Every nontrivial normal subgroup of $A_n$ contains a $3$-cycle for $n\ge 5$

Choose $1\mathrm{\ne }\sigma \in N$ having as many fixed points as possible; this is possible because $A_n$ is finite.

For any $3$-cycle $\tau$, the commutator $h=\tau \sigma {\tau }^{-1}{\sigma }^{-1}$ lies in $N$ by [F1]. Its support is contained in $\mathrm{supp}(\tau )\cup \sigma (\mathrm{supp}(\tau ))$.

If $\sigma$ itself is a $3$-cycle, there is nothing to prove. Suppose instead that $\sigma$ has a $3$-cycle $(a_1{a}_2{a}_3)$ and also moves a point $b$ outside it. The remaining disjoint cycles form a nonidentity even permutation, so [F4] shows that they move at least three points; hence $\sigma$ moves at least six points.

It remains that every nontrivial cycle of $\sigma$ is a transposition. Because $\sigma \in A_n$, [F4] makes their number $r$ even, so $r\ge 2$.

If $\sigma$ has a cycle $(a_1{a}_2{a}_3{a}_4\dots )$ of length at least $4$, take $\tau =(a_1{a}_2{a}_3)$. Direct use of [F3] gives $h=(a_1{a}_2{a}_4)$, a $3$-cycle in $N$.

Put $\tau =(a_1{a}_{2}b)$. Its support is not preserved by $\sigma$, so $h\mathrm{\ne }1$; step 1.2 shows that $h$ moves at most the five points $a_1,a_2,a_3,b,\sigma (b)$. Thus $h$ fixes more points than $\sigma$, contradicting step 1.1.

If $r\ge 3$, write two factors as $(ab)(cd)$ and take $\tau =(abc)$. Calculation gives $h=(ac)(bd)$, which is nonidentity and moves four points, fewer than the at least six moved by $\sigma$; this again contradicts step 1.1.

If $r=2$, write $\sigma =(ab)(cd)$. Since $n\ge 5$, choose a fixed point $e$ and take $\tau =(abe)$. Calculation gives $h=(aeb)$, a $3$-cycle in $N$.

The exhaustive cycle cases in [F2] show that either $\sigma$ itself, step 2.1, or step 2.4 supplies a $3$-cycle, while steps 2.2 and 2.3 exclude every other case. Therefore $N$ contains a $3$-cycle.