Mixing
If we divide a group and a proper subgroup of this group by the same normal subgroup, can the quotients be...
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Let $G$ be a group and $H$ be a proper subgroup of $G$. Let $N$ be a normal subgroup of both $G$ and $H$.
Question: Is it possible that $G/N = H/N$?
Motivation: For a finite extension $L/K$ of local fields, we have exact sequences
$$newcommand{ra}[1]{kern-1.5exxrightarrow{ #1 }phantom{}kern-1.5ex}
newcommand{ras}[1]{kern-1.5exxrightarrow{ smash{#1} }phantom{}kern-1.5ex}
newcommand{da}[1]{biggdownarrowraise.5exrlap{scriptstyle#1}}
begin{array}{c}
1& ra{} & I_L & ra{} & G_L & ra{} & operatorname{Gal}(L^{ur}/L) & ra{} & 1 & & \
& & da{} & & da{} & & da{} & & & & \
1 & ra{} & I_K & ra{} & G_K & ra{} & operatorname{Gal}(K^{ur}/K) & ra{} & 1\
end{array}$$
We always have $operatorname{Gal}(L^{ur}/L) = operatorname{Gal}(K^{ur}/K) simeq hat{mathbb{Z}}$. If $L/K$ is totally ramified, we have $I_L = I_K$ and I would love to conclude that $G_L = G_K$. But I am not sure whether this is possible or not.
Could you please help me resolving this problem? Thank you!
abstract-algebra group-theory normal-subgroups quotient-group
asked 2 days ago
Diglett
822420
add a comment |
up vote
1
down vote
favorite
Let $G$ be a group and $H$ be a proper subgroup of $G$. Let $N$ be a normal subgroup of both $G$ and $H$.
Question: Is it possible that $G/N = H/N$?
Motivation: For a finite extension $L/K$ of local fields, we have exact sequences
$$newcommand{ra}[1]{kern-1.5exxrightarrow{ #1 }phantom{}kern-1.5ex}
newcommand{ras}[1]{kern-1.5exxrightarrow{ smash{#1} }phantom{}kern-1.5ex}
newcommand{da}[1]{biggdownarrowraise.5exrlap{scriptstyle#1}}
begin{array}{c}
1& ra{} & I_L & ra{} & G_L & ra{} & operatorname{Gal}(L^{ur}/L) & ra{} & 1 & & \
& & da{} & & da{} & & da{} & & & & \
1 & ra{} & I_K & ra{} & G_K & ra{} & operatorname{Gal}(K^{ur}/K) & ra{} & 1\
end{array}$$
We always have $operatorname{Gal}(L^{ur}/L) = operatorname{Gal}(K^{ur}/K) simeq hat{mathbb{Z}}$. If $L/K$ is totally ramified, we have $I_L = I_K$ and I would love to conclude that $G_L = G_K$. But I am not sure whether this is possible or not.
Could you please help me resolving this problem? Thank you!
abstract-algebra group-theory normal-subgroups quotient-group
asked 2 days ago
Diglett
822420
2
You can have groups which are isomorphic to proper subgroups(integers, free group on two generators), so for N trivial you get the result. Can you apply the five lemma in the motivating example(or are the isomorhism you talk about not compatible with the exact sequences.
– Paul Plummer
2 days ago
1
Do you mean isomorphic or equal? For equality, the answer is trivially no. For isomorphism, you might as well assume that $N=1$, and then you are asking whether a group can be isomorphic to a proper subgroup, and then you can take an example like $mathbb{Z}$ and any of its proper non-trivial subgroups.
– verret
2 days ago
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $G$ be a group and $H$ be a proper subgroup of $G$. Let $N$ be a normal subgroup of both $G$ and $H$.
Question: Is it possible that $G/N = H/N$?
Motivation: For a finite extension $L/K$ of local fields, we have exact sequences
$$newcommand{ra}[1]{kern-1.5exxrightarrow{ #1 }phantom{}kern-1.5ex}
newcommand{ras}[1]{kern-1.5exxrightarrow{ smash{#1} }phantom{}kern-1.5ex}
newcommand{da}[1]{biggdownarrowraise.5exrlap{scriptstyle#1}}
begin{array}{c}
1& ra{} & I_L & ra{} & G_L & ra{} & operatorname{Gal}(L^{ur}/L) & ra{} & 1 & & \
& & da{} & & da{} & & da{} & & & & \
1 & ra{} & I_K & ra{} & G_K & ra{} & operatorname{Gal}(K^{ur}/K) & ra{} & 1\
end{array}$$
We always have $operatorname{Gal}(L^{ur}/L) = operatorname{Gal}(K^{ur}/K) simeq hat{mathbb{Z}}$. If $L/K$ is totally ramified, we have $I_L = I_K$ and I would love to conclude that $G_L = G_K$. But I am not sure whether this is possible or not.
Could you please help me resolving this problem? Thank you!
abstract-algebra group-theory normal-subgroups quotient-group
asked 2 days ago
Diglett
822420
Let $G$ be a group and $H$ be a proper subgroup of $G$. Let $N$ be a normal subgroup of both $G$ and $H$.
Question: Is it possible that $G/N = H/N$?
Motivation: For a finite extension $L/K$ of local fields, we have exact sequences
$$newcommand{ra}[1]{kern-1.5exxrightarrow{ #1 }phantom{}kern-1.5ex}
newcommand{ras}[1]{kern-1.5exxrightarrow{ smash{#1} }phantom{}kern-1.5ex}
newcommand{da}[1]{biggdownarrowraise.5exrlap{scriptstyle#1}}
begin{array}{c}
1& ra{} & I_L & ra{} & G_L & ra{} & operatorname{Gal}(L^{ur}/L) & ra{} & 1 & & \
& & da{} & & da{} & & da{} & & & & \
1 & ra{} & I_K & ra{} & G_K & ra{} & operatorname{Gal}(K^{ur}/K) & ra{} & 1\
end{array}$$
We always have $operatorname{Gal}(L^{ur}/L) = operatorname{Gal}(K^{ur}/K) simeq hat{mathbb{Z}}$. If $L/K$ is totally ramified, we have $I_L = I_K$ and I would love to conclude that $G_L = G_K$. But I am not sure whether this is possible or not.
Could you please help me resolving this problem? Thank you!
abstract-algebra group-theory normal-subgroups quotient-group
abstract-algebra group-theory normal-subgroups quotient-group
asked 2 days ago
Diglett
822420
asked 2 days ago
Diglett
822420
asked 2 days ago
Diglett
822420
asked 2 days ago
Diglett
822420
asked 2 days ago
Diglett
822420
822420
2
You can have groups which are isomorphic to proper subgroups(integers, free group on two generators), so for N trivial you get the result. Can you apply the five lemma in the motivating example(or are the isomorhism you talk about not compatible with the exact sequences.
– Paul Plummer
2 days ago
1
Do you mean isomorphic or equal? For equality, the answer is trivially no. For isomorphism, you might as well assume that $N=1$, and then you are asking whether a group can be isomorphic to a proper subgroup, and then you can take an example like $mathbb{Z}$ and any of its proper non-trivial subgroups.
– verret
2 days ago
add a comment |
2
You can have groups which are isomorphic to proper subgroups(integers, free group on two generators), so for N trivial you get the result. Can you apply the five lemma in the motivating example(or are the isomorhism you talk about not compatible with the exact sequences.
– Paul Plummer
2 days ago
1
Do you mean isomorphic or equal? For equality, the answer is trivially no. For isomorphism, you might as well assume that $N=1$, and then you are asking whether a group can be isomorphic to a proper subgroup, and then you can take an example like $mathbb{Z}$ and any of its proper non-trivial subgroups.
– verret
2 days ago
2
2
You can have groups which are isomorphic to proper subgroups(integers, free group on two generators), so for N trivial you get the result. Can you apply the five lemma in the motivating example(or are the isomorhism you talk about not compatible with the exact sequences.
– Paul Plummer
2 days ago
You can have groups which are isomorphic to proper subgroups(integers, free group on two generators), so for N trivial you get the result. Can you apply the five lemma in the motivating example(or are the isomorhism you talk about not compatible with the exact sequences.
– Paul Plummer
2 days ago
1
1
Do you mean isomorphic or equal? For equality, the answer is trivially no. For isomorphism, you might as well assume that $N=1$, and then you are asking whether a group can be isomorphic to a proper subgroup, and then you can take an example like $mathbb{Z}$ and any of its proper non-trivial subgroups.
– verret
2 days ago
Do you mean isomorphic or equal? For equality, the answer is trivially no. For isomorphism, you might as well assume that $N=1$, and then you are asking whether a group can be isomorphic to a proper subgroup, and then you can take an example like $mathbb{Z}$ and any of its proper non-trivial subgroups.
– verret
2 days ago
add a comment |
1 Answer
1
active
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votes
up vote
2
down vote
accepted
No. That $gin G$, and $G/N equiv H/N$ would imply that the coset $gN subset H$, so that $gin H$. This for all $g$.
answered 2 days ago
R.C.Cowsik
1653
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
No. That $gin G$, and $G/N equiv H/N$ would imply that the coset $gN subset H$, so that $gin H$. This for all $g$.
answered 2 days ago
R.C.Cowsik
1653
add a comment |
up vote
2
down vote
accepted
No. That $gin G$, and $G/N equiv H/N$ would imply that the coset $gN subset H$, so that $gin H$. This for all $g$.
answered 2 days ago
R.C.Cowsik
1653
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
No. That $gin G$, and $G/N equiv H/N$ would imply that the coset $gN subset H$, so that $gin H$. This for all $g$.
answered 2 days ago
R.C.Cowsik
1653
No. That $gin G$, and $G/N equiv H/N$ would imply that the coset $gN subset H$, so that $gin H$. This for all $g$.
answered 2 days ago
R.C.Cowsik
1653
answered 2 days ago
R.C.Cowsik
1653
answered 2 days ago
R.C.Cowsik
1653
answered 2 days ago
R.C.Cowsik
1653
1653
add a comment |
add a comment |
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You can have groups which are isomorphic to proper subgroups(integers, free group on two generators), so for N trivial you get the result. Can you apply the five lemma in the motivating example(or are the isomorhism you talk about not compatible with the exact sequences.
– Paul Plummer
2 days ago
1
Do you mean isomorphic or equal? For equality, the answer is trivially no. For isomorphism, you might as well assume that $N=1$, and then you are asking whether a group can be isomorphic to a proper subgroup, and then you can take an example like $mathbb{Z}$ and any of its proper non-trivial subgroups.
– verret
2 days ago