Mixing
Isomorphism theorem like theorem for cosets
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
I would like to a get a result similar to the third isomorphism theorem, but I want to deal with sets of cosets, rather than a quotient subgroup.
Let me formalize it.
Say $G$ is a group, with subgroups $Asubset Bsubset G$. We may assume $A$ is normal in $B$, but no other normality may be assumed.
Let $O$ be a set of sets of cosets $O=C_i_iin I$ where $C_i=Ac_ij_jin J_i$ for $c_ijin G$, that is $Osubset P(G/A)$ where $P(X)$ is the set of subsets, and $G/A$ is the space of $A$-cosets. We may assume that the elements of $O$ are pairwise disjoint, although I doubt it will be necessary.
Define $O^prime=D_i_iin I$ where $D_i=Bc_ij_jin J_i$. That is, we look at the same cosets, but this time with respect to $B$.
I would like to give a lower bound on the cardinality of $O^prime$.
I would think that the desired $|O^prime|geq frac1[B:A] |O|$ should hold, where $[B:A]$ is the index.
An attempt of a proof would be the mimic the proof of the third isomorphism theorem, which pretty much gives this result if everything was a normal subgroup.
I would define $f:Oto O^prime$ by $f(C_i)=D_i$ and would try to prove that $|f^-1(D_i)|leq [B:A]$. If $C_i$ are singletons, this is easy. If these are larger sets, it looks like it doesn't quite work, or at least I cant seem to be able to do it. Any help would be appreciated.
P.S: $[B:A]$ may be assumed to be finite. If necessary, $O$ may be assumed to be finite.
abstract-algebra group-theory quotient-spaces quotient-group
asked Jul 19 at 18:42
The way of life
709216
add a comment |Â
up vote
0
down vote
favorite
I would like to a get a result similar to the third isomorphism theorem, but I want to deal with sets of cosets, rather than a quotient subgroup.
Let me formalize it.
Say $G$ is a group, with subgroups $Asubset Bsubset G$. We may assume $A$ is normal in $B$, but no other normality may be assumed.
Let $O$ be a set of sets of cosets $O=C_i_iin I$ where $C_i=Ac_ij_jin J_i$ for $c_ijin G$, that is $Osubset P(G/A)$ where $P(X)$ is the set of subsets, and $G/A$ is the space of $A$-cosets. We may assume that the elements of $O$ are pairwise disjoint, although I doubt it will be necessary.
Define $O^prime=D_i_iin I$ where $D_i=Bc_ij_jin J_i$. That is, we look at the same cosets, but this time with respect to $B$.
I would like to give a lower bound on the cardinality of $O^prime$.
I would think that the desired $|O^prime|geq frac1[B:A] |O|$ should hold, where $[B:A]$ is the index.
An attempt of a proof would be the mimic the proof of the third isomorphism theorem, which pretty much gives this result if everything was a normal subgroup.
I would define $f:Oto O^prime$ by $f(C_i)=D_i$ and would try to prove that $|f^-1(D_i)|leq [B:A]$. If $C_i$ are singletons, this is easy. If these are larger sets, it looks like it doesn't quite work, or at least I cant seem to be able to do it. Any help would be appreciated.
P.S: $[B:A]$ may be assumed to be finite. If necessary, $O$ may be assumed to be finite.
abstract-algebra group-theory quotient-spaces quotient-group
asked Jul 19 at 18:42
The way of life
709216
Each coset of $A$ iin $G$ is a disjoint union of $|B:A|$ cosets of $B$ in $G$, which immediately implies $|f^-1(D_i)| le |B:A|$.
– Derek Holt
Jul 19 at 20:00
@DerekHolt Thanks for the comment. Yet, I cannot see why it follows immediately from this statement, as $D_i$ is a set of cosets rather than a coset. So for each element of $D_i$ there are indeed up to $[B:A]$ preimages, but couldn't this give raise to more than $[B:A]$ preimages of $D_i$? Something like $[B:A]^$ (choosing a preimage out of the possible ones, for each element)?
– The way of life
Jul 19 at 20:27
But your condition that the elements of $O$ are pairwise disjoint means that we can only use each coset once as a preimage.
– Derek Holt
Jul 19 at 22:02
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I would like to a get a result similar to the third isomorphism theorem, but I want to deal with sets of cosets, rather than a quotient subgroup.
Let me formalize it.
Say $G$ is a group, with subgroups $Asubset Bsubset G$. We may assume $A$ is normal in $B$, but no other normality may be assumed.
Let $O$ be a set of sets of cosets $O=C_i_iin I$ where $C_i=Ac_ij_jin J_i$ for $c_ijin G$, that is $Osubset P(G/A)$ where $P(X)$ is the set of subsets, and $G/A$ is the space of $A$-cosets. We may assume that the elements of $O$ are pairwise disjoint, although I doubt it will be necessary.
Define $O^prime=D_i_iin I$ where $D_i=Bc_ij_jin J_i$. That is, we look at the same cosets, but this time with respect to $B$.
I would like to give a lower bound on the cardinality of $O^prime$.
I would think that the desired $|O^prime|geq frac1[B:A] |O|$ should hold, where $[B:A]$ is the index.
An attempt of a proof would be the mimic the proof of the third isomorphism theorem, which pretty much gives this result if everything was a normal subgroup.
I would define $f:Oto O^prime$ by $f(C_i)=D_i$ and would try to prove that $|f^-1(D_i)|leq [B:A]$. If $C_i$ are singletons, this is easy. If these are larger sets, it looks like it doesn't quite work, or at least I cant seem to be able to do it. Any help would be appreciated.
P.S: $[B:A]$ may be assumed to be finite. If necessary, $O$ may be assumed to be finite.
abstract-algebra group-theory quotient-spaces quotient-group
asked Jul 19 at 18:42
The way of life
709216
I would like to a get a result similar to the third isomorphism theorem, but I want to deal with sets of cosets, rather than a quotient subgroup.
Let me formalize it.
Say $G$ is a group, with subgroups $Asubset Bsubset G$. We may assume $A$ is normal in $B$, but no other normality may be assumed.
Let $O$ be a set of sets of cosets $O=C_i_iin I$ where $C_i=Ac_ij_jin J_i$ for $c_ijin G$, that is $Osubset P(G/A)$ where $P(X)$ is the set of subsets, and $G/A$ is the space of $A$-cosets. We may assume that the elements of $O$ are pairwise disjoint, although I doubt it will be necessary.
Define $O^prime=D_i_iin I$ where $D_i=Bc_ij_jin J_i$. That is, we look at the same cosets, but this time with respect to $B$.
I would like to give a lower bound on the cardinality of $O^prime$.
I would think that the desired $|O^prime|geq frac1[B:A] |O|$ should hold, where $[B:A]$ is the index.
An attempt of a proof would be the mimic the proof of the third isomorphism theorem, which pretty much gives this result if everything was a normal subgroup.
I would define $f:Oto O^prime$ by $f(C_i)=D_i$ and would try to prove that $|f^-1(D_i)|leq [B:A]$. If $C_i$ are singletons, this is easy. If these are larger sets, it looks like it doesn't quite work, or at least I cant seem to be able to do it. Any help would be appreciated.
P.S: $[B:A]$ may be assumed to be finite. If necessary, $O$ may be assumed to be finite.
abstract-algebra group-theory quotient-spaces quotient-group
asked Jul 19 at 18:42
The way of life
709216
edited Jul 19 at 19:30
asked Jul 19 at 18:42
The way of life
709216
asked Jul 19 at 18:42
The way of life
709216
asked Jul 19 at 18:42
The way of life
709216
709216
Each coset of $A$ iin $G$ is a disjoint union of $|B:A|$ cosets of $B$ in $G$, which immediately implies $|f^-1(D_i)| le |B:A|$.
– Derek Holt
Jul 19 at 20:00
@DerekHolt Thanks for the comment. Yet, I cannot see why it follows immediately from this statement, as $D_i$ is a set of cosets rather than a coset. So for each element of $D_i$ there are indeed up to $[B:A]$ preimages, but couldn't this give raise to more than $[B:A]$ preimages of $D_i$? Something like $[B:A]^$ (choosing a preimage out of the possible ones, for each element)?
– The way of life
Jul 19 at 20:27
But your condition that the elements of $O$ are pairwise disjoint means that we can only use each coset once as a preimage.
– Derek Holt
Jul 19 at 22:02
add a comment |Â
Each coset of $A$ iin $G$ is a disjoint union of $|B:A|$ cosets of $B$ in $G$, which immediately implies $|f^-1(D_i)| le |B:A|$.
– Derek Holt
Jul 19 at 20:00
@DerekHolt Thanks for the comment. Yet, I cannot see why it follows immediately from this statement, as $D_i$ is a set of cosets rather than a coset. So for each element of $D_i$ there are indeed up to $[B:A]$ preimages, but couldn't this give raise to more than $[B:A]$ preimages of $D_i$? Something like $[B:A]^$ (choosing a preimage out of the possible ones, for each element)?
– The way of life
Jul 19 at 20:27
But your condition that the elements of $O$ are pairwise disjoint means that we can only use each coset once as a preimage.
– Derek Holt
Jul 19 at 22:02
Each coset of $A$ iin $G$ is a disjoint union of $|B:A|$ cosets of $B$ in $G$, which immediately implies $|f^-1(D_i)| le |B:A|$.
– Derek Holt
Jul 19 at 20:00
Each coset of $A$ iin $G$ is a disjoint union of $|B:A|$ cosets of $B$ in $G$, which immediately implies $|f^-1(D_i)| le |B:A|$.
– Derek Holt
Jul 19 at 20:00
@DerekHolt Thanks for the comment. Yet, I cannot see why it follows immediately from this statement, as $D_i$ is a set of cosets rather than a coset. So for each element of $D_i$ there are indeed up to $[B:A]$ preimages, but couldn't this give raise to more than $[B:A]$ preimages of $D_i$? Something like $[B:A]^$ (choosing a preimage out of the possible ones, for each element)?
– The way of life
Jul 19 at 20:27
@DerekHolt Thanks for the comment. Yet, I cannot see why it follows immediately from this statement, as $D_i$ is a set of cosets rather than a coset. So for each element of $D_i$ there are indeed up to $[B:A]$ preimages, but couldn't this give raise to more than $[B:A]$ preimages of $D_i$? Something like $[B:A]^$ (choosing a preimage out of the possible ones, for each element)?
– The way of life
Jul 19 at 20:27
But your condition that the elements of $O$ are pairwise disjoint means that we can only use each coset once as a preimage.
– Derek Holt
Jul 19 at 22:02
But your condition that the elements of $O$ are pairwise disjoint means that we can only use each coset once as a preimage.
– Derek Holt
Jul 19 at 22:02
add a comment |Â
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2856950%2fisomorphism-theorem-like-theorem-for-cosets%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Each coset of $A$ iin $G$ is a disjoint union of $|B:A|$ cosets of $B$ in $G$, which immediately implies $|f^-1(D_i)| le |B:A|$.
– Derek Holt
Jul 19 at 20:00
@DerekHolt Thanks for the comment. Yet, I cannot see why it follows immediately from this statement, as $D_i$ is a set of cosets rather than a coset. So for each element of $D_i$ there are indeed up to $[B:A]$ preimages, but couldn't this give raise to more than $[B:A]$ preimages of $D_i$? Something like $[B:A]^$ (choosing a preimage out of the possible ones, for each element)?
– The way of life
Jul 19 at 20:27
But your condition that the elements of $O$ are pairwise disjoint means that we can only use each coset once as a preimage.
– Derek Holt
Jul 19 at 22:02