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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

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

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  • $\begingroup$ ..@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
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    $\begingroup$ From the right side, of course. This is just composition of functions, and $(f\circ g)(x)$ means $f\bigl(g(x)\bigr)$. $\endgroup$ – 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
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    $\begingroup$ I am using the standard notation here. $\endgroup$ – José Carlos Santos Jan 9 at 14:27
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    $\begingroup$ 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
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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 $$aba^{-1} = (a(2)a(3))(a(1)a(4)) = (31)(25).$$

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