Order of taking quotients?
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Let $G$ be a group and $M,N$ be its normal subgroups. I am thinking about the relationship between $(G/M)/N$, $(G/N)/M$, $G/(M/N)$ and $(G/Mcup N)/(Mcap N)$ (they are not proper notions but I am thinking of some quotient groups of quotient groups).
Define the quotient operation $/:mathbf{Grp}timesmathbf{Grp}mapstomathbf{Grp}$ somehow properly. When is this operation associative, commutivative, etc. (with proper definitions)?
The original question is there is a group and many types of equivalence classes that will be all taken. I wonder the order of those classes matters.
group-theory category-theory quotient-group
asked Nov 16 at 5:04
erachang
969
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up vote
0
down vote
favorite
Let $G$ be a group and $M,N$ be its normal subgroups. I am thinking about the relationship between $(G/M)/N$, $(G/N)/M$, $G/(M/N)$ and $(G/Mcup N)/(Mcap N)$ (they are not proper notions but I am thinking of some quotient groups of quotient groups).
Define the quotient operation $/:mathbf{Grp}timesmathbf{Grp}mapstomathbf{Grp}$ somehow properly. When is this operation associative, commutivative, etc. (with proper definitions)?
The original question is there is a group and many types of equivalence classes that will be all taken. I wonder the order of those classes matters.
group-theory category-theory quotient-group
asked Nov 16 at 5:04
erachang
969
3
The domain is not $text{Grp} times text{Grp}$, it's pairs consisting of a group $G$ and a normal subgroup $N$ of $G$ (or perhaps a morphism into $G$).
– Qiaochu Yuan
Nov 16 at 5:35
An important point is that even if $N$ is only isomorphic to a normal subgroup of $G$, it may be isomorphic to two normal subgroups $N_0,N_1$ of $G$ with $G/N_0notcong G/N_1$, so $/$ doesn't even extend to a map on Grp $times$ Grp
– Robert Chamberlain
Nov 16 at 10:14
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $G$ be a group and $M,N$ be its normal subgroups. I am thinking about the relationship between $(G/M)/N$, $(G/N)/M$, $G/(M/N)$ and $(G/Mcup N)/(Mcap N)$ (they are not proper notions but I am thinking of some quotient groups of quotient groups).
Define the quotient operation $/:mathbf{Grp}timesmathbf{Grp}mapstomathbf{Grp}$ somehow properly. When is this operation associative, commutivative, etc. (with proper definitions)?
The original question is there is a group and many types of equivalence classes that will be all taken. I wonder the order of those classes matters.
group-theory category-theory quotient-group
asked Nov 16 at 5:04
erachang
969
Let $G$ be a group and $M,N$ be its normal subgroups. I am thinking about the relationship between $(G/M)/N$, $(G/N)/M$, $G/(M/N)$ and $(G/Mcup N)/(Mcap N)$ (they are not proper notions but I am thinking of some quotient groups of quotient groups).
Define the quotient operation $/:mathbf{Grp}timesmathbf{Grp}mapstomathbf{Grp}$ somehow properly. When is this operation associative, commutivative, etc. (with proper definitions)?
The original question is there is a group and many types of equivalence classes that will be all taken. I wonder the order of those classes matters.
group-theory category-theory quotient-group
group-theory category-theory quotient-group
asked Nov 16 at 5:04
erachang
969
asked Nov 16 at 5:04
erachang
969
asked Nov 16 at 5:04
erachang
969
asked Nov 16 at 5:04
erachang
969
asked Nov 16 at 5:04
erachang
969
969
3
The domain is not $text{Grp} times text{Grp}$, it's pairs consisting of a group $G$ and a normal subgroup $N$ of $G$ (or perhaps a morphism into $G$).
– Qiaochu Yuan
Nov 16 at 5:35
An important point is that even if $N$ is only isomorphic to a normal subgroup of $G$, it may be isomorphic to two normal subgroups $N_0,N_1$ of $G$ with $G/N_0notcong G/N_1$, so $/$ doesn't even extend to a map on Grp $times$ Grp
– Robert Chamberlain
Nov 16 at 10:14
add a comment |
3
The domain is not $text{Grp} times text{Grp}$, it's pairs consisting of a group $G$ and a normal subgroup $N$ of $G$ (or perhaps a morphism into $G$).
– Qiaochu Yuan
Nov 16 at 5:35
An important point is that even if $N$ is only isomorphic to a normal subgroup of $G$, it may be isomorphic to two normal subgroups $N_0,N_1$ of $G$ with $G/N_0notcong G/N_1$, so $/$ doesn't even extend to a map on Grp $times$ Grp
– Robert Chamberlain
Nov 16 at 10:14
3
3
The domain is not $text{Grp} times text{Grp}$, it's pairs consisting of a group $G$ and a normal subgroup $N$ of $G$ (or perhaps a morphism into $G$).
– Qiaochu Yuan
Nov 16 at 5:35
The domain is not $text{Grp} times text{Grp}$, it's pairs consisting of a group $G$ and a normal subgroup $N$ of $G$ (or perhaps a morphism into $G$).
– Qiaochu Yuan
Nov 16 at 5:35
An important point is that even if $N$ is only isomorphic to a normal subgroup of $G$, it may be isomorphic to two normal subgroups $N_0,N_1$ of $G$ with $G/N_0notcong G/N_1$, so $/$ doesn't even extend to a map on Grp $times$ Grp
– Robert Chamberlain
Nov 16 at 10:14
An important point is that even if $N$ is only isomorphic to a normal subgroup of $G$, it may be isomorphic to two normal subgroups $N_0,N_1$ of $G$ with $G/N_0notcong G/N_1$, so $/$ doesn't even extend to a map on Grp $times$ Grp
– Robert Chamberlain
Nov 16 at 10:14
add a comment |
1 Answer
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In a quotient $X/Y$ of groups the denominator $Y$ must be a normal subgroup of the numerator $X$. You could have nested situations: $N unlhd M unlhd G$, and then the (third) isomorphism theorem asserts $G/M cong (G/N) / (M/N)$. So what you are up to in your question does not make much sense.
answered Nov 16 at 10:07
Nicky Hekster
28k53254
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
In a quotient $X/Y$ of groups the denominator $Y$ must be a normal subgroup of the numerator $X$. You could have nested situations: $N unlhd M unlhd G$, and then the (third) isomorphism theorem asserts $G/M cong (G/N) / (M/N)$. So what you are up to in your question does not make much sense.
answered Nov 16 at 10:07
Nicky Hekster
28k53254
add a comment |
up vote
0
down vote
In a quotient $X/Y$ of groups the denominator $Y$ must be a normal subgroup of the numerator $X$. You could have nested situations: $N unlhd M unlhd G$, and then the (third) isomorphism theorem asserts $G/M cong (G/N) / (M/N)$. So what you are up to in your question does not make much sense.
answered Nov 16 at 10:07
Nicky Hekster
28k53254
add a comment |
up vote
0
down vote
up vote
0
down vote
In a quotient $X/Y$ of groups the denominator $Y$ must be a normal subgroup of the numerator $X$. You could have nested situations: $N unlhd M unlhd G$, and then the (third) isomorphism theorem asserts $G/M cong (G/N) / (M/N)$. So what you are up to in your question does not make much sense.
answered Nov 16 at 10:07
Nicky Hekster
28k53254
In a quotient $X/Y$ of groups the denominator $Y$ must be a normal subgroup of the numerator $X$. You could have nested situations: $N unlhd M unlhd G$, and then the (third) isomorphism theorem asserts $G/M cong (G/N) / (M/N)$. So what you are up to in your question does not make much sense.
answered Nov 16 at 10:07
Nicky Hekster
28k53254
edited Nov 16 at 10:24
answered Nov 16 at 10:07
Nicky Hekster
28k53254
answered Nov 16 at 10:07
Nicky Hekster
28k53254
answered Nov 16 at 10:07
Nicky Hekster
28k53254
28k53254
add a comment |
add a comment |
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The domain is not $text{Grp} times text{Grp}$, it's pairs consisting of a group $G$ and a normal subgroup $N$ of $G$ (or perhaps a morphism into $G$).
– Qiaochu Yuan
Nov 16 at 5:35
An important point is that even if $N$ is only isomorphic to a normal subgroup of $G$, it may be isomorphic to two normal subgroups $N_0,N_1$ of $G$ with $G/N_0notcong G/N_1$, so $/$ doesn't even extend to a map on Grp $times$ Grp
– Robert Chamberlain
Nov 16 at 10:14