Given two equations $(xax)^3 = bx$ and $x^2a = (xa)^{-1}$ in a nonabelian group, solve for $x$.












4












$begingroup$


This is for a basic non-commutative group.



My steps:



$(xax)^3 = bx$



$xaxxaxxax = bx$



$xax^2ax^2ax = bx$



$xax^2ax^2a = b$



$x^2a = (xax^2a)^{-1}b$



Now, substituting $x^2a$ into the second equation:



$(xax^2a)^{-1}b = (xa)^{-1}$



$b = (xax^2a)(xa)^{-1} $



$b = (xax^2a)(a^{-1}x^{-1})$



$b = xax$



Now back to the first equation:



$b^3 = bx$



$b^2 = x$



Apparently, this is not the correct answer according to my answer sheet.



Am I making a mistake somewhere?






abstract-algebra group-theory







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here.
    $endgroup$
    – Shaun
    Dec 8 '18 at 11:26










  • $begingroup$
    Your solution seems okay to me.
    $endgroup$
    – Shaun
    Dec 8 '18 at 11:33










  • $begingroup$
    Also, again, please edit the title.
    $endgroup$
    – Shaun
    Dec 8 '18 at 11:35










  • $begingroup$
    Answer sheet says $(ab)^{-1} = x$
    $endgroup$
    – David Davidson
    Dec 8 '18 at 11:36










  • $begingroup$
    I think the answer sheet might be wrong. I've checked your solution several times; as I said, it seems okay to me.
    $endgroup$
    – Shaun
    Dec 8 '18 at 11:52
















4












$begingroup$


This is for a basic non-commutative group.



My steps:



$(xax)^3 = bx$



$xaxxaxxax = bx$



$xax^2ax^2ax = bx$



$xax^2ax^2a = b$



$x^2a = (xax^2a)^{-1}b$



Now, substituting $x^2a$ into the second equation:



$(xax^2a)^{-1}b = (xa)^{-1}$



$b = (xax^2a)(xa)^{-1} $



$b = (xax^2a)(a^{-1}x^{-1})$



$b = xax$



Now back to the first equation:



$b^3 = bx$



$b^2 = x$



Apparently, this is not the correct answer according to my answer sheet.



Am I making a mistake somewhere?






abstract-algebra group-theory







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here.
    $endgroup$
    – Shaun
    Dec 8 '18 at 11:26










  • $begingroup$
    Your solution seems okay to me.
    $endgroup$
    – Shaun
    Dec 8 '18 at 11:33










  • $begingroup$
    Also, again, please edit the title.
    $endgroup$
    – Shaun
    Dec 8 '18 at 11:35










  • $begingroup$
    Answer sheet says $(ab)^{-1} = x$
    $endgroup$
    – David Davidson
    Dec 8 '18 at 11:36










  • $begingroup$
    I think the answer sheet might be wrong. I've checked your solution several times; as I said, it seems okay to me.
    $endgroup$
    – Shaun
    Dec 8 '18 at 11:52














4












4








4


1



$begingroup$


This is for a basic non-commutative group.



My steps:



$(xax)^3 = bx$



$xaxxaxxax = bx$



$xax^2ax^2ax = bx$



$xax^2ax^2a = b$



$x^2a = (xax^2a)^{-1}b$



Now, substituting $x^2a$ into the second equation:



$(xax^2a)^{-1}b = (xa)^{-1}$



$b = (xax^2a)(xa)^{-1} $



$b = (xax^2a)(a^{-1}x^{-1})$



$b = xax$



Now back to the first equation:



$b^3 = bx$



$b^2 = x$



Apparently, this is not the correct answer according to my answer sheet.



Am I making a mistake somewhere?






abstract-algebra group-theory







share|cite|improve this question











$endgroup$




This is for a basic non-commutative group.



My steps:



$(xax)^3 = bx$



$xaxxaxxax = bx$



$xax^2ax^2ax = bx$



$xax^2ax^2a = b$



$x^2a = (xax^2a)^{-1}b$



Now, substituting $x^2a$ into the second equation:



$(xax^2a)^{-1}b = (xa)^{-1}$



$b = (xax^2a)(xa)^{-1} $



$b = (xax^2a)(a^{-1}x^{-1})$



$b = xax$



Now back to the first equation:



$b^3 = bx$



$b^2 = x$



Apparently, this is not the correct answer according to my answer sheet.



Am I making a mistake somewhere?






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 Dec 8 '18 at 12:57









Namaste

1




1










asked Dec 8 '18 at 11:25









David DavidsonDavid Davidson

626




626








  • 1




    $begingroup$
    Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here.
    $endgroup$
    – Shaun
    Dec 8 '18 at 11:26










  • $begingroup$
    Your solution seems okay to me.
    $endgroup$
    – Shaun
    Dec 8 '18 at 11:33










  • $begingroup$
    Also, again, please edit the title.
    $endgroup$
    – Shaun
    Dec 8 '18 at 11:35










  • $begingroup$
    Answer sheet says $(ab)^{-1} = x$
    $endgroup$
    – David Davidson
    Dec 8 '18 at 11:36










  • $begingroup$
    I think the answer sheet might be wrong. I've checked your solution several times; as I said, it seems okay to me.
    $endgroup$
    – Shaun
    Dec 8 '18 at 11:52














  • 1




    $begingroup$
    Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here.
    $endgroup$
    – Shaun
    Dec 8 '18 at 11:26










  • $begingroup$
    Your solution seems okay to me.
    $endgroup$
    – Shaun
    Dec 8 '18 at 11:33










  • $begingroup$
    Also, again, please edit the title.
    $endgroup$
    – Shaun
    Dec 8 '18 at 11:35










  • $begingroup$
    Answer sheet says $(ab)^{-1} = x$
    $endgroup$
    – David Davidson
    Dec 8 '18 at 11:36










  • $begingroup$
    I think the answer sheet might be wrong. I've checked your solution several times; as I said, it seems okay to me.
    $endgroup$
    – Shaun
    Dec 8 '18 at 11:52








1




1




$begingroup$
Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here.
$endgroup$
– Shaun
Dec 8 '18 at 11:26




$begingroup$
Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here.
$endgroup$
– Shaun
Dec 8 '18 at 11:26












$begingroup$
Your solution seems okay to me.
$endgroup$
– Shaun
Dec 8 '18 at 11:33




$begingroup$
Your solution seems okay to me.
$endgroup$
– Shaun
Dec 8 '18 at 11:33












$begingroup$
Also, again, please edit the title.
$endgroup$
– Shaun
Dec 8 '18 at 11:35




$begingroup$
Also, again, please edit the title.
$endgroup$
– Shaun
Dec 8 '18 at 11:35












$begingroup$
Answer sheet says $(ab)^{-1} = x$
$endgroup$
– David Davidson
Dec 8 '18 at 11:36




$begingroup$
Answer sheet says $(ab)^{-1} = x$
$endgroup$
– David Davidson
Dec 8 '18 at 11:36












$begingroup$
I think the answer sheet might be wrong. I've checked your solution several times; as I said, it seems okay to me.
$endgroup$
– Shaun
Dec 8 '18 at 11:52




$begingroup$
I think the answer sheet might be wrong. I've checked your solution several times; as I said, it seems okay to me.
$endgroup$
– Shaun
Dec 8 '18 at 11:52










3 Answers
3






active

oldest

votes


















2












$begingroup$

You have $xaxxaxxax=bx$, so
$$
xa(x^2a)(x^2a)=b
$$
Now $a(x^2a)=a(xa)^{-1}=aa^{-1}x^{-1}=x^{-1}$ and therefore
$$
b=xx^{-1}a^{-1}x^{-1}=a^{-1}x^{-1}
$$
whence $x^{-1}=ab$ and $x=(ab)^{-1}$.



Can we say that $b^2=(ab)^{-1}$? This is equivalent to $b^3=a^{-1}$ or, as you proved that $b^3=bx$, to $bx=a^{-1}$, which is true.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    The solution in the answer sheet is equivalent; your solution is fine.



    Marking your work line by line . . .




    My steps:



    $(xax)^3 = bx$ $color{red}{quadcheckmarktext{Def}}$



    $xaxxaxxax = bx$ $color{red}{quadcheckmarktext{Def}}$



    $xax^2ax^2ax = bx$$color{red}{quadcheckmarktext{Rewrite}}$



    $xax^2ax^2a = b$ $color{red}{quadcheckmark (times x^{-1})}$



    $x^2a = (xax^2a)^{-1}b$ $color{red}{quadcheckmark (times (xax^2a)^{-1})}$



    Now, substituting $x^2a$ into the second equation:



    $(xax^2a)^{-1}b = (xa)^{-1}$ $color{red}{quadcheckmark (text{Sub})}$



    $b = (xax^2a)(xa)^{-1} $ $color{red}{quadcheckmark (times (xax^2a))}$



    $b = (xax^2a)(a^{-1}x^{-1})$ $color{red}{quadcheckmark (text{Use of 'inverse of product' lemma})}$



    $b = xax$ $color{red}{quadcheckmark (text{Use of inverses})}$



    Now back to the first equation:



    $b^3 = bx$ $color{red}{quadcheckmark (text{Sub})}$



    $b^2 = x$ $color{red}{quadcheckmark (times b^{-1})}$







    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Is it possible that both answers are correct? That $(ab)^{-1} = b^2$?
      $endgroup$
      – David Davidson
      Dec 8 '18 at 11:58












    • $begingroup$
      Yes, @DavidDavidson. I don't see how though. I've edited the question accordingly. I am going to improve this answer further in a short while.
      $endgroup$
      – Shaun
      Dec 8 '18 at 12:00










    • $begingroup$
      @DavidDavidson Is that clear now?
      $endgroup$
      – Shaun
      Dec 8 '18 at 12:09










    • $begingroup$
      The steps from the answer sheet seem to be: $(xax)^3 = bx$ | $xax^2ax^2ax = bx$ | $xa(xa)^{-1}(xa)^{-1}x = bx$ | $xaa^{-1}x^{-1}a^{-1}x^{-1}x = bx$ | $a^{-1}x^{-1}=b$ | $b^{-1}a^{-1} = x$ | $(ab)^{-1} = x$ |
      $endgroup$
      – David Davidson
      Dec 8 '18 at 12:09












    • $begingroup$
      Then your answer sheet is correct too. You're both right, @DavidDavidson.
      $endgroup$
      – Shaun
      Dec 8 '18 at 12:23





















    -2












    $begingroup$


    1. The following is solution in the image attached from the link.
      solution Image on paper


    Form the above solution x = b-1a-1. substituting this in second equation given. a = b-3. which the solution b^2 can be written in infinite ways if you include a like b3ab2, b6a2b2 etc etc.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      (xa)−1 is not equal to (a−1x−1). a need not be invertible.
      $endgroup$
      – Raghu Veer
      Dec 8 '18 at 12:44










    • $begingroup$
      No, the solution is correct.
      $endgroup$
      – Shaun
      Dec 8 '18 at 12:46










    • $begingroup$
      Here $a$ is an element of the group. It is invertible by definition.
      $endgroup$
      – Shaun
      Dec 8 '18 at 12:47








    • 1




      $begingroup$
      Yeah, right. Thanks for the explanation
      $endgroup$
      – Raghu Veer
      Dec 8 '18 at 12:48










    • $begingroup$
      It is well known that for any $g, h$ in any group $G$, we have $(gh)^{-1}=h^{-1}g^{-1}$.
      $endgroup$
      – Shaun
      Dec 8 '18 at 12:49











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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    You have $xaxxaxxax=bx$, so
    $$
    xa(x^2a)(x^2a)=b
    $$
    Now $a(x^2a)=a(xa)^{-1}=aa^{-1}x^{-1}=x^{-1}$ and therefore
    $$
    b=xx^{-1}a^{-1}x^{-1}=a^{-1}x^{-1}
    $$
    whence $x^{-1}=ab$ and $x=(ab)^{-1}$.



    Can we say that $b^2=(ab)^{-1}$? This is equivalent to $b^3=a^{-1}$ or, as you proved that $b^3=bx$, to $bx=a^{-1}$, which is true.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      You have $xaxxaxxax=bx$, so
      $$
      xa(x^2a)(x^2a)=b
      $$
      Now $a(x^2a)=a(xa)^{-1}=aa^{-1}x^{-1}=x^{-1}$ and therefore
      $$
      b=xx^{-1}a^{-1}x^{-1}=a^{-1}x^{-1}
      $$
      whence $x^{-1}=ab$ and $x=(ab)^{-1}$.



      Can we say that $b^2=(ab)^{-1}$? This is equivalent to $b^3=a^{-1}$ or, as you proved that $b^3=bx$, to $bx=a^{-1}$, which is true.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        You have $xaxxaxxax=bx$, so
        $$
        xa(x^2a)(x^2a)=b
        $$
        Now $a(x^2a)=a(xa)^{-1}=aa^{-1}x^{-1}=x^{-1}$ and therefore
        $$
        b=xx^{-1}a^{-1}x^{-1}=a^{-1}x^{-1}
        $$
        whence $x^{-1}=ab$ and $x=(ab)^{-1}$.



        Can we say that $b^2=(ab)^{-1}$? This is equivalent to $b^3=a^{-1}$ or, as you proved that $b^3=bx$, to $bx=a^{-1}$, which is true.






        share|cite|improve this answer









        $endgroup$



        You have $xaxxaxxax=bx$, so
        $$
        xa(x^2a)(x^2a)=b
        $$
        Now $a(x^2a)=a(xa)^{-1}=aa^{-1}x^{-1}=x^{-1}$ and therefore
        $$
        b=xx^{-1}a^{-1}x^{-1}=a^{-1}x^{-1}
        $$
        whence $x^{-1}=ab$ and $x=(ab)^{-1}$.



        Can we say that $b^2=(ab)^{-1}$? This is equivalent to $b^3=a^{-1}$ or, as you proved that $b^3=bx$, to $bx=a^{-1}$, which is true.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 8 '18 at 13:39









        egregegreg

        184k1486205




        184k1486205























            2












            $begingroup$

            The solution in the answer sheet is equivalent; your solution is fine.



            Marking your work line by line . . .




            My steps:



            $(xax)^3 = bx$ $color{red}{quadcheckmarktext{Def}}$



            $xaxxaxxax = bx$ $color{red}{quadcheckmarktext{Def}}$



            $xax^2ax^2ax = bx$$color{red}{quadcheckmarktext{Rewrite}}$



            $xax^2ax^2a = b$ $color{red}{quadcheckmark (times x^{-1})}$



            $x^2a = (xax^2a)^{-1}b$ $color{red}{quadcheckmark (times (xax^2a)^{-1})}$



            Now, substituting $x^2a$ into the second equation:



            $(xax^2a)^{-1}b = (xa)^{-1}$ $color{red}{quadcheckmark (text{Sub})}$



            $b = (xax^2a)(xa)^{-1} $ $color{red}{quadcheckmark (times (xax^2a))}$



            $b = (xax^2a)(a^{-1}x^{-1})$ $color{red}{quadcheckmark (text{Use of 'inverse of product' lemma})}$



            $b = xax$ $color{red}{quadcheckmark (text{Use of inverses})}$



            Now back to the first equation:



            $b^3 = bx$ $color{red}{quadcheckmark (text{Sub})}$



            $b^2 = x$ $color{red}{quadcheckmark (times b^{-1})}$







            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Is it possible that both answers are correct? That $(ab)^{-1} = b^2$?
              $endgroup$
              – David Davidson
              Dec 8 '18 at 11:58












            • $begingroup$
              Yes, @DavidDavidson. I don't see how though. I've edited the question accordingly. I am going to improve this answer further in a short while.
              $endgroup$
              – Shaun
              Dec 8 '18 at 12:00










            • $begingroup$
              @DavidDavidson Is that clear now?
              $endgroup$
              – Shaun
              Dec 8 '18 at 12:09










            • $begingroup$
              The steps from the answer sheet seem to be: $(xax)^3 = bx$ | $xax^2ax^2ax = bx$ | $xa(xa)^{-1}(xa)^{-1}x = bx$ | $xaa^{-1}x^{-1}a^{-1}x^{-1}x = bx$ | $a^{-1}x^{-1}=b$ | $b^{-1}a^{-1} = x$ | $(ab)^{-1} = x$ |
              $endgroup$
              – David Davidson
              Dec 8 '18 at 12:09












            • $begingroup$
              Then your answer sheet is correct too. You're both right, @DavidDavidson.
              $endgroup$
              – Shaun
              Dec 8 '18 at 12:23


















            2












            $begingroup$

            The solution in the answer sheet is equivalent; your solution is fine.



            Marking your work line by line . . .




            My steps:



            $(xax)^3 = bx$ $color{red}{quadcheckmarktext{Def}}$



            $xaxxaxxax = bx$ $color{red}{quadcheckmarktext{Def}}$



            $xax^2ax^2ax = bx$$color{red}{quadcheckmarktext{Rewrite}}$



            $xax^2ax^2a = b$ $color{red}{quadcheckmark (times x^{-1})}$



            $x^2a = (xax^2a)^{-1}b$ $color{red}{quadcheckmark (times (xax^2a)^{-1})}$



            Now, substituting $x^2a$ into the second equation:



            $(xax^2a)^{-1}b = (xa)^{-1}$ $color{red}{quadcheckmark (text{Sub})}$



            $b = (xax^2a)(xa)^{-1} $ $color{red}{quadcheckmark (times (xax^2a))}$



            $b = (xax^2a)(a^{-1}x^{-1})$ $color{red}{quadcheckmark (text{Use of 'inverse of product' lemma})}$



            $b = xax$ $color{red}{quadcheckmark (text{Use of inverses})}$



            Now back to the first equation:



            $b^3 = bx$ $color{red}{quadcheckmark (text{Sub})}$



            $b^2 = x$ $color{red}{quadcheckmark (times b^{-1})}$







            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Is it possible that both answers are correct? That $(ab)^{-1} = b^2$?
              $endgroup$
              – David Davidson
              Dec 8 '18 at 11:58












            • $begingroup$
              Yes, @DavidDavidson. I don't see how though. I've edited the question accordingly. I am going to improve this answer further in a short while.
              $endgroup$
              – Shaun
              Dec 8 '18 at 12:00










            • $begingroup$
              @DavidDavidson Is that clear now?
              $endgroup$
              – Shaun
              Dec 8 '18 at 12:09










            • $begingroup$
              The steps from the answer sheet seem to be: $(xax)^3 = bx$ | $xax^2ax^2ax = bx$ | $xa(xa)^{-1}(xa)^{-1}x = bx$ | $xaa^{-1}x^{-1}a^{-1}x^{-1}x = bx$ | $a^{-1}x^{-1}=b$ | $b^{-1}a^{-1} = x$ | $(ab)^{-1} = x$ |
              $endgroup$
              – David Davidson
              Dec 8 '18 at 12:09












            • $begingroup$
              Then your answer sheet is correct too. You're both right, @DavidDavidson.
              $endgroup$
              – Shaun
              Dec 8 '18 at 12:23
















            2












            2








            2





            $begingroup$

            The solution in the answer sheet is equivalent; your solution is fine.



            Marking your work line by line . . .




            My steps:



            $(xax)^3 = bx$ $color{red}{quadcheckmarktext{Def}}$



            $xaxxaxxax = bx$ $color{red}{quadcheckmarktext{Def}}$



            $xax^2ax^2ax = bx$$color{red}{quadcheckmarktext{Rewrite}}$



            $xax^2ax^2a = b$ $color{red}{quadcheckmark (times x^{-1})}$



            $x^2a = (xax^2a)^{-1}b$ $color{red}{quadcheckmark (times (xax^2a)^{-1})}$



            Now, substituting $x^2a$ into the second equation:



            $(xax^2a)^{-1}b = (xa)^{-1}$ $color{red}{quadcheckmark (text{Sub})}$



            $b = (xax^2a)(xa)^{-1} $ $color{red}{quadcheckmark (times (xax^2a))}$



            $b = (xax^2a)(a^{-1}x^{-1})$ $color{red}{quadcheckmark (text{Use of 'inverse of product' lemma})}$



            $b = xax$ $color{red}{quadcheckmark (text{Use of inverses})}$



            Now back to the first equation:



            $b^3 = bx$ $color{red}{quadcheckmark (text{Sub})}$



            $b^2 = x$ $color{red}{quadcheckmark (times b^{-1})}$







            share|cite|improve this answer











            $endgroup$



            The solution in the answer sheet is equivalent; your solution is fine.



            Marking your work line by line . . .




            My steps:



            $(xax)^3 = bx$ $color{red}{quadcheckmarktext{Def}}$



            $xaxxaxxax = bx$ $color{red}{quadcheckmarktext{Def}}$



            $xax^2ax^2ax = bx$$color{red}{quadcheckmarktext{Rewrite}}$



            $xax^2ax^2a = b$ $color{red}{quadcheckmark (times x^{-1})}$



            $x^2a = (xax^2a)^{-1}b$ $color{red}{quadcheckmark (times (xax^2a)^{-1})}$



            Now, substituting $x^2a$ into the second equation:



            $(xax^2a)^{-1}b = (xa)^{-1}$ $color{red}{quadcheckmark (text{Sub})}$



            $b = (xax^2a)(xa)^{-1} $ $color{red}{quadcheckmark (times (xax^2a))}$



            $b = (xax^2a)(a^{-1}x^{-1})$ $color{red}{quadcheckmark (text{Use of 'inverse of product' lemma})}$



            $b = xax$ $color{red}{quadcheckmark (text{Use of inverses})}$



            Now back to the first equation:



            $b^3 = bx$ $color{red}{quadcheckmark (text{Sub})}$



            $b^2 = x$ $color{red}{quadcheckmark (times b^{-1})}$








            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 9 '18 at 2:54

























            answered Dec 8 '18 at 11:54









            ShaunShaun

            9,442113684




            9,442113684












            • $begingroup$
              Is it possible that both answers are correct? That $(ab)^{-1} = b^2$?
              $endgroup$
              – David Davidson
              Dec 8 '18 at 11:58












            • $begingroup$
              Yes, @DavidDavidson. I don't see how though. I've edited the question accordingly. I am going to improve this answer further in a short while.
              $endgroup$
              – Shaun
              Dec 8 '18 at 12:00










            • $begingroup$
              @DavidDavidson Is that clear now?
              $endgroup$
              – Shaun
              Dec 8 '18 at 12:09










            • $begingroup$
              The steps from the answer sheet seem to be: $(xax)^3 = bx$ | $xax^2ax^2ax = bx$ | $xa(xa)^{-1}(xa)^{-1}x = bx$ | $xaa^{-1}x^{-1}a^{-1}x^{-1}x = bx$ | $a^{-1}x^{-1}=b$ | $b^{-1}a^{-1} = x$ | $(ab)^{-1} = x$ |
              $endgroup$
              – David Davidson
              Dec 8 '18 at 12:09












            • $begingroup$
              Then your answer sheet is correct too. You're both right, @DavidDavidson.
              $endgroup$
              – Shaun
              Dec 8 '18 at 12:23




















            • $begingroup$
              Is it possible that both answers are correct? That $(ab)^{-1} = b^2$?
              $endgroup$
              – David Davidson
              Dec 8 '18 at 11:58












            • $begingroup$
              Yes, @DavidDavidson. I don't see how though. I've edited the question accordingly. I am going to improve this answer further in a short while.
              $endgroup$
              – Shaun
              Dec 8 '18 at 12:00










            • $begingroup$
              @DavidDavidson Is that clear now?
              $endgroup$
              – Shaun
              Dec 8 '18 at 12:09










            • $begingroup$
              The steps from the answer sheet seem to be: $(xax)^3 = bx$ | $xax^2ax^2ax = bx$ | $xa(xa)^{-1}(xa)^{-1}x = bx$ | $xaa^{-1}x^{-1}a^{-1}x^{-1}x = bx$ | $a^{-1}x^{-1}=b$ | $b^{-1}a^{-1} = x$ | $(ab)^{-1} = x$ |
              $endgroup$
              – David Davidson
              Dec 8 '18 at 12:09












            • $begingroup$
              Then your answer sheet is correct too. You're both right, @DavidDavidson.
              $endgroup$
              – Shaun
              Dec 8 '18 at 12:23


















            $begingroup$
            Is it possible that both answers are correct? That $(ab)^{-1} = b^2$?
            $endgroup$
            – David Davidson
            Dec 8 '18 at 11:58






            $begingroup$
            Is it possible that both answers are correct? That $(ab)^{-1} = b^2$?
            $endgroup$
            – David Davidson
            Dec 8 '18 at 11:58














            $begingroup$
            Yes, @DavidDavidson. I don't see how though. I've edited the question accordingly. I am going to improve this answer further in a short while.
            $endgroup$
            – Shaun
            Dec 8 '18 at 12:00




            $begingroup$
            Yes, @DavidDavidson. I don't see how though. I've edited the question accordingly. I am going to improve this answer further in a short while.
            $endgroup$
            – Shaun
            Dec 8 '18 at 12:00












            $begingroup$
            @DavidDavidson Is that clear now?
            $endgroup$
            – Shaun
            Dec 8 '18 at 12:09




            $begingroup$
            @DavidDavidson Is that clear now?
            $endgroup$
            – Shaun
            Dec 8 '18 at 12:09












            $begingroup$
            The steps from the answer sheet seem to be: $(xax)^3 = bx$ | $xax^2ax^2ax = bx$ | $xa(xa)^{-1}(xa)^{-1}x = bx$ | $xaa^{-1}x^{-1}a^{-1}x^{-1}x = bx$ | $a^{-1}x^{-1}=b$ | $b^{-1}a^{-1} = x$ | $(ab)^{-1} = x$ |
            $endgroup$
            – David Davidson
            Dec 8 '18 at 12:09






            $begingroup$
            The steps from the answer sheet seem to be: $(xax)^3 = bx$ | $xax^2ax^2ax = bx$ | $xa(xa)^{-1}(xa)^{-1}x = bx$ | $xaa^{-1}x^{-1}a^{-1}x^{-1}x = bx$ | $a^{-1}x^{-1}=b$ | $b^{-1}a^{-1} = x$ | $(ab)^{-1} = x$ |
            $endgroup$
            – David Davidson
            Dec 8 '18 at 12:09














            $begingroup$
            Then your answer sheet is correct too. You're both right, @DavidDavidson.
            $endgroup$
            – Shaun
            Dec 8 '18 at 12:23






            $begingroup$
            Then your answer sheet is correct too. You're both right, @DavidDavidson.
            $endgroup$
            – Shaun
            Dec 8 '18 at 12:23













            -2












            $begingroup$


            1. The following is solution in the image attached from the link.
              solution Image on paper


            Form the above solution x = b-1a-1. substituting this in second equation given. a = b-3. which the solution b^2 can be written in infinite ways if you include a like b3ab2, b6a2b2 etc etc.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              (xa)−1 is not equal to (a−1x−1). a need not be invertible.
              $endgroup$
              – Raghu Veer
              Dec 8 '18 at 12:44










            • $begingroup$
              No, the solution is correct.
              $endgroup$
              – Shaun
              Dec 8 '18 at 12:46










            • $begingroup$
              Here $a$ is an element of the group. It is invertible by definition.
              $endgroup$
              – Shaun
              Dec 8 '18 at 12:47








            • 1




              $begingroup$
              Yeah, right. Thanks for the explanation
              $endgroup$
              – Raghu Veer
              Dec 8 '18 at 12:48










            • $begingroup$
              It is well known that for any $g, h$ in any group $G$, we have $(gh)^{-1}=h^{-1}g^{-1}$.
              $endgroup$
              – Shaun
              Dec 8 '18 at 12:49
















            -2












            $begingroup$


            1. The following is solution in the image attached from the link.
              solution Image on paper


            Form the above solution x = b-1a-1. substituting this in second equation given. a = b-3. which the solution b^2 can be written in infinite ways if you include a like b3ab2, b6a2b2 etc etc.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              (xa)−1 is not equal to (a−1x−1). a need not be invertible.
              $endgroup$
              – Raghu Veer
              Dec 8 '18 at 12:44










            • $begingroup$
              No, the solution is correct.
              $endgroup$
              – Shaun
              Dec 8 '18 at 12:46










            • $begingroup$
              Here $a$ is an element of the group. It is invertible by definition.
              $endgroup$
              – Shaun
              Dec 8 '18 at 12:47








            • 1




              $begingroup$
              Yeah, right. Thanks for the explanation
              $endgroup$
              – Raghu Veer
              Dec 8 '18 at 12:48










            • $begingroup$
              It is well known that for any $g, h$ in any group $G$, we have $(gh)^{-1}=h^{-1}g^{-1}$.
              $endgroup$
              – Shaun
              Dec 8 '18 at 12:49














            -2












            -2








            -2





            $begingroup$


            1. The following is solution in the image attached from the link.
              solution Image on paper


            Form the above solution x = b-1a-1. substituting this in second equation given. a = b-3. which the solution b^2 can be written in infinite ways if you include a like b3ab2, b6a2b2 etc etc.






            share|cite|improve this answer











            $endgroup$




            1. The following is solution in the image attached from the link.
              solution Image on paper


            Form the above solution x = b-1a-1. substituting this in second equation given. a = b-3. which the solution b^2 can be written in infinite ways if you include a like b3ab2, b6a2b2 etc etc.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 8 '18 at 13:00

























            answered Dec 8 '18 at 12:35









            Raghu VeerRaghu Veer

            11




            11












            • $begingroup$
              (xa)−1 is not equal to (a−1x−1). a need not be invertible.
              $endgroup$
              – Raghu Veer
              Dec 8 '18 at 12:44










            • $begingroup$
              No, the solution is correct.
              $endgroup$
              – Shaun
              Dec 8 '18 at 12:46










            • $begingroup$
              Here $a$ is an element of the group. It is invertible by definition.
              $endgroup$
              – Shaun
              Dec 8 '18 at 12:47








            • 1




              $begingroup$
              Yeah, right. Thanks for the explanation
              $endgroup$
              – Raghu Veer
              Dec 8 '18 at 12:48










            • $begingroup$
              It is well known that for any $g, h$ in any group $G$, we have $(gh)^{-1}=h^{-1}g^{-1}$.
              $endgroup$
              – Shaun
              Dec 8 '18 at 12:49


















            • $begingroup$
              (xa)−1 is not equal to (a−1x−1). a need not be invertible.
              $endgroup$
              – Raghu Veer
              Dec 8 '18 at 12:44










            • $begingroup$
              No, the solution is correct.
              $endgroup$
              – Shaun
              Dec 8 '18 at 12:46










            • $begingroup$
              Here $a$ is an element of the group. It is invertible by definition.
              $endgroup$
              – Shaun
              Dec 8 '18 at 12:47








            • 1




              $begingroup$
              Yeah, right. Thanks for the explanation
              $endgroup$
              – Raghu Veer
              Dec 8 '18 at 12:48










            • $begingroup$
              It is well known that for any $g, h$ in any group $G$, we have $(gh)^{-1}=h^{-1}g^{-1}$.
              $endgroup$
              – Shaun
              Dec 8 '18 at 12:49
















            $begingroup$
            (xa)−1 is not equal to (a−1x−1). a need not be invertible.
            $endgroup$
            – Raghu Veer
            Dec 8 '18 at 12:44




            $begingroup$
            (xa)−1 is not equal to (a−1x−1). a need not be invertible.
            $endgroup$
            – Raghu Veer
            Dec 8 '18 at 12:44












            $begingroup$
            No, the solution is correct.
            $endgroup$
            – Shaun
            Dec 8 '18 at 12:46




            $begingroup$
            No, the solution is correct.
            $endgroup$
            – Shaun
            Dec 8 '18 at 12:46












            $begingroup$
            Here $a$ is an element of the group. It is invertible by definition.
            $endgroup$
            – Shaun
            Dec 8 '18 at 12:47






            $begingroup$
            Here $a$ is an element of the group. It is invertible by definition.
            $endgroup$
            – Shaun
            Dec 8 '18 at 12:47






            1




            1




            $begingroup$
            Yeah, right. Thanks for the explanation
            $endgroup$
            – Raghu Veer
            Dec 8 '18 at 12:48




            $begingroup$
            Yeah, right. Thanks for the explanation
            $endgroup$
            – Raghu Veer
            Dec 8 '18 at 12:48












            $begingroup$
            It is well known that for any $g, h$ in any group $G$, we have $(gh)^{-1}=h^{-1}g^{-1}$.
            $endgroup$
            – Shaun
            Dec 8 '18 at 12:49




            $begingroup$
            It is well known that for any $g, h$ in any group $G$, we have $(gh)^{-1}=h^{-1}g^{-1}$.
            $endgroup$
            – Shaun
            Dec 8 '18 at 12:49


















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