If a normal subgroup $N$ of $A_n$ contains any $3$-cycle, then $N = A_n$

Question

Let $n\geq 3$, if a normal subgroup $N$ of $A_n$ contains any $3$-cycle, then $N = A_n$.

What I have done

If $n\geq 5$ the result follows by these lemmas:

(1) Let $n\geq3$, then every element of $A_n$ is a product of $3$-cycles and;

(2) if $n\geq 5 $, then any $3$-cycle are conjugate in $A_n$.

Now for $n=3$, we have that $A_3$ is a cyclic group generated by $(1 2 3)$ or $(1 3 2)$. Thus, if $(1 2 3)$ or $(1 3 2)\in N$, then $N=A_3$

My problem is when $n=4$. I know that $M=\{(1),(12)(34),(13)(24),(14)(23)\}\cong \mathbb{Z}_2\times \mathbb{Z}_2$ is a normal subgroup of $A_4$, but $M$ does't contain a $3$-cycle. How can I go on to show that if $N$ contains a $3$-cycle then $N=A_4$?

Answer 1

Hint $$ (124)^{-1} (123) (124) = (134) \\ (134)^{-1} (123) (134) = (142) \\ (123)^{-1} (134) (123) = (243) \\ $$

If $N$ is normal and contains a 3 cycle, WLOG $(123)$, you can deduce from the above that $N$ conatins all eight $3$-cycles. From here it is easy.