Coset of subgroup

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We are given the group $G=mathbbZ_8$ , an element $a$ of $G$ and a non-trivial subgroup $H$ of $G$ such that $Sigma=a,5 $ is a coset of $H$ in $G$. Then $a$ is one of these numbers:



$$7,0,5,6,1.$$



I think that $a$ is $5$ since a coset contains only numbers that are multiple the one to the other, am I right?










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    We are given the group $G=mathbbZ_8$ , an element $a$ of $G$ and a non-trivial subgroup $H$ of $G$ such that $Sigma=a,5 $ is a coset of $H$ in $G$. Then $a$ is one of these numbers:



    $$7,0,5,6,1.$$



    I think that $a$ is $5$ since a coset contains only numbers that are multiple the one to the other, am I right?










    share|cite|improve this question























      up vote
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      down vote

      favorite









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      We are given the group $G=mathbbZ_8$ , an element $a$ of $G$ and a non-trivial subgroup $H$ of $G$ such that $Sigma=a,5 $ is a coset of $H$ in $G$. Then $a$ is one of these numbers:



      $$7,0,5,6,1.$$



      I think that $a$ is $5$ since a coset contains only numbers that are multiple the one to the other, am I right?










      share|cite|improve this question













      We are given the group $G=mathbbZ_8$ , an element $a$ of $G$ and a non-trivial subgroup $H$ of $G$ such that $Sigma=a,5 $ is a coset of $H$ in $G$. Then $a$ is one of these numbers:



      $$7,0,5,6,1.$$



      I think that $a$ is $5$ since a coset contains only numbers that are multiple the one to the other, am I right?







      linear-algebra






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      asked Sep 10 at 10:45









      Evinda

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          No, you are not right. All cosets have the same number of elements. So, if $Sigma$ only has a single element, all other cosets have a single element too and that can only happen if $H=0$. But we are assuming that $H$ is a non-trivial subgroup.



          Hint: The answer is $a=1$. Can you prove it?






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          • $left|Sigmaright|=1$ only happens when $H=0$, not when $H=G$. So for $a=5$ we have a coset of the trivial subgroup.
            – Christoph
            Sep 10 at 10:57











          • @Christoph I've edited my answer. Thank you.
            – José Carlos Santos
            Sep 10 at 10:57

















          up vote
          1
          down vote













          Here are the intermediate steps I would do to get through your problem:



          1. How many elements does $H$ have?

          2. Which subgroups of $G$ have that many elements?

          3. For each of those subgroups, what are all its cosets?

          4. Which of those cosets contain $5$?

          5. What other elements are together with $5$ in any of those cosets?





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            2 Answers
            2






            active

            oldest

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            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

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            up vote
            2
            down vote













            No, you are not right. All cosets have the same number of elements. So, if $Sigma$ only has a single element, all other cosets have a single element too and that can only happen if $H=0$. But we are assuming that $H$ is a non-trivial subgroup.



            Hint: The answer is $a=1$. Can you prove it?






            share|cite|improve this answer






















            • $left|Sigmaright|=1$ only happens when $H=0$, not when $H=G$. So for $a=5$ we have a coset of the trivial subgroup.
              – Christoph
              Sep 10 at 10:57











            • @Christoph I've edited my answer. Thank you.
              – José Carlos Santos
              Sep 10 at 10:57














            up vote
            2
            down vote













            No, you are not right. All cosets have the same number of elements. So, if $Sigma$ only has a single element, all other cosets have a single element too and that can only happen if $H=0$. But we are assuming that $H$ is a non-trivial subgroup.



            Hint: The answer is $a=1$. Can you prove it?






            share|cite|improve this answer






















            • $left|Sigmaright|=1$ only happens when $H=0$, not when $H=G$. So for $a=5$ we have a coset of the trivial subgroup.
              – Christoph
              Sep 10 at 10:57











            • @Christoph I've edited my answer. Thank you.
              – José Carlos Santos
              Sep 10 at 10:57












            up vote
            2
            down vote










            up vote
            2
            down vote









            No, you are not right. All cosets have the same number of elements. So, if $Sigma$ only has a single element, all other cosets have a single element too and that can only happen if $H=0$. But we are assuming that $H$ is a non-trivial subgroup.



            Hint: The answer is $a=1$. Can you prove it?






            share|cite|improve this answer














            No, you are not right. All cosets have the same number of elements. So, if $Sigma$ only has a single element, all other cosets have a single element too and that can only happen if $H=0$. But we are assuming that $H$ is a non-trivial subgroup.



            Hint: The answer is $a=1$. Can you prove it?







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Sep 10 at 10:57

























            answered Sep 10 at 10:49









            José Carlos Santos

            124k17101186




            124k17101186











            • $left|Sigmaright|=1$ only happens when $H=0$, not when $H=G$. So for $a=5$ we have a coset of the trivial subgroup.
              – Christoph
              Sep 10 at 10:57











            • @Christoph I've edited my answer. Thank you.
              – José Carlos Santos
              Sep 10 at 10:57
















            • $left|Sigmaright|=1$ only happens when $H=0$, not when $H=G$. So for $a=5$ we have a coset of the trivial subgroup.
              – Christoph
              Sep 10 at 10:57











            • @Christoph I've edited my answer. Thank you.
              – José Carlos Santos
              Sep 10 at 10:57















            $left|Sigmaright|=1$ only happens when $H=0$, not when $H=G$. So for $a=5$ we have a coset of the trivial subgroup.
            – Christoph
            Sep 10 at 10:57





            $left|Sigmaright|=1$ only happens when $H=0$, not when $H=G$. So for $a=5$ we have a coset of the trivial subgroup.
            – Christoph
            Sep 10 at 10:57













            @Christoph I've edited my answer. Thank you.
            – José Carlos Santos
            Sep 10 at 10:57




            @Christoph I've edited my answer. Thank you.
            – José Carlos Santos
            Sep 10 at 10:57










            up vote
            1
            down vote













            Here are the intermediate steps I would do to get through your problem:



            1. How many elements does $H$ have?

            2. Which subgroups of $G$ have that many elements?

            3. For each of those subgroups, what are all its cosets?

            4. Which of those cosets contain $5$?

            5. What other elements are together with $5$ in any of those cosets?





            share|cite|improve this answer
























              up vote
              1
              down vote













              Here are the intermediate steps I would do to get through your problem:



              1. How many elements does $H$ have?

              2. Which subgroups of $G$ have that many elements?

              3. For each of those subgroups, what are all its cosets?

              4. Which of those cosets contain $5$?

              5. What other elements are together with $5$ in any of those cosets?





              share|cite|improve this answer






















                up vote
                1
                down vote










                up vote
                1
                down vote









                Here are the intermediate steps I would do to get through your problem:



                1. How many elements does $H$ have?

                2. Which subgroups of $G$ have that many elements?

                3. For each of those subgroups, what are all its cosets?

                4. Which of those cosets contain $5$?

                5. What other elements are together with $5$ in any of those cosets?





                share|cite|improve this answer












                Here are the intermediate steps I would do to get through your problem:



                1. How many elements does $H$ have?

                2. Which subgroups of $G$ have that many elements?

                3. For each of those subgroups, what are all its cosets?

                4. Which of those cosets contain $5$?

                5. What other elements are together with $5$ in any of those cosets?






                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Sep 10 at 11:02









                Arthur

                102k797178




                102k797178



























                     

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