Cyclic quotient group of permutation group $\mathbb{S}_n$

Question

I have a group-theoretic lemma that I am having difficulty with:

$\mathbb{S}_n$ has a cyclic quotient group of prime order $p$ iff $p=2$ and $n\geq2$

Proof: $n=2$ is trivial so suppose $n\geq3$.

Let $N\trianglelefteq G$ and suppose $\mathbb{S}_n/N\cong \mathbb{Z}_p$ which is Abelian. By nature of $N$, $ghg^{-1}h^{-1}\in N$ for $g,h\in\mathbb{S}_n$.

jump in logic

Let $g,h$ be 2-cycles i.e. $g=(ab)$ and $h=(ac)$ for a,b,c distinct. Thus, $$ghg^{-1}h^{-1}=(bca)$$ is a 3-cycle and all possible 3-cycles can be obtained this way. Therefore $N$ contains all 3-cycles. Since 3-cycles generate $\mathbb{A}_n$, $N \subseteq \mathbb{A}_n$

jump in logic

$\therefore p=2$ since $|\mathbb{S}_n/\mathbb{A}_n|=2$

Question 1: Why are we picking 2-cycles in particular? Are they picked because they construct the 3-cycles to be used later in the proof?

Question 2: How are concluding that $\mathbb{A}_n\subseteq N$? If we are not using this, how is the jump in logic explained?

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P.S. I cannot use the property that $\mathbb{A}_n$ is simple for $n\neq4$

Answer 1

This is trivial because of the following. So assume there exists a surjection $G\rightarrow \mathbb Z_p$ where $p\neq2$. Then since the order of any transposition is $2$ it goes to $0$ in the image. But transpositions generate $S_n;n\geq2$ and hence $Im$ $G=0$ which contradicts surjectivity.

Answer 2

Thanks to the insight provided by Soumik Ghosh's answer and the extremely clear explanation by Mark, I have managed to translate the quite clever proof into something that I can understand quite simply.

By our hypothesis, $\pi:\mathbb{S}_n\to \mathbb{S}_n/N \cong \mathbb{Z}_p$, for $p \neq 2$.

Since $\mathbb{S}_n$ is generated by transpositions, then any permutation $s=t_1t_2\cdots t_n$ where $t_i$ are transpositions.

However, for $N \trianglelefteq \mathbb{S}_n$, $\pi:\mathbb{S}_n\to \mathbb{S}_n/N$ is a surjective homomorphism so $\forall g\in\mathbb{Z}_p$, $\exists s\in\mathbb{S}_n$ s.t. $\pi(s)=g$.

This implies that $g^2=(\pi(s))^2=\pi(t_1)\cdots\pi(t_n)\pi(t_1)\cdots\pi(t_n)$.

But $\pi$ is abelian so $g^2=\pi(t_1^2)\cdots\pi(t_n^2)=\pi(1)=1$. But every element g is of order $p\neq2$. Contradiction!