There is no $\alpha$ such that $\alpha(123) \alpha^{-1}=(13)(578)$

Question

I have a question about permutations. I'm trying to prove there is no permutation $\alpha$ such that $\alpha(123) \alpha^{-1}=(13)(578)$.

I think I would have to use that $\alpha(123) \alpha^{-1} = (\alpha(1)\alpha(2)\alpha(3))$.

Is the fact that $(13)(578)$ is a disjoint 2-cycle permutation and $(\alpha(1)\alpha(2)\alpha(3))$ is one-cycle enough to argue there is no $\alpha$ that satisfies the condition?

Answer 1

In general: if two elements in a group are conjugate then they have the same order: $g^{-1}xg=y$ implies $o(x)=o(y)$. This is easy to prove and I leave it to you Cure. Now $o(123)=3$, but $o((13)(578))=6$.

Answer 2

The parities differ. Note, that the parity of a $3$-cycle is even.

$$sgn(\alpha(123) \alpha^{-1}) = sgn (123) = 1 \color{red}{\neq} -1 = sgn(13)\cdot sgn(578) = sgn((13)(578))$$