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Let $G$ be a group and $A subseteq G$. Assume $aAmid a in G$ form a partition of $G$. For some $a_0in G$, $H=a_0A$ is a subgroup of $G$.



The 2019 Stack Overflow Developer Survey Results Are InUnion of more than $2$ subgroups can be a subgroup for some groupQuestion about Fraleigh book, semidirect products and group action.Let $G$ be an abelian group;$H$, $K$ are finite cyclic subgroups. Show that $G$ contains a cyclic subgroup of order $lcm(r, s)$.partition of a group and cosetsShow that the equation $x^2ax=a^-1$ is solvable for $x$ in a group $G$Some Subgroup of Dihedral Group is NormalLet $G$ be a group and $K = x^2 mid xin G$Prove that for the additive group (Z, +) of integers every subgroup is of the form kZ.partition of a group to have normal subgroupShowing that a factor group of upper triangular matrices is abelian










6












$begingroup$


This is an extra problem that my Algebra teacher suggested working through for practice.



Let $G$ be a group and let $A subseteq G$. Assume that $K = aAmid a in G $ form a partition of G. Show that there exists $a_0in G$ such that H = $a_0$A is a subgroup of G and $aAmid a in G = bHmid b in G$.



I know that if $K$ is a partition of $G$, then either $aA = bA$ or $aA cap bA = emptyset $ for all $a,b in G$ and that there must exist $a_0in A$ such that for $x in G$, $a_0a = x$ which is equivalent to $a_0 = xa^-1$, since $G$ is a group. This format is similar to $H = a_0A$. I'm struggling with forming this subgroup $H$ and showing it's a subgroup. As for the second part, I think I could just show that each set is a subset of the other.



I am new to Abstract Algebra and struggle with the concepts. I am looking for very straightforward, simple suggestions. Thank you!






abstract-algebra group-theory







share|cite|improve this question











$endgroup$







  • 1




    $begingroup$
    For any $a_0 in G$, $a_0 G subset G = eG = a_0 a_0^-1 G subset a_0 G$, so $a_0 G = G$.
    $endgroup$
    – jawheele
    Apr 6 at 16:33















6












$begingroup$


This is an extra problem that my Algebra teacher suggested working through for practice.



Let $G$ be a group and let $A subseteq G$. Assume that $K = aAmid a in G $ form a partition of G. Show that there exists $a_0in G$ such that H = $a_0$A is a subgroup of G and $aAmid a in G = bHmid b in G$.



I know that if $K$ is a partition of $G$, then either $aA = bA$ or $aA cap bA = emptyset $ for all $a,b in G$ and that there must exist $a_0in A$ such that for $x in G$, $a_0a = x$ which is equivalent to $a_0 = xa^-1$, since $G$ is a group. This format is similar to $H = a_0A$. I'm struggling with forming this subgroup $H$ and showing it's a subgroup. As for the second part, I think I could just show that each set is a subset of the other.



I am new to Abstract Algebra and struggle with the concepts. I am looking for very straightforward, simple suggestions. Thank you!






abstract-algebra group-theory







share|cite|improve this question











$endgroup$







  • 1




    $begingroup$
    For any $a_0 in G$, $a_0 G subset G = eG = a_0 a_0^-1 G subset a_0 G$, so $a_0 G = G$.
    $endgroup$
    – jawheele
    Apr 6 at 16:33













6












6








6





$begingroup$


This is an extra problem that my Algebra teacher suggested working through for practice.



Let $G$ be a group and let $A subseteq G$. Assume that $K = aAmid a in G $ form a partition of G. Show that there exists $a_0in G$ such that H = $a_0$A is a subgroup of G and $aAmid a in G = bHmid b in G$.



I know that if $K$ is a partition of $G$, then either $aA = bA$ or $aA cap bA = emptyset $ for all $a,b in G$ and that there must exist $a_0in A$ such that for $x in G$, $a_0a = x$ which is equivalent to $a_0 = xa^-1$, since $G$ is a group. This format is similar to $H = a_0A$. I'm struggling with forming this subgroup $H$ and showing it's a subgroup. As for the second part, I think I could just show that each set is a subset of the other.



I am new to Abstract Algebra and struggle with the concepts. I am looking for very straightforward, simple suggestions. Thank you!






abstract-algebra group-theory







share|cite|improve this question











$endgroup$




This is an extra problem that my Algebra teacher suggested working through for practice.



Let $G$ be a group and let $A subseteq G$. Assume that $K = aAmid a in G $ form a partition of G. Show that there exists $a_0in G$ such that H = $a_0$A is a subgroup of G and $aAmid a in G = bHmid b in G$.



I know that if $K$ is a partition of $G$, then either $aA = bA$ or $aA cap bA = emptyset $ for all $a,b in G$ and that there must exist $a_0in A$ such that for $x in G$, $a_0a = x$ which is equivalent to $a_0 = xa^-1$, since $G$ is a group. This format is similar to $H = a_0A$. I'm struggling with forming this subgroup $H$ and showing it's a subgroup. As for the second part, I think I could just show that each set is a subset of the other.



I am new to Abstract Algebra and struggle with the concepts. I am looking for very straightforward, simple suggestions. Thank you!






abstract-algebra group-theory




abstract-algebra group-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Apr 6 at 16:54







dorkichar

















asked Apr 6 at 16:22









dorkichardorkichar

1024




1024







  • 1




    $begingroup$
    For any $a_0 in G$, $a_0 G subset G = eG = a_0 a_0^-1 G subset a_0 G$, so $a_0 G = G$.
    $endgroup$
    – jawheele
    Apr 6 at 16:33












  • 1




    $begingroup$
    For any $a_0 in G$, $a_0 G subset G = eG = a_0 a_0^-1 G subset a_0 G$, so $a_0 G = G$.
    $endgroup$
    – jawheele
    Apr 6 at 16:33







1




1




$begingroup$
For any $a_0 in G$, $a_0 G subset G = eG = a_0 a_0^-1 G subset a_0 G$, so $a_0 G = G$.
$endgroup$
– jawheele
Apr 6 at 16:33




$begingroup$
For any $a_0 in G$, $a_0 G subset G = eG = a_0 a_0^-1 G subset a_0 G$, so $a_0 G = G$.
$endgroup$
– jawheele
Apr 6 at 16:33










2 Answers
2






active

oldest

votes


















1












$begingroup$

Hint: Any partition $P$ of $G$ defines an equivalence relation $sim_P$ (and vice versa). Consider $$[e]_sim_K=gin Gmid gsim_K e,$$ where $e$ is the identity of $G$.



Note: $e$ is not necessarily in $A$.






share|cite|improve this answer











$endgroup$












  • $begingroup$
    Let me know if you need a further hint, @dorkichar.
    $endgroup$
    – Shaun
    Apr 6 at 17:21


















1












$begingroup$

Here’s the trick we want to use:




Let $gin G$ be arbitrary. Then $gin H$ if and only if $gH = H$ if and only if $gH cap H ne emptyset$.




You should prove this! Use the fact that the translates of $H$ form a partition of $G$.



Now, the first step in proving $H$ is a subgroup is showing that for any $hin H$, its inverse $h^-1$ is also in $H$. Explain why $ein h^-1H$, and why this proves that $h^-1in H$.



The second step in proving $H$ is a subgroup is showing that for any $g,hin H$, their product $gh$ is also in $H$. Using the step above, prove that $gin (gh)H$ and explain why this means that $ghin H$.






share|cite|improve this answer









$endgroup$













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    2 Answers
    2






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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    Hint: Any partition $P$ of $G$ defines an equivalence relation $sim_P$ (and vice versa). Consider $$[e]_sim_K=gin Gmid gsim_K e,$$ where $e$ is the identity of $G$.



    Note: $e$ is not necessarily in $A$.






    share|cite|improve this answer











    $endgroup$












    • $begingroup$
      Let me know if you need a further hint, @dorkichar.
      $endgroup$
      – Shaun
      Apr 6 at 17:21















    1












    $begingroup$

    Hint: Any partition $P$ of $G$ defines an equivalence relation $sim_P$ (and vice versa). Consider $$[e]_sim_K=gin Gmid gsim_K e,$$ where $e$ is the identity of $G$.



    Note: $e$ is not necessarily in $A$.






    share|cite|improve this answer











    $endgroup$












    • $begingroup$
      Let me know if you need a further hint, @dorkichar.
      $endgroup$
      – Shaun
      Apr 6 at 17:21













    1












    1








    1





    $begingroup$

    Hint: Any partition $P$ of $G$ defines an equivalence relation $sim_P$ (and vice versa). Consider $$[e]_sim_K=gin Gmid gsim_K e,$$ where $e$ is the identity of $G$.



    Note: $e$ is not necessarily in $A$.






    share|cite|improve this answer











    $endgroup$



    Hint: Any partition $P$ of $G$ defines an equivalence relation $sim_P$ (and vice versa). Consider $$[e]_sim_K=gin Gmid gsim_K e,$$ where $e$ is the identity of $G$.



    Note: $e$ is not necessarily in $A$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Apr 6 at 17:29

























    answered Apr 6 at 17:07









    ShaunShaun

    10.5k113687




    10.5k113687











    • $begingroup$
      Let me know if you need a further hint, @dorkichar.
      $endgroup$
      – Shaun
      Apr 6 at 17:21
















    • $begingroup$
      Let me know if you need a further hint, @dorkichar.
      $endgroup$
      – Shaun
      Apr 6 at 17:21















    $begingroup$
    Let me know if you need a further hint, @dorkichar.
    $endgroup$
    – Shaun
    Apr 6 at 17:21




    $begingroup$
    Let me know if you need a further hint, @dorkichar.
    $endgroup$
    – Shaun
    Apr 6 at 17:21











    1












    $begingroup$

    Here’s the trick we want to use:




    Let $gin G$ be arbitrary. Then $gin H$ if and only if $gH = H$ if and only if $gH cap H ne emptyset$.




    You should prove this! Use the fact that the translates of $H$ form a partition of $G$.



    Now, the first step in proving $H$ is a subgroup is showing that for any $hin H$, its inverse $h^-1$ is also in $H$. Explain why $ein h^-1H$, and why this proves that $h^-1in H$.



    The second step in proving $H$ is a subgroup is showing that for any $g,hin H$, their product $gh$ is also in $H$. Using the step above, prove that $gin (gh)H$ and explain why this means that $ghin H$.






    share|cite|improve this answer









    $endgroup$

















      1












      $begingroup$

      Here’s the trick we want to use:




      Let $gin G$ be arbitrary. Then $gin H$ if and only if $gH = H$ if and only if $gH cap H ne emptyset$.




      You should prove this! Use the fact that the translates of $H$ form a partition of $G$.



      Now, the first step in proving $H$ is a subgroup is showing that for any $hin H$, its inverse $h^-1$ is also in $H$. Explain why $ein h^-1H$, and why this proves that $h^-1in H$.



      The second step in proving $H$ is a subgroup is showing that for any $g,hin H$, their product $gh$ is also in $H$. Using the step above, prove that $gin (gh)H$ and explain why this means that $ghin H$.






      share|cite|improve this answer









      $endgroup$















        1












        1








        1





        $begingroup$

        Here’s the trick we want to use:




        Let $gin G$ be arbitrary. Then $gin H$ if and only if $gH = H$ if and only if $gH cap H ne emptyset$.




        You should prove this! Use the fact that the translates of $H$ form a partition of $G$.



        Now, the first step in proving $H$ is a subgroup is showing that for any $hin H$, its inverse $h^-1$ is also in $H$. Explain why $ein h^-1H$, and why this proves that $h^-1in H$.



        The second step in proving $H$ is a subgroup is showing that for any $g,hin H$, their product $gh$ is also in $H$. Using the step above, prove that $gin (gh)H$ and explain why this means that $ghin H$.






        share|cite|improve this answer









        $endgroup$



        Here’s the trick we want to use:




        Let $gin G$ be arbitrary. Then $gin H$ if and only if $gH = H$ if and only if $gH cap H ne emptyset$.




        You should prove this! Use the fact that the translates of $H$ form a partition of $G$.



        Now, the first step in proving $H$ is a subgroup is showing that for any $hin H$, its inverse $h^-1$ is also in $H$. Explain why $ein h^-1H$, and why this proves that $h^-1in H$.



        The second step in proving $H$ is a subgroup is showing that for any $g,hin H$, their product $gh$ is also in $H$. Using the step above, prove that $gin (gh)H$ and explain why this means that $ghin H$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Apr 6 at 17:53









        Santana AftonSantana Afton

        3,0992730




        3,0992730



























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