I am trying to understand a solution to an problem that involves proving that if $ N \lhd S_n $ such that $\frac{S_n}{N}$ is cyclic, then $ N = S_n $ or $ A_n$
I understand why the order of $\frac{S_n}{N} = 1$ or $2$ and also that $o(\frac{S_n}{N}) = 1 \Rightarrow N=S_n$
But in proving that $o(\frac{S_n}{N}) = 2 \Rightarrow N=A_n$, it says $g^2\in N, $ $ \forall g \in S_N $ and I am trying to understand why that is the case.
$o(S_n/N)=2$ means the subgroup $N$ has index 2 in $S_n$. But $A_n$ is the only subgroup of index $2$ in $S_n$, so $N=A_n$
Also if $g\in S_n$, then $gN \in S_n/N$ and so $(gN)^2=N$, since $S_n/N$ has order $2$. Which means $g^2N=N$ and hence $g^2 \in N$