Make a 3-cycle with conjugation

Question

As a part of proof that $A_n$, $n >= 5$ has no normal subgroups. I need to prove that it's not possible to make a cycle with conjugation:

Let permutation $\pi = (1 2)(3 4)$. Is it possible to find such $g$ that $g \pi g^{-1}$ is a 3-cycle? $g \in S_n, n >= 5$ Please provide proof.

Answer 1

$$g(1\,2)(3\,4)g^{-1}=(g(1)\,g(2))(g(3)\,g(4))$$ and so is a product of two $2$-cycles.

Answer 2

Conjugation by $g$ may be interpreted as relabelling the elements of $\{1,2,\ldots,n\}$ according to $g$. It just gives the elements that permutations act on new names, but doesn't change any of the "important" properties of the permutations. (More rigorously: Conjugation by a fixed element $g$ is an isomorphism.)

Specifically, no matter what $g$ is, $g\pi g^{-1}$ will always be a product of two disjoint $2$-cycles.