Closed subgroups of abelian groups












1












$begingroup$


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 $mathbb R^2/S^1$?










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$endgroup$








  • 1




    $begingroup$
    Isn't $G=mathbb{R}$ and $H=mathbb{Z}$ an example of the non-iso you want? All you need to show is that $mathbb{R}$ is not iso to $S^1 times mathbb{Z}$. That's easy.
    $endgroup$
    – Randall
    2 hours ago


















1












$begingroup$


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 $mathbb R^2/S^1$?










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Isn't $G=mathbb{R}$ and $H=mathbb{Z}$ an example of the non-iso you want? All you need to show is that $mathbb{R}$ is not iso to $S^1 times mathbb{Z}$. That's easy.
    $endgroup$
    – Randall
    2 hours ago
















1












1








1





$begingroup$


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 $mathbb R^2/S^1$?










share|cite|improve this question











$endgroup$




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






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share|cite|improve this question













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share|cite|improve this question








edited 3 hours ago









Clayton

19.6k33288




19.6k33288










asked 4 hours ago









Amrat AAmrat A

335111




335111








  • 1




    $begingroup$
    Isn't $G=mathbb{R}$ and $H=mathbb{Z}$ an example of the non-iso you want? All you need to show is that $mathbb{R}$ is not iso to $S^1 times mathbb{Z}$. That's easy.
    $endgroup$
    – Randall
    2 hours ago
















  • 1




    $begingroup$
    Isn't $G=mathbb{R}$ and $H=mathbb{Z}$ an example of the non-iso you want? All you need to show is that $mathbb{R}$ is not iso to $S^1 times mathbb{Z}$. That's easy.
    $endgroup$
    – Randall
    2 hours ago










1




1




$begingroup$
Isn't $G=mathbb{R}$ and $H=mathbb{Z}$ an example of the non-iso you want? All you need to show is that $mathbb{R}$ is not iso to $S^1 times mathbb{Z}$. That's easy.
$endgroup$
– Randall
2 hours ago






$begingroup$
Isn't $G=mathbb{R}$ and $H=mathbb{Z}$ an example of the non-iso you want? All you need to show is that $mathbb{R}$ is not iso to $S^1 times mathbb{Z}$. That's easy.
$endgroup$
– Randall
2 hours ago












2 Answers
2






active

oldest

votes


















1












$begingroup$

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 $mathbb{R}^mtimes (S^1)^n$ for some $n$. In fact, given $G$ you can read off $n$ and $m$ as $n=mathrm{rank}(pi_1(G))$ 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, $mathrm{rank}(pi_1(G))=mathrm{rank}(pi_1(H))+mathrm{rank}(pi_1(G/H))$. 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



$$mathrm{rk}(pi_1(G))=mathrm{rk}(pi_1(Htimes (G/H))=mathrm{rk}(pi_1(G))+mathrm{rk}(pi_1(G/H))$$



and



$$mathrm{dim}(G)-mathrm{rk}(pi_1(G))=dim(Gtimes (G/H))-mathrm{rk}(pi_1(Htimes (G/H))$$



The first equality holds by remark about the long exact sequence on homotopy groups from the fibration. The second is given as follows:



$$begin{aligned}dim(G)-mathrm{rk}(pi_1(G)) &= dim(H)+dim(G/H)-(mathrm{rk}(pi_1(H))+mathrm{rk}(pi_1(G/H))\ &= dim(Gtimes G/H))-mathrm{rank}(pi_1(Gtimes (G/H)))end{aligned}$$





(Below is for the non-abelian situation)
Here's a simple interesting example.



Take $mathrm{GL}_2(mathbb{C})$ with its center $Z:={lambda I_2:lambdainmathbb{C}^times}$. Then, $mathrm{GL}_2(mathbb{C})/Zcong mathrm{PGL}_2(mathbb{C})$. To see that $mathrm{GL}_2(mathbb{C})notcong Ztimesmathrm{PGL}_2(mathbb{C})$ note that the derived (i.e. commutative) subgroup of the former is $mathrm{SL}_2(mathbb{C})$ whereas the latter is $mathrm{PGL}_2(mathbb{C})$. Of course, these groups aren't isomorphic as the former is simply connected and the latter is not.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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



















2












$begingroup$

Take $G = mathbb{R}$ and $H=mathbb{Z}$. The quotient $G/H$ is the circle $S^1$. The question is now to compare $mathbb{R}$ to $S^1 times mathbb{Z}$. 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 $mathbb{R}$ is connected but $S^1 times mathbb{Z}$ is not (it's a stack of circles). Algebraically they're also distinct by looking at elements of order $2$ ($mathbb{R}$ has none, $S^1 times mathbb{Z}$ has at least one).






share|cite|improve this answer









$endgroup$









  • 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












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2 Answers
2






active

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2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

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 $mathbb{R}^mtimes (S^1)^n$ for some $n$. In fact, given $G$ you can read off $n$ and $m$ as $n=mathrm{rank}(pi_1(G))$ 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, $mathrm{rank}(pi_1(G))=mathrm{rank}(pi_1(H))+mathrm{rank}(pi_1(G/H))$. 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



$$mathrm{rk}(pi_1(G))=mathrm{rk}(pi_1(Htimes (G/H))=mathrm{rk}(pi_1(G))+mathrm{rk}(pi_1(G/H))$$



and



$$mathrm{dim}(G)-mathrm{rk}(pi_1(G))=dim(Gtimes (G/H))-mathrm{rk}(pi_1(Htimes (G/H))$$



The first equality holds by remark about the long exact sequence on homotopy groups from the fibration. The second is given as follows:



$$begin{aligned}dim(G)-mathrm{rk}(pi_1(G)) &= dim(H)+dim(G/H)-(mathrm{rk}(pi_1(H))+mathrm{rk}(pi_1(G/H))\ &= dim(Gtimes G/H))-mathrm{rank}(pi_1(Gtimes (G/H)))end{aligned}$$





(Below is for the non-abelian situation)
Here's a simple interesting example.



Take $mathrm{GL}_2(mathbb{C})$ with its center $Z:={lambda I_2:lambdainmathbb{C}^times}$. Then, $mathrm{GL}_2(mathbb{C})/Zcong mathrm{PGL}_2(mathbb{C})$. To see that $mathrm{GL}_2(mathbb{C})notcong Ztimesmathrm{PGL}_2(mathbb{C})$ note that the derived (i.e. commutative) subgroup of the former is $mathrm{SL}_2(mathbb{C})$ whereas the latter is $mathrm{PGL}_2(mathbb{C})$. Of course, these groups aren't isomorphic as the former is simply connected and the latter is not.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
















1












$begingroup$

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 $mathbb{R}^mtimes (S^1)^n$ for some $n$. In fact, given $G$ you can read off $n$ and $m$ as $n=mathrm{rank}(pi_1(G))$ 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, $mathrm{rank}(pi_1(G))=mathrm{rank}(pi_1(H))+mathrm{rank}(pi_1(G/H))$. 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



$$mathrm{rk}(pi_1(G))=mathrm{rk}(pi_1(Htimes (G/H))=mathrm{rk}(pi_1(G))+mathrm{rk}(pi_1(G/H))$$



and



$$mathrm{dim}(G)-mathrm{rk}(pi_1(G))=dim(Gtimes (G/H))-mathrm{rk}(pi_1(Htimes (G/H))$$



The first equality holds by remark about the long exact sequence on homotopy groups from the fibration. The second is given as follows:



$$begin{aligned}dim(G)-mathrm{rk}(pi_1(G)) &= dim(H)+dim(G/H)-(mathrm{rk}(pi_1(H))+mathrm{rk}(pi_1(G/H))\ &= dim(Gtimes G/H))-mathrm{rank}(pi_1(Gtimes (G/H)))end{aligned}$$





(Below is for the non-abelian situation)
Here's a simple interesting example.



Take $mathrm{GL}_2(mathbb{C})$ with its center $Z:={lambda I_2:lambdainmathbb{C}^times}$. Then, $mathrm{GL}_2(mathbb{C})/Zcong mathrm{PGL}_2(mathbb{C})$. To see that $mathrm{GL}_2(mathbb{C})notcong Ztimesmathrm{PGL}_2(mathbb{C})$ note that the derived (i.e. commutative) subgroup of the former is $mathrm{SL}_2(mathbb{C})$ whereas the latter is $mathrm{PGL}_2(mathbb{C})$. Of course, these groups aren't isomorphic as the former is simply connected and the latter is not.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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














1












1








1





$begingroup$

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 $mathbb{R}^mtimes (S^1)^n$ for some $n$. In fact, given $G$ you can read off $n$ and $m$ as $n=mathrm{rank}(pi_1(G))$ 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, $mathrm{rank}(pi_1(G))=mathrm{rank}(pi_1(H))+mathrm{rank}(pi_1(G/H))$. 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



$$mathrm{rk}(pi_1(G))=mathrm{rk}(pi_1(Htimes (G/H))=mathrm{rk}(pi_1(G))+mathrm{rk}(pi_1(G/H))$$



and



$$mathrm{dim}(G)-mathrm{rk}(pi_1(G))=dim(Gtimes (G/H))-mathrm{rk}(pi_1(Htimes (G/H))$$



The first equality holds by remark about the long exact sequence on homotopy groups from the fibration. The second is given as follows:



$$begin{aligned}dim(G)-mathrm{rk}(pi_1(G)) &= dim(H)+dim(G/H)-(mathrm{rk}(pi_1(H))+mathrm{rk}(pi_1(G/H))\ &= dim(Gtimes G/H))-mathrm{rank}(pi_1(Gtimes (G/H)))end{aligned}$$





(Below is for the non-abelian situation)
Here's a simple interesting example.



Take $mathrm{GL}_2(mathbb{C})$ with its center $Z:={lambda I_2:lambdainmathbb{C}^times}$. Then, $mathrm{GL}_2(mathbb{C})/Zcong mathrm{PGL}_2(mathbb{C})$. To see that $mathrm{GL}_2(mathbb{C})notcong Ztimesmathrm{PGL}_2(mathbb{C})$ note that the derived (i.e. commutative) subgroup of the former is $mathrm{SL}_2(mathbb{C})$ whereas the latter is $mathrm{PGL}_2(mathbb{C})$. Of course, these groups aren't isomorphic as the former is simply connected and the latter is not.






share|cite|improve this answer











$endgroup$



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 $mathbb{R}^mtimes (S^1)^n$ for some $n$. In fact, given $G$ you can read off $n$ and $m$ as $n=mathrm{rank}(pi_1(G))$ 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, $mathrm{rank}(pi_1(G))=mathrm{rank}(pi_1(H))+mathrm{rank}(pi_1(G/H))$. 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



$$mathrm{rk}(pi_1(G))=mathrm{rk}(pi_1(Htimes (G/H))=mathrm{rk}(pi_1(G))+mathrm{rk}(pi_1(G/H))$$



and



$$mathrm{dim}(G)-mathrm{rk}(pi_1(G))=dim(Gtimes (G/H))-mathrm{rk}(pi_1(Htimes (G/H))$$



The first equality holds by remark about the long exact sequence on homotopy groups from the fibration. The second is given as follows:



$$begin{aligned}dim(G)-mathrm{rk}(pi_1(G)) &= dim(H)+dim(G/H)-(mathrm{rk}(pi_1(H))+mathrm{rk}(pi_1(G/H))\ &= dim(Gtimes G/H))-mathrm{rank}(pi_1(Gtimes (G/H)))end{aligned}$$





(Below is for the non-abelian situation)
Here's a simple interesting example.



Take $mathrm{GL}_2(mathbb{C})$ with its center $Z:={lambda I_2:lambdainmathbb{C}^times}$. Then, $mathrm{GL}_2(mathbb{C})/Zcong mathrm{PGL}_2(mathbb{C})$. To see that $mathrm{GL}_2(mathbb{C})notcong Ztimesmathrm{PGL}_2(mathbb{C})$ note that the derived (i.e. commutative) subgroup of the former is $mathrm{SL}_2(mathbb{C})$ whereas the latter is $mathrm{PGL}_2(mathbb{C})$. Of course, these groups aren't isomorphic as the former is simply connected and the latter is not.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 1 hour ago

























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


















  • $begingroup$
    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











2












$begingroup$

Take $G = mathbb{R}$ and $H=mathbb{Z}$. The quotient $G/H$ is the circle $S^1$. The question is now to compare $mathbb{R}$ to $S^1 times mathbb{Z}$. 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 $mathbb{R}$ is connected but $S^1 times mathbb{Z}$ is not (it's a stack of circles). Algebraically they're also distinct by looking at elements of order $2$ ($mathbb{R}$ has none, $S^1 times mathbb{Z}$ has at least one).






share|cite|improve this answer









$endgroup$









  • 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
















2












$begingroup$

Take $G = mathbb{R}$ and $H=mathbb{Z}$. The quotient $G/H$ is the circle $S^1$. The question is now to compare $mathbb{R}$ to $S^1 times mathbb{Z}$. 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 $mathbb{R}$ is connected but $S^1 times mathbb{Z}$ is not (it's a stack of circles). Algebraically they're also distinct by looking at elements of order $2$ ($mathbb{R}$ has none, $S^1 times mathbb{Z}$ has at least one).






share|cite|improve this answer









$endgroup$









  • 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














2












2








2





$begingroup$

Take $G = mathbb{R}$ and $H=mathbb{Z}$. The quotient $G/H$ is the circle $S^1$. The question is now to compare $mathbb{R}$ to $S^1 times mathbb{Z}$. 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 $mathbb{R}$ is connected but $S^1 times mathbb{Z}$ is not (it's a stack of circles). Algebraically they're also distinct by looking at elements of order $2$ ($mathbb{R}$ has none, $S^1 times mathbb{Z}$ has at least one).






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$endgroup$



Take $G = mathbb{R}$ and $H=mathbb{Z}$. The quotient $G/H$ is the circle $S^1$. The question is now to compare $mathbb{R}$ to $S^1 times mathbb{Z}$. 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 $mathbb{R}$ is connected but $S^1 times mathbb{Z}$ is not (it's a stack of circles). Algebraically they're also distinct by looking at elements of order $2$ ($mathbb{R}$ has none, $S^1 times mathbb{Z}$ has at least one).







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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














  • 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


















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