Is it true that if $H_1$ and $H_2$ are isomorphic cyclic subgroups of $G$, then $G/H_1cong G/H_2$?












1














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 disprove 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?










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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
    Nov 25 at 8:11












  • @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
    Nov 25 at 8:24










  • 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
    Nov 25 at 8:33










  • @Derek: really? Then how do you refer to the transitive $G$-set with stabilizer $H$?
    – Qiaochu Yuan
    Nov 25 at 9:09












  • 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
    Nov 25 at 10:55
















1














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 disprove 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?










share|cite|improve this question
























  • 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
    Nov 25 at 8:11












  • @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
    Nov 25 at 8:24










  • 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
    Nov 25 at 8:33










  • @Derek: really? Then how do you refer to the transitive $G$-set with stabilizer $H$?
    – Qiaochu Yuan
    Nov 25 at 9:09












  • 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
    Nov 25 at 10:55














1












1








1


1





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 disprove 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?










share|cite|improve this question















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 disprove 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






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edited Dec 4 at 6:26









the_fox

2,32911430




2,32911430










asked Nov 25 at 0:25









user573497

16119




16119












  • 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
    Nov 25 at 8:11












  • @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
    Nov 25 at 8:24










  • 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
    Nov 25 at 8:33










  • @Derek: really? Then how do you refer to the transitive $G$-set with stabilizer $H$?
    – Qiaochu Yuan
    Nov 25 at 9:09












  • 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
    Nov 25 at 10:55


















  • 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
    Nov 25 at 8:11












  • @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
    Nov 25 at 8:24










  • 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
    Nov 25 at 8:33










  • @Derek: really? Then how do you refer to the transitive $G$-set with stabilizer $H$?
    – Qiaochu Yuan
    Nov 25 at 9:09












  • 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
    Nov 25 at 10:55
















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
Nov 25 at 8:11






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
Nov 25 at 8:11














@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
Nov 25 at 8:24




@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
Nov 25 at 8:24












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
Nov 25 at 8:33




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
Nov 25 at 8:33












@Derek: really? Then how do you refer to the transitive $G$-set with stabilizer $H$?
– Qiaochu Yuan
Nov 25 at 9:09






@Derek: really? Then how do you refer to the transitive $G$-set with stabilizer $H$?
– Qiaochu Yuan
Nov 25 at 9:09














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
Nov 25 at 10:55




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
Nov 25 at 10:55










2 Answers
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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 $G/K cong C_2 times C_2$.






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    2














    Did you try cyclic subgroups of the additive group of integers $mathbb{Z}$?






    share|cite|improve this answer





















    • But what about integers mod n? I edited my question and I found very interesting that observation about the isomorphism about quotient groups.
      – user573497
      Nov 25 at 0:34











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    2 Answers
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    2 Answers
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    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 $G/K cong C_2 times C_2$.






    share|cite|improve this answer


























      8














      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 $G/K cong C_2 times C_2$.






      share|cite|improve this answer
























        8












        8








        8






        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 $G/K cong C_2 times C_2$.






        share|cite|improve this answer












        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 $G/K cong C_2 times C_2$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 25 at 0:33









        the_fox

        2,32911430




        2,32911430























            2














            Did you try cyclic subgroups of the additive group of integers $mathbb{Z}$?






            share|cite|improve this answer





















            • But what about integers mod n? I edited my question and I found very interesting that observation about the isomorphism about quotient groups.
              – user573497
              Nov 25 at 0:34
















            2














            Did you try cyclic subgroups of the additive group of integers $mathbb{Z}$?






            share|cite|improve this answer





















            • But what about integers mod n? I edited my question and I found very interesting that observation about the isomorphism about quotient groups.
              – user573497
              Nov 25 at 0:34














            2












            2








            2






            Did you try cyclic subgroups of the additive group of integers $mathbb{Z}$?






            share|cite|improve this answer












            Did you try cyclic subgroups of the additive group of integers $mathbb{Z}$?







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 25 at 0:32









            Bartosz Malman

            8011620




            8011620












            • But what about integers mod n? I edited my question and I found very interesting that observation about the isomorphism about quotient groups.
              – user573497
              Nov 25 at 0:34


















            • But what about integers mod n? I edited my question and I found very interesting that observation about the isomorphism about quotient groups.
              – user573497
              Nov 25 at 0:34
















            But what about integers mod n? I edited my question and I found very interesting that observation about the isomorphism about quotient groups.
            – user573497
            Nov 25 at 0:34




            But what about integers mod n? I edited my question and I found very interesting that observation about the isomorphism about quotient groups.
            – user573497
            Nov 25 at 0:34


















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