Difference between containing a cycle of length 3 and containing an element having a cycle of length 3

Question

The whole question is to prove that if $n \geq 5$ and $N$ is a nontrivial normal subgroup of $A_n$ then $N$ = $A_n$. the problem is broken down into three parts. (a) is showing that $N$ contains an element having a cycle of length $\geq 3$ and (b) is showing that N contains a cycle of length 3. What is the difference? I have been given hints to prove both, but I don't know how to use them. For a, I was told to use the description of conjugate permutations given in the proposition "If $f$ is an irreducible polynomial over a subfield $F$ of $\mathbb{R}$ and $deg(f) =n \geq 2$ is prime and $f$ has precisely $n-2$ real roots. Then $Gal_F(f)$ is isomorphic to the full permutation group $S_n$. For b I was told to consider $f=(123...)(..)..$ take a cycle $g=(12x)$ and compute the commutator $f^{-1}gfg^{-1}$. Any assistance is appreciated. I mostly don't really see what the difference between a and b is so if someone could explain that, that would be great. I've seen elsewhere that $A_n$ is generated by 3 cycles and then $A_n = N$ as a result, but I haven't been able to figure out or find anything related to (a)

Answer 1

(b) means that $N$ contains an element like $(3~4~5)$. One cycle of length 3 (and all other cycles of length 1).

(a) means that $N$ contains an element like $(1~2~3)(4~5~6)$ or $(1~2)(3~4~5)$. At least one cycle of length 3, and the other disjoint cycles can be anything.