Group Theory, Permutations.

Question

Let $x := (175)(2436)$ and $y := (1234567)$ Compute the elements $x^{-1}yx$ and $y^{-1}xy$?

And if $z := (1326745)$ for which $u$ is there an element $v$ such that $v^{-1}zv=u$. If $u$ exists find it.

Thanks in advance.

Answer 1

Let $\sigma=(a_1,a_2,...,a_k)$ be cycle and $\tau$ be a any permutation then,

$\tau^{-1}\sigma\tau=(\tau^{-1}_1,\tau^{-1}_2,...,\tau^{-1}_k)$ where $\tau^{-1}_i=\tau^{-1}(a_i)$. You can find this formula in almost all algebra books. (it is used for proving that under conjugation cycle type does not change.)

if $\sigma$ consist of more than one cycle; $\sigma=\sigma_1\sigma_2$ use this trick;

$\tau^{-1}\sigma\tau=\tau^{-1}\sigma_1\tau\tau^{-1}\sigma_2\tau$.

Now, it is enough to compute; $x^{-1}=(1, 5, 7) (2, 6, 3, 4)$ and

$x^{-1}(1)=5,x^{-1}(2)=6...$ at the and you will see that $x^{-1}yx=(1, 5, 6, 4, 2, 7, 3)$.

I leaved it to compute $y^{-1}xy$.

Secand part of your question;

$z= (1326745)$ , by above argument only conjugates of $z$ are ones having a cycle with length $7$, i.e. $(1,2,3,4,5,6,7)$ is one of the conjugate of $z$ but $(1,2)(3,4,5,6,7)$ is not conjugate of $z$. (you can see that conjuge class of $z$ has $(7-1)!=6!$ elements. )