compute $aba^{−1}$












0












$begingroup$


With the usual notations, compute $aba^{−1}$
in $S_5$ and express it as the
product of disjoint cycles, where
$a = (1 2 3)(4 5)$ and $b = (2 3)(1 4).$



My attempt : $ab$ = $(1345)$ and $a^{-1} = (321)(54)$



now now i got $aba^{-1} = (12)(35)$



is its correct ?



Any hint/ solution will be appreciated






abstract-algebra







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    With the usual notations, compute $aba^{−1}$
    in $S_5$ and express it as the
    product of disjoint cycles, where
    $a = (1 2 3)(4 5)$ and $b = (2 3)(1 4).$



    My attempt : $ab$ = $(1345)$ and $a^{-1} = (321)(54)$



    now now i got $aba^{-1} = (12)(35)$



    is its correct ?



    Any hint/ solution will be appreciated






    abstract-algebra







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      With the usual notations, compute $aba^{−1}$
      in $S_5$ and express it as the
      product of disjoint cycles, where
      $a = (1 2 3)(4 5)$ and $b = (2 3)(1 4).$



      My attempt : $ab$ = $(1345)$ and $a^{-1} = (321)(54)$



      now now i got $aba^{-1} = (12)(35)$



      is its correct ?



      Any hint/ solution will be appreciated






      abstract-algebra







      share|cite|improve this question









      $endgroup$




      With the usual notations, compute $aba^{−1}$
      in $S_5$ and express it as the
      product of disjoint cycles, where
      $a = (1 2 3)(4 5)$ and $b = (2 3)(1 4).$



      My attempt : $ab$ = $(1345)$ and $a^{-1} = (321)(54)$



      now now i got $aba^{-1} = (12)(35)$



      is its correct ?



      Any hint/ solution will be appreciated






      abstract-algebra




      abstract-algebra






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 9 at 14:15









      jasminejasmine

      1,941420




      1,941420






















          2 Answers
          2






          active

          oldest

          votes


















          2












          $begingroup$

          No, it is not correct, since $ab=(1 5 4 2)$. Actually,$$aba^{-1}=(1 3)(2 5).$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            ..@Jose carlos sir u take from right side or left side ? i mean count from left side or right side ?
            $endgroup$
            – jasmine
            Jan 9 at 14:24








          • 1




            $begingroup$
            From the right side, of course. This is just composition of functions, and $(fcirc g)(x)$ means $fbigl(g(x)bigr)$.
            $endgroup$
            – José Carlos Santos
            Jan 9 at 14:25










          • $begingroup$
            But in Herstein/Artin they count from left side im confused ? as gallian count from right side which one is applicable ??
            $endgroup$
            – jasmine
            Jan 9 at 14:26






          • 1




            $begingroup$
            I am using the standard notation here.
            $endgroup$
            – José Carlos Santos
            Jan 9 at 14:27






          • 1




            $begingroup$
            How to compute $(1 2)(1 3)$? Let us see whae this does to $1$. First, $(1 3)$ maps $1$ into $3$ and then $(1 2)$ maps $3$ into itself. So, $(1 2)(1 3)$ maps $1$ into $3$. And what about $3$? First, $(1 3)$ maps $3$ into $1$ and then $(1 2)$ maps $1$ into $2$. So, $(1 2)(1 3)$ maps $3$ into $2$. And what about $2$? First, $(1 3)$ maps $2$ into itself and then $(1 2)$ maps $2$ into $1$. So, $(1 2)(1 3)$ maps $2$ into $1$. Therefore, $(1 2)(1 3)=(1 3 2)$.
            $endgroup$
            – José Carlos Santos
            Jan 9 at 14:36



















          1












          $begingroup$

          Computing conjugates in $mathfrak S_5$ is very easy. If $c=(i_1 , i_2 dots i_k)$ is a cycle then $sigma c sigma^{-1} = (sigma(i_1) , sigma(i_2) dots sigma(i_k))$. In your case, this yields $$aba^{-1} = (a(2)a(3))(a(1)a(4)) = (31)(25).$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            how $a(2) =3$?..
            $endgroup$
            – jasmine
            Jan 9 at 14:22






          • 1




            $begingroup$
            You need to remember that the product between cycles is simply composition of applications. On the one hand you have $(45)(2)=2$, and on the other hand $(123)(2)=3$, so $a(2)=(123)(45)(2)=(123)(2)=3$.
            $endgroup$
            – A. Bailleul
            Jan 9 at 14:25














          Your Answer





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          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$

          No, it is not correct, since $ab=(1 5 4 2)$. Actually,$$aba^{-1}=(1 3)(2 5).$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            ..@Jose carlos sir u take from right side or left side ? i mean count from left side or right side ?
            $endgroup$
            – jasmine
            Jan 9 at 14:24








          • 1




            $begingroup$
            From the right side, of course. This is just composition of functions, and $(fcirc g)(x)$ means $fbigl(g(x)bigr)$.
            $endgroup$
            – José Carlos Santos
            Jan 9 at 14:25










          • $begingroup$
            But in Herstein/Artin they count from left side im confused ? as gallian count from right side which one is applicable ??
            $endgroup$
            – jasmine
            Jan 9 at 14:26






          • 1




            $begingroup$
            I am using the standard notation here.
            $endgroup$
            – José Carlos Santos
            Jan 9 at 14:27






          • 1




            $begingroup$
            How to compute $(1 2)(1 3)$? Let us see whae this does to $1$. First, $(1 3)$ maps $1$ into $3$ and then $(1 2)$ maps $3$ into itself. So, $(1 2)(1 3)$ maps $1$ into $3$. And what about $3$? First, $(1 3)$ maps $3$ into $1$ and then $(1 2)$ maps $1$ into $2$. So, $(1 2)(1 3)$ maps $3$ into $2$. And what about $2$? First, $(1 3)$ maps $2$ into itself and then $(1 2)$ maps $2$ into $1$. So, $(1 2)(1 3)$ maps $2$ into $1$. Therefore, $(1 2)(1 3)=(1 3 2)$.
            $endgroup$
            – José Carlos Santos
            Jan 9 at 14:36
















          2












          $begingroup$

          No, it is not correct, since $ab=(1 5 4 2)$. Actually,$$aba^{-1}=(1 3)(2 5).$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            ..@Jose carlos sir u take from right side or left side ? i mean count from left side or right side ?
            $endgroup$
            – jasmine
            Jan 9 at 14:24








          • 1




            $begingroup$
            From the right side, of course. This is just composition of functions, and $(fcirc g)(x)$ means $fbigl(g(x)bigr)$.
            $endgroup$
            – José Carlos Santos
            Jan 9 at 14:25










          • $begingroup$
            But in Herstein/Artin they count from left side im confused ? as gallian count from right side which one is applicable ??
            $endgroup$
            – jasmine
            Jan 9 at 14:26






          • 1




            $begingroup$
            I am using the standard notation here.
            $endgroup$
            – José Carlos Santos
            Jan 9 at 14:27






          • 1




            $begingroup$
            How to compute $(1 2)(1 3)$? Let us see whae this does to $1$. First, $(1 3)$ maps $1$ into $3$ and then $(1 2)$ maps $3$ into itself. So, $(1 2)(1 3)$ maps $1$ into $3$. And what about $3$? First, $(1 3)$ maps $3$ into $1$ and then $(1 2)$ maps $1$ into $2$. So, $(1 2)(1 3)$ maps $3$ into $2$. And what about $2$? First, $(1 3)$ maps $2$ into itself and then $(1 2)$ maps $2$ into $1$. So, $(1 2)(1 3)$ maps $2$ into $1$. Therefore, $(1 2)(1 3)=(1 3 2)$.
            $endgroup$
            – José Carlos Santos
            Jan 9 at 14:36














          2












          2








          2





          $begingroup$

          No, it is not correct, since $ab=(1 5 4 2)$. Actually,$$aba^{-1}=(1 3)(2 5).$$






          share|cite|improve this answer









          $endgroup$



          No, it is not correct, since $ab=(1 5 4 2)$. Actually,$$aba^{-1}=(1 3)(2 5).$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 9 at 14:18









          José Carlos SantosJosé Carlos Santos

          172k22132239




          172k22132239












          • $begingroup$
            ..@Jose carlos sir u take from right side or left side ? i mean count from left side or right side ?
            $endgroup$
            – jasmine
            Jan 9 at 14:24








          • 1




            $begingroup$
            From the right side, of course. This is just composition of functions, and $(fcirc g)(x)$ means $fbigl(g(x)bigr)$.
            $endgroup$
            – José Carlos Santos
            Jan 9 at 14:25










          • $begingroup$
            But in Herstein/Artin they count from left side im confused ? as gallian count from right side which one is applicable ??
            $endgroup$
            – jasmine
            Jan 9 at 14:26






          • 1




            $begingroup$
            I am using the standard notation here.
            $endgroup$
            – José Carlos Santos
            Jan 9 at 14:27






          • 1




            $begingroup$
            How to compute $(1 2)(1 3)$? Let us see whae this does to $1$. First, $(1 3)$ maps $1$ into $3$ and then $(1 2)$ maps $3$ into itself. So, $(1 2)(1 3)$ maps $1$ into $3$. And what about $3$? First, $(1 3)$ maps $3$ into $1$ and then $(1 2)$ maps $1$ into $2$. So, $(1 2)(1 3)$ maps $3$ into $2$. And what about $2$? First, $(1 3)$ maps $2$ into itself and then $(1 2)$ maps $2$ into $1$. So, $(1 2)(1 3)$ maps $2$ into $1$. Therefore, $(1 2)(1 3)=(1 3 2)$.
            $endgroup$
            – José Carlos Santos
            Jan 9 at 14:36


















          • $begingroup$
            ..@Jose carlos sir u take from right side or left side ? i mean count from left side or right side ?
            $endgroup$
            – jasmine
            Jan 9 at 14:24








          • 1




            $begingroup$
            From the right side, of course. This is just composition of functions, and $(fcirc g)(x)$ means $fbigl(g(x)bigr)$.
            $endgroup$
            – José Carlos Santos
            Jan 9 at 14:25










          • $begingroup$
            But in Herstein/Artin they count from left side im confused ? as gallian count from right side which one is applicable ??
            $endgroup$
            – jasmine
            Jan 9 at 14:26






          • 1




            $begingroup$
            I am using the standard notation here.
            $endgroup$
            – José Carlos Santos
            Jan 9 at 14:27






          • 1




            $begingroup$
            How to compute $(1 2)(1 3)$? Let us see whae this does to $1$. First, $(1 3)$ maps $1$ into $3$ and then $(1 2)$ maps $3$ into itself. So, $(1 2)(1 3)$ maps $1$ into $3$. And what about $3$? First, $(1 3)$ maps $3$ into $1$ and then $(1 2)$ maps $1$ into $2$. So, $(1 2)(1 3)$ maps $3$ into $2$. And what about $2$? First, $(1 3)$ maps $2$ into itself and then $(1 2)$ maps $2$ into $1$. So, $(1 2)(1 3)$ maps $2$ into $1$. Therefore, $(1 2)(1 3)=(1 3 2)$.
            $endgroup$
            – José Carlos Santos
            Jan 9 at 14:36
















          $begingroup$
          ..@Jose carlos sir u take from right side or left side ? i mean count from left side or right side ?
          $endgroup$
          – jasmine
          Jan 9 at 14:24






          $begingroup$
          ..@Jose carlos sir u take from right side or left side ? i mean count from left side or right side ?
          $endgroup$
          – jasmine
          Jan 9 at 14:24






          1




          1




          $begingroup$
          From the right side, of course. This is just composition of functions, and $(fcirc g)(x)$ means $fbigl(g(x)bigr)$.
          $endgroup$
          – José Carlos Santos
          Jan 9 at 14:25




          $begingroup$
          From the right side, of course. This is just composition of functions, and $(fcirc g)(x)$ means $fbigl(g(x)bigr)$.
          $endgroup$
          – José Carlos Santos
          Jan 9 at 14:25












          $begingroup$
          But in Herstein/Artin they count from left side im confused ? as gallian count from right side which one is applicable ??
          $endgroup$
          – jasmine
          Jan 9 at 14:26




          $begingroup$
          But in Herstein/Artin they count from left side im confused ? as gallian count from right side which one is applicable ??
          $endgroup$
          – jasmine
          Jan 9 at 14:26




          1




          1




          $begingroup$
          I am using the standard notation here.
          $endgroup$
          – José Carlos Santos
          Jan 9 at 14:27




          $begingroup$
          I am using the standard notation here.
          $endgroup$
          – José Carlos Santos
          Jan 9 at 14:27




          1




          1




          $begingroup$
          How to compute $(1 2)(1 3)$? Let us see whae this does to $1$. First, $(1 3)$ maps $1$ into $3$ and then $(1 2)$ maps $3$ into itself. So, $(1 2)(1 3)$ maps $1$ into $3$. And what about $3$? First, $(1 3)$ maps $3$ into $1$ and then $(1 2)$ maps $1$ into $2$. So, $(1 2)(1 3)$ maps $3$ into $2$. And what about $2$? First, $(1 3)$ maps $2$ into itself and then $(1 2)$ maps $2$ into $1$. So, $(1 2)(1 3)$ maps $2$ into $1$. Therefore, $(1 2)(1 3)=(1 3 2)$.
          $endgroup$
          – José Carlos Santos
          Jan 9 at 14:36




          $begingroup$
          How to compute $(1 2)(1 3)$? Let us see whae this does to $1$. First, $(1 3)$ maps $1$ into $3$ and then $(1 2)$ maps $3$ into itself. So, $(1 2)(1 3)$ maps $1$ into $3$. And what about $3$? First, $(1 3)$ maps $3$ into $1$ and then $(1 2)$ maps $1$ into $2$. So, $(1 2)(1 3)$ maps $3$ into $2$. And what about $2$? First, $(1 3)$ maps $2$ into itself and then $(1 2)$ maps $2$ into $1$. So, $(1 2)(1 3)$ maps $2$ into $1$. Therefore, $(1 2)(1 3)=(1 3 2)$.
          $endgroup$
          – José Carlos Santos
          Jan 9 at 14:36











          1












          $begingroup$

          Computing conjugates in $mathfrak S_5$ is very easy. If $c=(i_1 , i_2 dots i_k)$ is a cycle then $sigma c sigma^{-1} = (sigma(i_1) , sigma(i_2) dots sigma(i_k))$. In your case, this yields $$aba^{-1} = (a(2)a(3))(a(1)a(4)) = (31)(25).$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            how $a(2) =3$?..
            $endgroup$
            – jasmine
            Jan 9 at 14:22






          • 1




            $begingroup$
            You need to remember that the product between cycles is simply composition of applications. On the one hand you have $(45)(2)=2$, and on the other hand $(123)(2)=3$, so $a(2)=(123)(45)(2)=(123)(2)=3$.
            $endgroup$
            – A. Bailleul
            Jan 9 at 14:25


















          1












          $begingroup$

          Computing conjugates in $mathfrak S_5$ is very easy. If $c=(i_1 , i_2 dots i_k)$ is a cycle then $sigma c sigma^{-1} = (sigma(i_1) , sigma(i_2) dots sigma(i_k))$. In your case, this yields $$aba^{-1} = (a(2)a(3))(a(1)a(4)) = (31)(25).$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            how $a(2) =3$?..
            $endgroup$
            – jasmine
            Jan 9 at 14:22






          • 1




            $begingroup$
            You need to remember that the product between cycles is simply composition of applications. On the one hand you have $(45)(2)=2$, and on the other hand $(123)(2)=3$, so $a(2)=(123)(45)(2)=(123)(2)=3$.
            $endgroup$
            – A. Bailleul
            Jan 9 at 14:25
















          1












          1








          1





          $begingroup$

          Computing conjugates in $mathfrak S_5$ is very easy. If $c=(i_1 , i_2 dots i_k)$ is a cycle then $sigma c sigma^{-1} = (sigma(i_1) , sigma(i_2) dots sigma(i_k))$. In your case, this yields $$aba^{-1} = (a(2)a(3))(a(1)a(4)) = (31)(25).$$






          share|cite|improve this answer









          $endgroup$



          Computing conjugates in $mathfrak S_5$ is very easy. If $c=(i_1 , i_2 dots i_k)$ is a cycle then $sigma c sigma^{-1} = (sigma(i_1) , sigma(i_2) dots sigma(i_k))$. In your case, this yields $$aba^{-1} = (a(2)a(3))(a(1)a(4)) = (31)(25).$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 9 at 14:18









          A. BailleulA. Bailleul

          1587




          1587












          • $begingroup$
            how $a(2) =3$?..
            $endgroup$
            – jasmine
            Jan 9 at 14:22






          • 1




            $begingroup$
            You need to remember that the product between cycles is simply composition of applications. On the one hand you have $(45)(2)=2$, and on the other hand $(123)(2)=3$, so $a(2)=(123)(45)(2)=(123)(2)=3$.
            $endgroup$
            – A. Bailleul
            Jan 9 at 14:25




















          • $begingroup$
            how $a(2) =3$?..
            $endgroup$
            – jasmine
            Jan 9 at 14:22






          • 1




            $begingroup$
            You need to remember that the product between cycles is simply composition of applications. On the one hand you have $(45)(2)=2$, and on the other hand $(123)(2)=3$, so $a(2)=(123)(45)(2)=(123)(2)=3$.
            $endgroup$
            – A. Bailleul
            Jan 9 at 14:25


















          $begingroup$
          how $a(2) =3$?..
          $endgroup$
          – jasmine
          Jan 9 at 14:22




          $begingroup$
          how $a(2) =3$?..
          $endgroup$
          – jasmine
          Jan 9 at 14:22




          1




          1




          $begingroup$
          You need to remember that the product between cycles is simply composition of applications. On the one hand you have $(45)(2)=2$, and on the other hand $(123)(2)=3$, so $a(2)=(123)(45)(2)=(123)(2)=3$.
          $endgroup$
          – A. Bailleul
          Jan 9 at 14:25






          $begingroup$
          You need to remember that the product between cycles is simply composition of applications. On the one hand you have $(45)(2)=2$, and on the other hand $(123)(2)=3$, so $a(2)=(123)(45)(2)=(123)(2)=3$.
          $endgroup$
          – A. Bailleul
          Jan 9 at 14:25




















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          .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
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