Showing that $S = b in G : a^-1 b a = b $ is a subgroup of $G$. [closed]

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Let's say we have a group $G$ and $a$ is an element of $G$. I want to show that the set $S = b in G : a^-1 b a = b $ is a group. The hint is to find inverses.




abstract-algebra group-theory number-theory




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closed as off-topic by José Carlos Santos, Claude Leibovici, barto, Brahadeesh, rain Aug 15 at 18:29


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – José Carlos Santos, Claude Leibovici, barto, Brahadeesh, rain
If this question can be reworded to fit the rules in the help center, please edit the question.












  • The identity is 1. Take the inverse of $a^-1ba=b$ both sides for the inverse part.
    – Ginger88895
    Feb 24 at 13:01










  • $a^-1ba=b$ iff $ b in C(a)$. Now C(a) is a subgroup of G(why?)
    – Sadil Khan
    Aug 14 at 9:05














up vote
-1
down vote

favorite












Let's say we have a group $G$ and $a$ is an element of $G$. I want to show that the set $S = b in G : a^-1 b a = b $ is a group. The hint is to find inverses.




abstract-algebra group-theory number-theory




share|cite|improve this question














closed as off-topic by José Carlos Santos, Claude Leibovici, barto, Brahadeesh, rain Aug 15 at 18:29


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – José Carlos Santos, Claude Leibovici, barto, Brahadeesh, rain
If this question can be reworded to fit the rules in the help center, please edit the question.












  • The identity is 1. Take the inverse of $a^-1ba=b$ both sides for the inverse part.
    – Ginger88895
    Feb 24 at 13:01










  • $a^-1ba=b$ iff $ b in C(a)$. Now C(a) is a subgroup of G(why?)
    – Sadil Khan
    Aug 14 at 9:05












up vote
-1
down vote

favorite









up vote
-1
down vote

favorite











Let's say we have a group $G$ and $a$ is an element of $G$. I want to show that the set $S = b in G : a^-1 b a = b $ is a group. The hint is to find inverses.




abstract-algebra group-theory number-theory




share|cite|improve this question














Let's say we have a group $G$ and $a$ is an element of $G$. I want to show that the set $S = b in G : a^-1 b a = b $ is a group. The hint is to find inverses.





abstract-algebra group-theory number-theory





share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 15 at 10:08









Brahadeesh

4,06131549




4,06131549










asked Feb 24 at 12:57









rain

39118




39118




closed as off-topic by José Carlos Santos, Claude Leibovici, barto, Brahadeesh, rain Aug 15 at 18:29


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – José Carlos Santos, Claude Leibovici, barto, Brahadeesh, rain
If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by José Carlos Santos, Claude Leibovici, barto, Brahadeesh, rain Aug 15 at 18:29


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – José Carlos Santos, Claude Leibovici, barto, Brahadeesh, rain
If this question can be reworded to fit the rules in the help center, please edit the question.











  • The identity is 1. Take the inverse of $a^-1ba=b$ both sides for the inverse part.
    – Ginger88895
    Feb 24 at 13:01










  • $a^-1ba=b$ iff $ b in C(a)$. Now C(a) is a subgroup of G(why?)
    – Sadil Khan
    Aug 14 at 9:05
















  • The identity is 1. Take the inverse of $a^-1ba=b$ both sides for the inverse part.
    – Ginger88895
    Feb 24 at 13:01










  • $a^-1ba=b$ iff $ b in C(a)$. Now C(a) is a subgroup of G(why?)
    – Sadil Khan
    Aug 14 at 9:05















The identity is 1. Take the inverse of $a^-1ba=b$ both sides for the inverse part.
– Ginger88895
Feb 24 at 13:01




The identity is 1. Take the inverse of $a^-1ba=b$ both sides for the inverse part.
– Ginger88895
Feb 24 at 13:01












$a^-1ba=b$ iff $ b in C(a)$. Now C(a) is a subgroup of G(why?)
– Sadil Khan
Aug 14 at 9:05




$a^-1ba=b$ iff $ b in C(a)$. Now C(a) is a subgroup of G(why?)
– Sadil Khan
Aug 14 at 9:05










1 Answer
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Assuming that $bin S$ (i.e., that $a^-1ba=b$), you want to show that the inverse $b^-1$ of $b$ (which is known to exist in the group $G$) is also in $S$. That is, you want to show $a^-1b^-1a=b^-1$:
$$a^-1b^-1a = a^-1b^-1a =cdot 1=a^-1b^-1acdot colorredbcdot b^-1 = underbracea^-1b^-1acdot colorreda^-1ba_textall cancelscdot b^-1=b^-1$$






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    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    4
    down vote



    accepted










    Assuming that $bin S$ (i.e., that $a^-1ba=b$), you want to show that the inverse $b^-1$ of $b$ (which is known to exist in the group $G$) is also in $S$. That is, you want to show $a^-1b^-1a=b^-1$:
    $$a^-1b^-1a = a^-1b^-1a =cdot 1=a^-1b^-1acdot colorredbcdot b^-1 = underbracea^-1b^-1acdot colorreda^-1ba_textall cancelscdot b^-1=b^-1$$






    share|cite|improve this answer
























      up vote
      4
      down vote



      accepted










      Assuming that $bin S$ (i.e., that $a^-1ba=b$), you want to show that the inverse $b^-1$ of $b$ (which is known to exist in the group $G$) is also in $S$. That is, you want to show $a^-1b^-1a=b^-1$:
      $$a^-1b^-1a = a^-1b^-1a =cdot 1=a^-1b^-1acdot colorredbcdot b^-1 = underbracea^-1b^-1acdot colorreda^-1ba_textall cancelscdot b^-1=b^-1$$






      share|cite|improve this answer






















        up vote
        4
        down vote



        accepted







        up vote
        4
        down vote



        accepted






        Assuming that $bin S$ (i.e., that $a^-1ba=b$), you want to show that the inverse $b^-1$ of $b$ (which is known to exist in the group $G$) is also in $S$. That is, you want to show $a^-1b^-1a=b^-1$:
        $$a^-1b^-1a = a^-1b^-1a =cdot 1=a^-1b^-1acdot colorredbcdot b^-1 = underbracea^-1b^-1acdot colorreda^-1ba_textall cancelscdot b^-1=b^-1$$






        share|cite|improve this answer












        Assuming that $bin S$ (i.e., that $a^-1ba=b$), you want to show that the inverse $b^-1$ of $b$ (which is known to exist in the group $G$) is also in $S$. That is, you want to show $a^-1b^-1a=b^-1$:
        $$a^-1b^-1a = a^-1b^-1a =cdot 1=a^-1b^-1acdot colorredbcdot b^-1 = underbracea^-1b^-1acdot colorreda^-1ba_textall cancelscdot b^-1=b^-1$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Feb 24 at 13:02









        Hagen von Eitzen

        265k21258477




        265k21258477












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