compute $aba^{−1}$

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

No, it is not correct, since $$ab=(1\ \ 5\ \ 4\ \ 2)$$. Actually,$$aba^{-1}=(1\ \ 3)(2\ \ 5).$$
• From the right side, of course. This is just composition of functions, and $(f\circ g)(x)$ means $f\bigl(g(x)\bigr)$. – José Carlos Santos Jan 9 at 14:25
• 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, $(1\ \ 2)(1\ \ 3)=(1\ \ 3\ \ 2)$. – José Carlos Santos Jan 9 at 14:36
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 $$aba^{-1} = (a(2)a(3))(a(1)a(4)) = (31)(25).$$
• how $a(2) =3$?.. – jasmine Jan 9 at 14:22
• 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 $a(2)=(123)(45)(2)=(123)(2)=3$. – A. Bailleul Jan 9 at 14:25