as desired
EDIT: Here are more details. To show that $Gcong Htimes (G/H)$ it suffices to show that
$$mathrmrk(pi_1(G))=mathrmrk(pi_1(Htimes (G/H))=mathrmrk(pi_1(G))+mathrmrk(pi_1(G/H))$$
and
$$mathrmdim(G)-mathrmrk(pi_1(G))=dim(Gtimes (G/H))-mathrmrk(pi_1(Htimes (G/H))$$
The first equality holds by remark about the long exact sequence on homotopy groups from the fibration. The second is given as follows:
$$beginaligneddim(G)-mathrmrk(pi_1(G)) &= dim(H)+dim(G/H)-(mathrmrk(pi_1(H))+mathrmrk(pi_1(G/H))\ &= dim(Gtimes G/H))-mathrmrank(pi_1(Gtimes (G/H)))endaligned$$
(Below is for the non-abelian situation)
Here's a simple interesting example.
Take $mathrmGL_2(mathbbC)$ with its center $Z:=lambda I_2:lambdainmathbbC^times$. Then, $mathrmGL_2(mathbbC)/Zcong mathrmPGL_2(mathbbC)$. To see that $mathrmGL_2(mathbbC)notcong ZtimesmathrmPGL_2(mathbbC)$ note that the derived (i.e. commutative) subgroup of the former is $mathrmSL_2(mathbbC)$ whereas the latter is $mathrmPGL_2(mathbbC)$. Of course, these groups aren't isomorphic as the former is simply connected and the latter is not.
answered 3 hours ago
Alex YoucisAlex Youcis
36k775115
edited 1 hour ago
answered 3 hours ago
Alex YoucisAlex Youcis
36k775115
answered 3 hours ago
Alex YoucisAlex Youcis
36k775115
answered 3 hours ago
Alex YoucisAlex Youcis
36k775115
36k775115
$begingroup$
Thank you very much Alex. So the bundle $Gto G/H$ is always trivial!
$endgroup$
– Amrat A
3 hours ago
$begingroup$
@AmratA No problem. Did you see the updated affirmative answer to the abelian situation?
$endgroup$
– Alex Youcis
3 hours ago
$begingroup$
Oh yes, I just did. Thanks again!
$endgroup$
– Amrat A
3 hours ago
1
$begingroup$
@AmratA This is not true. Be careful, I didn't even necessarily claim that the fibration is trivial in my proof. I just proved that abstractly $Gcong Htimes (G/H)$, not that the sequence splits.
$endgroup$
– Alex Youcis
3 hours ago
1
$begingroup$
@AmratA Updated.
$endgroup$
– Alex Youcis
3 hours ago
|
show 5 more comments
$begingroup$
Thank you very much Alex. So the bundle $Gto G/H$ is always trivial!
$endgroup$
– Amrat A
3 hours ago
$begingroup$
@AmratA No problem. Did you see the updated affirmative answer to the abelian situation?
$endgroup$
– Alex Youcis
3 hours ago
$begingroup$
Oh yes, I just did. Thanks again!
$endgroup$
– Amrat A
3 hours ago
1
$begingroup$
@AmratA This is not true. Be careful, I didn't even necessarily claim that the fibration is trivial in my proof. I just proved that abstractly $Gcong Htimes (G/H)$, not that the sequence splits.
$endgroup$
– Alex Youcis
3 hours ago
1
$begingroup$
@AmratA Updated.
$endgroup$
– Alex Youcis
3 hours ago
$begingroup$
Thank you very much Alex. So the bundle $Gto G/H$ is always trivial!
$endgroup$
– Amrat A
3 hours ago
$begingroup$
Thank you very much Alex. So the bundle $Gto G/H$ is always trivial!
$endgroup$
– Amrat A
3 hours ago
$begingroup$
@AmratA No problem. Did you see the updated affirmative answer to the abelian situation?
$endgroup$
– Alex Youcis
3 hours ago
$begingroup$
@AmratA No problem. Did you see the updated affirmative answer to the abelian situation?
$endgroup$
– Alex Youcis
3 hours ago
$begingroup$
Oh yes, I just did. Thanks again!
$endgroup$
– Amrat A
3 hours ago
$begingroup$
Oh yes, I just did. Thanks again!
$endgroup$
– Amrat A
3 hours ago
1
1
$begingroup$
@AmratA This is not true. Be careful, I didn't even necessarily claim that the fibration is trivial in my proof. I just proved that abstractly $Gcong Htimes (G/H)$, not that the sequence splits.
$endgroup$
– Alex Youcis
3 hours ago
$begingroup$
@AmratA This is not true. Be careful, I didn't even necessarily claim that the fibration is trivial in my proof. I just proved that abstractly $Gcong Htimes (G/H)$, not that the sequence splits.
$endgroup$
– Alex Youcis
3 hours ago
1
1
$begingroup$
@AmratA Updated.
$endgroup$
– Alex Youcis
3 hours ago
$begingroup$
@AmratA Updated.
$endgroup$
– Alex Youcis
3 hours ago
|
show 5 more comments
$begingroup$
Take $G = mathbbR$ and $H=mathbbZ$. The quotient $G/H$ is the circle $S^1$. The question is now to compare $mathbbR$ to $S^1 times mathbbZ$. Now, whether you interpret $ncong$ as "not topologically iso" or "not group iso" doesn't matter, as this is a counterexample to both at once. Topologically they are distinct as $mathbbR$ is connected but $S^1 times mathbbZ$ is not (it's a stack of circles). Algebraically they're also distinct by looking at elements of order $2$ ($mathbbR$ has none, $S^1 times mathbbZ$ has at least one).
answered 1 hour ago
RandallRandall
10.7k11431
$endgroup$
2
$begingroup$
As I said in response to your comment, I generally only think about connected groups, so made that assumption (perhaps unfairly for the OP). The disconnected case as you've mentioned is quite obviously no. I edited my post to reflect this.
$endgroup$
– Alex Youcis
1 hour ago
add a comment |
$begingroup$
Take $G = mathbbR$ and $H=mathbbZ$. The quotient $G/H$ is the circle $S^1$. The question is now to compare $mathbbR$ to $S^1 times mathbbZ$. Now, whether you interpret $ncong$ as "not topologically iso" or "not group iso" doesn't matter, as this is a counterexample to both at once. Topologically they are distinct as $mathbbR$ is connected but $S^1 times mathbbZ$ is not (it's a stack of circles). Algebraically they're also distinct by looking at elements of order $2$ ($mathbbR$ has none, $S^1 times mathbbZ$ has at least one).
answered 1 hour ago
RandallRandall
10.7k11431
$endgroup$
2
$begingroup$
As I said in response to your comment, I generally only think about connected groups, so made that assumption (perhaps unfairly for the OP). The disconnected case as you've mentioned is quite obviously no. I edited my post to reflect this.
$endgroup$
– Alex Youcis
1 hour ago
add a comment |
$begingroup$
Take $G = mathbbR$ and $H=mathbbZ$. The quotient $G/H$ is the circle $S^1$. The question is now to compare $mathbbR$ to $S^1 times mathbbZ$. Now, whether you interpret $ncong$ as "not topologically iso" or "not group iso" doesn't matter, as this is a counterexample to both at once. Topologically they are distinct as $mathbbR$ is connected but $S^1 times mathbbZ$ is not (it's a stack of circles). Algebraically they're also distinct by looking at elements of order $2$ ($mathbbR$ has none, $S^1 times mathbbZ$ has at least one).
answered 1 hour ago
RandallRandall
10.7k11431
$endgroup$
Take $G = mathbbR$ and $H=mathbbZ$. The quotient $G/H$ is the circle $S^1$. The question is now to compare $mathbbR$ to $S^1 times mathbbZ$. Now, whether you interpret $ncong$ as "not topologically iso" or "not group iso" doesn't matter, as this is a counterexample to both at once. Topologically they are distinct as $mathbbR$ is connected but $S^1 times mathbbZ$ is not (it's a stack of circles). Algebraically they're also distinct by looking at elements of order $2$ ($mathbbR$ has none, $S^1 times mathbbZ$ has at least one).
answered 1 hour ago
RandallRandall
10.7k11431
answered 1 hour ago
RandallRandall
10.7k11431
answered 1 hour ago
RandallRandall
10.7k11431
answered 1 hour ago
RandallRandall
10.7k11431
10.7k11431
2
$begingroup$
As I said in response to your comment, I generally only think about connected groups, so made that assumption (perhaps unfairly for the OP). The disconnected case as you've mentioned is quite obviously no. I edited my post to reflect this.
$endgroup$
– Alex Youcis
1 hour ago
add a comment |
2
$begingroup$
As I said in response to your comment, I generally only think about connected groups, so made that assumption (perhaps unfairly for the OP). The disconnected case as you've mentioned is quite obviously no. I edited my post to reflect this.
$endgroup$
– Alex Youcis
1 hour ago
2
2
$begingroup$
As I said in response to your comment, I generally only think about connected groups, so made that assumption (perhaps unfairly for the OP). The disconnected case as you've mentioned is quite obviously no. I edited my post to reflect this.
$endgroup$
– Alex Youcis
1 hour ago
$begingroup$
As I said in response to your comment, I generally only think about connected groups, so made that assumption (perhaps unfairly for the OP). The disconnected case as you've mentioned is quite obviously no. I edited my post to reflect this.
$endgroup$
– Alex Youcis
1 hour ago
add a comment |
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$begingroup$
Isn't $G=mathbbR$ and $H=mathbbZ$ an example of the non-iso you want? All you need to show is that $mathbbR$ is not iso to $S^1 times mathbbZ$. That's easy.
$endgroup$
– Randall
2 hours ago