MHB Permutation question concerning cycle shapes

[Confusedalways]

Consider the subset of $S_4$ defined by

$$K_4=\{(1)(2)(3)(4),(12)(34),(13)(24),(14)(23)\}$$

Show that for all $f \in K_4$ and all $h \in S_4$, we have $h^{-1}fh \in K_4$

I showed all the possible cycle shapes of h and am trying to show that $h^{-1}fh$ must always have cycle shape $(2,2)$, excluding the case of identity permutation.

Just don't know where to go from here

 

[Euge]

Hi, Confusedalways! Welcome. (Wave)

You're on the right track. Consider the following fact. If $\sigma = (a_1 \cdots a_r)$ is a cycle in $S_n$ and $\tau\in S_n$, then $\tau \sigma \tau^{-1} = (\tau(a_1)\cdots \tau(a_r))$. Take for instance $(12)(34)$. For all $h\in S_4$, $$h(12)(34)h^{-1} = h(12)h^{-1}h(34)h^{-1} = (h(1)\;h(2))\,(h(3)\;h(4))$$

so $h(12)(34)h^{-1}$ has cycle structure $(2,2)$. The same argument applies to the others.