Is it true that if $H_1$ and $H_2$ are isomorphic cyclic subgroups of $G$, then $G/H_1cong G/H_2$?
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I have a question: it is true that if $H_1$ and $H_2$ are cyclic groups that are isomorphic, then $G/H_1$ is isomorphic to $G/H_2$? I know that if I remove the condition "cyclic groups", the given statement is false and there are numerous counterexamples that disproves it, but I don't know if my statement is true and I don't how to create a counterexample or to prove it. If it is true, can you give me a hint about how can this be proven?
For example, I just have shown that $Z_{12}/ langle 2 rangle$ is isomorphic to $Z_{12}/Z_6$ which is isomorphic to $Z_3$ (since both $langle 2 rangle$ and $Z_6$ are isomorphic). How this can be generalized?
group-theory normal-subgroups
edited 2 days ago
Asaf Karagila♦
300k32420750
asked 2 days ago
user573497
15919
add a comment |
up vote
1
down vote
favorite
I have a question: it is true that if $H_1$ and $H_2$ are cyclic groups that are isomorphic, then $G/H_1$ is isomorphic to $G/H_2$? I know that if I remove the condition "cyclic groups", the given statement is false and there are numerous counterexamples that disproves it, but I don't know if my statement is true and I don't how to create a counterexample or to prove it. If it is true, can you give me a hint about how can this be proven?
For example, I just have shown that $Z_{12}/ langle 2 rangle$ is isomorphic to $Z_{12}/Z_6$ which is isomorphic to $Z_3$ (since both $langle 2 rangle$ and $Z_6$ are isomorphic). How this can be generalized?
group-theory normal-subgroups
edited 2 days ago
Asaf Karagila♦
300k32420750
asked 2 days ago
user573497
15919
You haven't specified whether $G$ is abelian or whether your subgroups are normal or not, so it's ambiguous what you mean by "isomorphic" in your question. If $G$ is nonabelian the natural reading would be "isomorphic as $G$-sets," but it sounds like you intend $G$ to be abelian and to ask for an isomorphism of abelian groups.
– Qiaochu Yuan
yesterday
@QiaochuYuan Many people working in group theory (including myself) only write $G/N$ when $N$ is a normal subgroup of $G$. And the fact that reference is made to isomorphisms between two quotients strongly suggests that this convention is eing used here.
– Derek Holt
yesterday
I agree with Derek. In this particular case though, I think the OP has simply neglected to mention that $H_1$, $H_2$ are assumed to be normal subgroups of the group.
– the_fox
yesterday
@Derek: really? Then how do you refer to the transitive $G$-set with stabilizer $H$?
– Qiaochu Yuan
yesterday
I don't have a specific notation for that. But in any case that would (mildly) conflict with its meaning as a quotient group, because a group is not the same thing as a $G$-set.
– Derek Holt
yesterday
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I have a question: it is true that if $H_1$ and $H_2$ are cyclic groups that are isomorphic, then $G/H_1$ is isomorphic to $G/H_2$? I know that if I remove the condition "cyclic groups", the given statement is false and there are numerous counterexamples that disproves it, but I don't know if my statement is true and I don't how to create a counterexample or to prove it. If it is true, can you give me a hint about how can this be proven?
For example, I just have shown that $Z_{12}/ langle 2 rangle$ is isomorphic to $Z_{12}/Z_6$ which is isomorphic to $Z_3$ (since both $langle 2 rangle$ and $Z_6$ are isomorphic). How this can be generalized?
group-theory normal-subgroups
edited 2 days ago
Asaf Karagila♦
300k32420750
asked 2 days ago
user573497
15919
I have a question: it is true that if $H_1$ and $H_2$ are cyclic groups that are isomorphic, then $G/H_1$ is isomorphic to $G/H_2$? I know that if I remove the condition "cyclic groups", the given statement is false and there are numerous counterexamples that disproves it, but I don't know if my statement is true and I don't how to create a counterexample or to prove it. If it is true, can you give me a hint about how can this be proven?
For example, I just have shown that $Z_{12}/ langle 2 rangle$ is isomorphic to $Z_{12}/Z_6$ which is isomorphic to $Z_3$ (since both $langle 2 rangle$ and $Z_6$ are isomorphic). How this can be generalized?
group-theory normal-subgroups
group-theory normal-subgroups
edited 2 days ago
Asaf Karagila♦
300k32420750
asked 2 days ago
user573497
15919
edited 2 days ago
Asaf Karagila♦
300k32420750
asked 2 days ago
user573497
15919
edited 2 days ago
Asaf Karagila♦
300k32420750
edited 2 days ago
Asaf Karagila♦
300k32420750
edited 2 days ago
Asaf Karagila♦
300k32420750
300k32420750
asked 2 days ago
user573497
15919
asked 2 days ago
user573497
15919
asked 2 days ago
user573497
15919
15919
You haven't specified whether $G$ is abelian or whether your subgroups are normal or not, so it's ambiguous what you mean by "isomorphic" in your question. If $G$ is nonabelian the natural reading would be "isomorphic as $G$-sets," but it sounds like you intend $G$ to be abelian and to ask for an isomorphism of abelian groups.
– Qiaochu Yuan
yesterday
@QiaochuYuan Many people working in group theory (including myself) only write $G/N$ when $N$ is a normal subgroup of $G$. And the fact that reference is made to isomorphisms between two quotients strongly suggests that this convention is eing used here.
– Derek Holt
yesterday
I agree with Derek. In this particular case though, I think the OP has simply neglected to mention that $H_1$, $H_2$ are assumed to be normal subgroups of the group.
– the_fox
yesterday
@Derek: really? Then how do you refer to the transitive $G$-set with stabilizer $H$?
– Qiaochu Yuan
yesterday
I don't have a specific notation for that. But in any case that would (mildly) conflict with its meaning as a quotient group, because a group is not the same thing as a $G$-set.
– Derek Holt
yesterday
add a comment |
You haven't specified whether $G$ is abelian or whether your subgroups are normal or not, so it's ambiguous what you mean by "isomorphic" in your question. If $G$ is nonabelian the natural reading would be "isomorphic as $G$-sets," but it sounds like you intend $G$ to be abelian and to ask for an isomorphism of abelian groups.
– Qiaochu Yuan
yesterday
@QiaochuYuan Many people working in group theory (including myself) only write $G/N$ when $N$ is a normal subgroup of $G$. And the fact that reference is made to isomorphisms between two quotients strongly suggests that this convention is eing used here.
– Derek Holt
yesterday
I agree with Derek. In this particular case though, I think the OP has simply neglected to mention that $H_1$, $H_2$ are assumed to be normal subgroups of the group.
– the_fox
yesterday
@Derek: really? Then how do you refer to the transitive $G$-set with stabilizer $H$?
– Qiaochu Yuan
yesterday
I don't have a specific notation for that. But in any case that would (mildly) conflict with its meaning as a quotient group, because a group is not the same thing as a $G$-set.
– Derek Holt
yesterday
You haven't specified whether $G$ is abelian or whether your subgroups are normal or not, so it's ambiguous what you mean by "isomorphic" in your question. If $G$ is nonabelian the natural reading would be "isomorphic as $G$-sets," but it sounds like you intend $G$ to be abelian and to ask for an isomorphism of abelian groups.
– Qiaochu Yuan
yesterday
You haven't specified whether $G$ is abelian or whether your subgroups are normal or not, so it's ambiguous what you mean by "isomorphic" in your question. If $G$ is nonabelian the natural reading would be "isomorphic as $G$-sets," but it sounds like you intend $G$ to be abelian and to ask for an isomorphism of abelian groups.
– Qiaochu Yuan
yesterday
@QiaochuYuan Many people working in group theory (including myself) only write $G/N$ when $N$ is a normal subgroup of $G$. And the fact that reference is made to isomorphisms between two quotients strongly suggests that this convention is eing used here.
– Derek Holt
yesterday
@QiaochuYuan Many people working in group theory (including myself) only write $G/N$ when $N$ is a normal subgroup of $G$. And the fact that reference is made to isomorphisms between two quotients strongly suggests that this convention is eing used here.
– Derek Holt
yesterday
I agree with Derek. In this particular case though, I think the OP has simply neglected to mention that $H_1$, $H_2$ are assumed to be normal subgroups of the group.
– the_fox
yesterday
I agree with Derek. In this particular case though, I think the OP has simply neglected to mention that $H_1$, $H_2$ are assumed to be normal subgroups of the group.
– the_fox
yesterday
@Derek: really? Then how do you refer to the transitive $G$-set with stabilizer $H$?
– Qiaochu Yuan
yesterday
@Derek: really? Then how do you refer to the transitive $G$-set with stabilizer $H$?
– Qiaochu Yuan
yesterday
I don't have a specific notation for that. But in any case that would (mildly) conflict with its meaning as a quotient group, because a group is not the same thing as a $G$-set.
– Derek Holt
yesterday
I don't have a specific notation for that. But in any case that would (mildly) conflict with its meaning as a quotient group, because a group is not the same thing as a $G$-set.
– Derek Holt
yesterday
add a comment |
2 Answers
2
active
oldest
votes
up vote
8
down vote
No, that's not true. Take $G = C_2 times C_4$ and note that since $G$ is abelian, every subgroup of $G$ is normal in $G$. Let $H = C_2$ be the first direct factor of $G$ and $K$ be the unique subgroup of order $2$ of $C_4$. Obviously, $H cong K$ since both have order $2$, but $G/H cong C_4$ while $G/K cong C_2 times C_2$.
answered 2 days ago
the_fox
2,1761429
add a comment |
up vote
2
down vote
Did you try cyclic subgroups of the additive group of integers $mathbb{Z}$?
answered 2 days ago
Bartosz Malman
7181520
But what about integers mod n? I edited my question and I found very interesting that observation about the isomorphism about quotient groups.
– user573497
2 days ago
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
8
down vote
No, that's not true. Take $G = C_2 times C_4$ and note that since $G$ is abelian, every subgroup of $G$ is normal in $G$. Let $H = C_2$ be the first direct factor of $G$ and $K$ be the unique subgroup of order $2$ of $C_4$. Obviously, $H cong K$ since both have order $2$, but $G/H cong C_4$ while $G/K cong C_2 times C_2$.
answered 2 days ago
the_fox
2,1761429
add a comment |
up vote
8
down vote
No, that's not true. Take $G = C_2 times C_4$ and note that since $G$ is abelian, every subgroup of $G$ is normal in $G$. Let $H = C_2$ be the first direct factor of $G$ and $K$ be the unique subgroup of order $2$ of $C_4$. Obviously, $H cong K$ since both have order $2$, but $G/H cong C_4$ while $G/K cong C_2 times C_2$.
answered 2 days ago
the_fox
2,1761429
add a comment |
up vote
8
down vote
up vote
8
down vote
No, that's not true. Take $G = C_2 times C_4$ and note that since $G$ is abelian, every subgroup of $G$ is normal in $G$. Let $H = C_2$ be the first direct factor of $G$ and $K$ be the unique subgroup of order $2$ of $C_4$. Obviously, $H cong K$ since both have order $2$, but $G/H cong C_4$ while $G/K cong C_2 times C_2$.
answered 2 days ago
the_fox
2,1761429
No, that's not true. Take $G = C_2 times C_4$ and note that since $G$ is abelian, every subgroup of $G$ is normal in $G$. Let $H = C_2$ be the first direct factor of $G$ and $K$ be the unique subgroup of order $2$ of $C_4$. Obviously, $H cong K$ since both have order $2$, but $G/H cong C_4$ while $G/K cong C_2 times C_2$.
answered 2 days ago
the_fox
2,1761429
answered 2 days ago
the_fox
2,1761429
answered 2 days ago
the_fox
2,1761429
answered 2 days ago
the_fox
2,1761429
2,1761429
add a comment |
add a comment |
up vote
2
down vote
Did you try cyclic subgroups of the additive group of integers $mathbb{Z}$?
answered 2 days ago
Bartosz Malman
7181520
But what about integers mod n? I edited my question and I found very interesting that observation about the isomorphism about quotient groups.
– user573497
2 days ago
add a comment |
up vote
2
down vote
Did you try cyclic subgroups of the additive group of integers $mathbb{Z}$?
answered 2 days ago
Bartosz Malman
7181520
But what about integers mod n? I edited my question and I found very interesting that observation about the isomorphism about quotient groups.
– user573497
2 days ago
add a comment |
up vote
2
down vote
up vote
2
down vote
Did you try cyclic subgroups of the additive group of integers $mathbb{Z}$?
answered 2 days ago
Bartosz Malman
7181520
Did you try cyclic subgroups of the additive group of integers $mathbb{Z}$?
answered 2 days ago
Bartosz Malman
7181520
answered 2 days ago
Bartosz Malman
7181520
answered 2 days ago
Bartosz Malman
7181520
answered 2 days ago
Bartosz Malman
7181520
7181520
But what about integers mod n? I edited my question and I found very interesting that observation about the isomorphism about quotient groups.
– user573497
2 days ago
add a comment |
But what about integers mod n? I edited my question and I found very interesting that observation about the isomorphism about quotient groups.
– user573497
2 days ago
But what about integers mod n? I edited my question and I found very interesting that observation about the isomorphism about quotient groups.
– user573497
2 days ago
But what about integers mod n? I edited my question and I found very interesting that observation about the isomorphism about quotient groups.
– user573497
2 days ago
add a comment |
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You haven't specified whether $G$ is abelian or whether your subgroups are normal or not, so it's ambiguous what you mean by "isomorphic" in your question. If $G$ is nonabelian the natural reading would be "isomorphic as $G$-sets," but it sounds like you intend $G$ to be abelian and to ask for an isomorphism of abelian groups.
– Qiaochu Yuan
yesterday
@QiaochuYuan Many people working in group theory (including myself) only write $G/N$ when $N$ is a normal subgroup of $G$. And the fact that reference is made to isomorphisms between two quotients strongly suggests that this convention is eing used here.
– Derek Holt
yesterday
I agree with Derek. In this particular case though, I think the OP has simply neglected to mention that $H_1$, $H_2$ are assumed to be normal subgroups of the group.
– the_fox
yesterday
@Derek: really? Then how do you refer to the transitive $G$-set with stabilizer $H$?
– Qiaochu Yuan
yesterday
I don't have a specific notation for that. But in any case that would (mildly) conflict with its meaning as a quotient group, because a group is not the same thing as a $G$-set.
– Derek Holt
yesterday