What property guarantees normal subgroups $H, K$ of $G$ imply that$ G/H$ isomorphic to $G/K$?
1
$begingroup$
I was surprised by this question which provides an example of a finite group $G$ where normal isomorphic subgroups $H, Ktriangleleft G$ do not imply isomorphic quotient groups, i.e. $G/Hnotcong G/K$.
Question: What condition must be met in order for the quotient groups to be isomorphic?
Trying to sort this out intuitively, I think like this:
- A finite group is fully described by its multiplication table and isomorphic groups have the same table.
- Still we evidently can't say "Let $Jtriangleleft G$ be a subgroup with table $T$; form the quotient group $G/J$".
- It breaks down when trying to carry out coset multiplication in $G/J$.
- Because coset multiplication depends not only on multiplication in $J$ as specified by $T$ but also on multiplication of elements in $J$ with elements in the embedding group $G$.
So an attempt at an answer would be that $forall gin G, hin H:gh=gpsi(h)$ where $psi$ is an isomorphism between $H$ and $K$. I am trying to express that isomorphic elements of the subgroups interact similarly with elements of the embedding group. Is this the right conclusion?
group-theory
edited Jan 16 at 4:31
Andrews
1,3112423
asked Jan 15 at 22:01
DanielDaniel
1061
$endgroup$
add a comment |
1
$begingroup$
I was surprised by this question which provides an example of a finite group $G$ where normal isomorphic subgroups $H, Ktriangleleft G$ do not imply isomorphic quotient groups, i.e. $G/Hnotcong G/K$.
Question: What condition must be met in order for the quotient groups to be isomorphic?
Trying to sort this out intuitively, I think like this:
- A finite group is fully described by its multiplication table and isomorphic groups have the same table.
- Still we evidently can't say "Let $Jtriangleleft G$ be a subgroup with table $T$; form the quotient group $G/J$".
- It breaks down when trying to carry out coset multiplication in $G/J$.
- Because coset multiplication depends not only on multiplication in $J$ as specified by $T$ but also on multiplication of elements in $J$ with elements in the embedding group $G$.
So an attempt at an answer would be that $forall gin G, hin H:gh=gpsi(h)$ where $psi$ is an isomorphism between $H$ and $K$. I am trying to express that isomorphic elements of the subgroups interact similarly with elements of the embedding group. Is this the right conclusion?
group-theory
edited Jan 16 at 4:31
Andrews
1,3112423
asked Jan 15 at 22:01
DanielDaniel
1061
$endgroup$
4
$begingroup$
I don't know about necessary assumptions, but a nice sufficient one is that there be an isomorphism of $G$ sending $H$ to $K$; or more generally a surjective morphism $Gto G$ such that $pi^{-1}(K) = H$.
$endgroup$
– Max
Jan 15 at 22:09
$begingroup$
$gh = gpsi(h) Rightarrow h=psi(h)$.
$endgroup$
– Derek Holt
Jan 15 at 22:30
2
$begingroup$
$H,K$ don't need to be isomorphic. Given two normal subgroups $H,K$ of same order and a function $f : G to G$ such that $f(aH) subset f(a)K$ and $f(ab)in f(a)f(b)K$ then $F : G/H to G/K, F(aH) = f(a)K$ is an homomorphism. If $f$ is a permutation of $G$ then $F$ is surjective so it is an isomorphism $G/H to G/K$.
$endgroup$
– reuns
Jan 16 at 5:09
1
$begingroup$
I suggest that as a preliminary project you ask the same question about finite abelian $p$-groups for a fixed prime $p$. Perhaps explore what can be said if $H=mathbb{Z}_{p^3}$ and $K=(mathbb{Z}_p)^3$ (I think smaller cases are trivial). I think you'll find that there's not much that can be said except "$G/Hcong G/K$".
$endgroup$
– ancientmathematician
Jan 16 at 8:09
$begingroup$
Thanks to all of you who responded. That set me off in a good direction.
$endgroup$
– Daniel
Jan 16 at 17:18
add a comment |
1
1
1
0
$begingroup$
I was surprised by this question which provides an example of a finite group $G$ where normal isomorphic subgroups $H, Ktriangleleft G$ do not imply isomorphic quotient groups, i.e. $G/Hnotcong G/K$.
Question: What condition must be met in order for the quotient groups to be isomorphic?
Trying to sort this out intuitively, I think like this:
- A finite group is fully described by its multiplication table and isomorphic groups have the same table.
- Still we evidently can't say "Let $Jtriangleleft G$ be a subgroup with table $T$; form the quotient group $G/J$".
- It breaks down when trying to carry out coset multiplication in $G/J$.
- Because coset multiplication depends not only on multiplication in $J$ as specified by $T$ but also on multiplication of elements in $J$ with elements in the embedding group $G$.
So an attempt at an answer would be that $forall gin G, hin H:gh=gpsi(h)$ where $psi$ is an isomorphism between $H$ and $K$. I am trying to express that isomorphic elements of the subgroups interact similarly with elements of the embedding group. Is this the right conclusion?
group-theory
edited Jan 16 at 4:31
Andrews
1,3112423
asked Jan 15 at 22:01
DanielDaniel
1061
$endgroup$
I was surprised by this question which provides an example of a finite group $G$ where normal isomorphic subgroups $H, Ktriangleleft G$ do not imply isomorphic quotient groups, i.e. $G/Hnotcong G/K$.
Question: What condition must be met in order for the quotient groups to be isomorphic?
Trying to sort this out intuitively, I think like this:
- A finite group is fully described by its multiplication table and isomorphic groups have the same table.
- Still we evidently can't say "Let $Jtriangleleft G$ be a subgroup with table $T$; form the quotient group $G/J$".
- It breaks down when trying to carry out coset multiplication in $G/J$.
- Because coset multiplication depends not only on multiplication in $J$ as specified by $T$ but also on multiplication of elements in $J$ with elements in the embedding group $G$.
So an attempt at an answer would be that $forall gin G, hin H:gh=gpsi(h)$ where $psi$ is an isomorphism between $H$ and $K$. I am trying to express that isomorphic elements of the subgroups interact similarly with elements of the embedding group. Is this the right conclusion?
group-theory
group-theory
edited Jan 16 at 4:31
Andrews
1,3112423
asked Jan 15 at 22:01
DanielDaniel
1061
edited Jan 16 at 4:31
Andrews
1,3112423
asked Jan 15 at 22:01
DanielDaniel
1061
edited Jan 16 at 4:31
Andrews
1,3112423
edited Jan 16 at 4:31
Andrews
1,3112423
edited Jan 16 at 4:31
Andrews
1,3112423
1,3112423
asked Jan 15 at 22:01
DanielDaniel
1061
asked Jan 15 at 22:01
DanielDaniel
1061
asked Jan 15 at 22:01
DanielDaniel
1061
1061
4
$begingroup$
I don't know about necessary assumptions, but a nice sufficient one is that there be an isomorphism of $G$ sending $H$ to $K$; or more generally a surjective morphism $Gto G$ such that $pi^{-1}(K) = H$.
$endgroup$
– Max
Jan 15 at 22:09
$begingroup$
$gh = gpsi(h) Rightarrow h=psi(h)$.
$endgroup$
– Derek Holt
Jan 15 at 22:30
2
$begingroup$
$H,K$ don't need to be isomorphic. Given two normal subgroups $H,K$ of same order and a function $f : G to G$ such that $f(aH) subset f(a)K$ and $f(ab)in f(a)f(b)K$ then $F : G/H to G/K, F(aH) = f(a)K$ is an homomorphism. If $f$ is a permutation of $G$ then $F$ is surjective so it is an isomorphism $G/H to G/K$.
$endgroup$
– reuns
Jan 16 at 5:09
1
$begingroup$
I suggest that as a preliminary project you ask the same question about finite abelian $p$-groups for a fixed prime $p$. Perhaps explore what can be said if $H=mathbb{Z}_{p^3}$ and $K=(mathbb{Z}_p)^3$ (I think smaller cases are trivial). I think you'll find that there's not much that can be said except "$G/Hcong G/K$".
$endgroup$
– ancientmathematician
Jan 16 at 8:09
$begingroup$
Thanks to all of you who responded. That set me off in a good direction.
$endgroup$
– Daniel
Jan 16 at 17:18
add a comment |
4
$begingroup$
I don't know about necessary assumptions, but a nice sufficient one is that there be an isomorphism of $G$ sending $H$ to $K$; or more generally a surjective morphism $Gto G$ such that $pi^{-1}(K) = H$.
$endgroup$
– Max
Jan 15 at 22:09
$begingroup$
$gh = gpsi(h) Rightarrow h=psi(h)$.
$endgroup$
– Derek Holt
Jan 15 at 22:30
2
$begingroup$
$H,K$ don't need to be isomorphic. Given two normal subgroups $H,K$ of same order and a function $f : G to G$ such that $f(aH) subset f(a)K$ and $f(ab)in f(a)f(b)K$ then $F : G/H to G/K, F(aH) = f(a)K$ is an homomorphism. If $f$ is a permutation of $G$ then $F$ is surjective so it is an isomorphism $G/H to G/K$.
$endgroup$
– reuns
Jan 16 at 5:09
1
$begingroup$
I suggest that as a preliminary project you ask the same question about finite abelian $p$-groups for a fixed prime $p$. Perhaps explore what can be said if $H=mathbb{Z}_{p^3}$ and $K=(mathbb{Z}_p)^3$ (I think smaller cases are trivial). I think you'll find that there's not much that can be said except "$G/Hcong G/K$".
$endgroup$
– ancientmathematician
Jan 16 at 8:09
$begingroup$
Thanks to all of you who responded. That set me off in a good direction.
$endgroup$
– Daniel
Jan 16 at 17:18
4
4
$begingroup$
I don't know about necessary assumptions, but a nice sufficient one is that there be an isomorphism of $G$ sending $H$ to $K$; or more generally a surjective morphism $Gto G$ such that $pi^{-1}(K) = H$.
$endgroup$
– Max
Jan 15 at 22:09
$begingroup$
I don't know about necessary assumptions, but a nice sufficient one is that there be an isomorphism of $G$ sending $H$ to $K$; or more generally a surjective morphism $Gto G$ such that $pi^{-1}(K) = H$.
$endgroup$
– Max
Jan 15 at 22:09
$begingroup$
$gh = gpsi(h) Rightarrow h=psi(h)$.
$endgroup$
– Derek Holt
Jan 15 at 22:30
$begingroup$
$gh = gpsi(h) Rightarrow h=psi(h)$.
$endgroup$
– Derek Holt
Jan 15 at 22:30
2
2
$begingroup$
$H,K$ don't need to be isomorphic. Given two normal subgroups $H,K$ of same order and a function $f : G to G$ such that $f(aH) subset f(a)K$ and $f(ab)in f(a)f(b)K$ then $F : G/H to G/K, F(aH) = f(a)K$ is an homomorphism. If $f$ is a permutation of $G$ then $F$ is surjective so it is an isomorphism $G/H to G/K$.
$endgroup$
– reuns
Jan 16 at 5:09
$begingroup$
$H,K$ don't need to be isomorphic. Given two normal subgroups $H,K$ of same order and a function $f : G to G$ such that $f(aH) subset f(a)K$ and $f(ab)in f(a)f(b)K$ then $F : G/H to G/K, F(aH) = f(a)K$ is an homomorphism. If $f$ is a permutation of $G$ then $F$ is surjective so it is an isomorphism $G/H to G/K$.
$endgroup$
– reuns
Jan 16 at 5:09
1
1
$begingroup$
I suggest that as a preliminary project you ask the same question about finite abelian $p$-groups for a fixed prime $p$. Perhaps explore what can be said if $H=mathbb{Z}_{p^3}$ and $K=(mathbb{Z}_p)^3$ (I think smaller cases are trivial). I think you'll find that there's not much that can be said except "$G/Hcong G/K$".
$endgroup$
– ancientmathematician
Jan 16 at 8:09
$begingroup$
I suggest that as a preliminary project you ask the same question about finite abelian $p$-groups for a fixed prime $p$. Perhaps explore what can be said if $H=mathbb{Z}_{p^3}$ and $K=(mathbb{Z}_p)^3$ (I think smaller cases are trivial). I think you'll find that there's not much that can be said except "$G/Hcong G/K$".
$endgroup$
– ancientmathematician
Jan 16 at 8:09
$begingroup$
Thanks to all of you who responded. That set me off in a good direction.
$endgroup$
– Daniel
Jan 16 at 17:18
$begingroup$
Thanks to all of you who responded. That set me off in a good direction.
$endgroup$
– Daniel
Jan 16 at 17:18
add a comment |
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4
$begingroup$
I don't know about necessary assumptions, but a nice sufficient one is that there be an isomorphism of $G$ sending $H$ to $K$; or more generally a surjective morphism $Gto G$ such that $pi^{-1}(K) = H$.
$endgroup$
– Max
Jan 15 at 22:09
$begingroup$
$gh = gpsi(h) Rightarrow h=psi(h)$.
$endgroup$
– Derek Holt
Jan 15 at 22:30
2
$begingroup$
$H,K$ don't need to be isomorphic. Given two normal subgroups $H,K$ of same order and a function $f : G to G$ such that $f(aH) subset f(a)K$ and $f(ab)in f(a)f(b)K$ then $F : G/H to G/K, F(aH) = f(a)K$ is an homomorphism. If $f$ is a permutation of $G$ then $F$ is surjective so it is an isomorphism $G/H to G/K$.
$endgroup$
– reuns
Jan 16 at 5:09
1
$begingroup$
I suggest that as a preliminary project you ask the same question about finite abelian $p$-groups for a fixed prime $p$. Perhaps explore what can be said if $H=mathbb{Z}_{p^3}$ and $K=(mathbb{Z}_p)^3$ (I think smaller cases are trivial). I think you'll find that there's not much that can be said except "$G/Hcong G/K$".
$endgroup$
– ancientmathematician
Jan 16 at 8:09
$begingroup$
Thanks to all of you who responded. That set me off in a good direction.
$endgroup$
– Daniel
Jan 16 at 17:18