In an ultimate goal to prove that the alternating group $\mathbb{A}_n$ is simple for $n\geq 5$, one of the lemma's I'm asked to prove is this:
Let $n\geq 5$ and let $N$ be an arbitrary nontrivial normal subgroup of $\mathbb{A}_n$. If $N$ contains an element whose cycle decomposition has a cycle of length $t\geq 4$, show that $N=\mathbb{A}_n$.
I have already showed that $\mathbb{A}_n$ is generated by the set $\{(r s k): r,s\neq k \in \{1, \dots n\} \}$ and that any normal subgroup of $\mathbb{A}_n$ that contains a 3-cycle is equal to $\mathbb{A}_n$.
I was given a hint to let $g=(a_1 a_2 \dots a_t)h$ and consider the commuter $[g,c]$ where $c=(a_1 a_3 a_2)$.
I showed that the commutator suggested is equal to the cycle $(a_1 a_4 a_3)$, which, if in $N$, would generate $\mathbb{A}_n$. But, how do I know that this commutator should be in $N$?
For any group $ G $ and a normal subgroup $ N $, given $ g \in G $ and $ n \in N $, the commutator $ [g, n] \in N $. This is because $ [g, n] = gng^{-1} n^{-1} $, and both $ gng^{-1} $ (normality is used here) and $ n^{-1} $ are in $ N $, therefore their product is also.