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Here is my attempt at this proof:

We are given that $n\geq 3$. Suppose $G=S_n$. We are to prove that $G$ cannot have a normal subgroup of order $2$. So, suppose $H$ is a normal subgroup of $G$, and that $|H|=2$. We shall prove by induction.

(Base Case): Let $n=3$. Then, $G=S_n=S_3$. By Lagrange's Theorem, we have $|G|/|H|=\frac{6}{2}=3\cong\mathbb{Z}_3$. Although $\mathbb{Z}_3$ is abelian, and all of its subgroups are normal--it is a simple group. In other words, it only has the trivial normal subgroups. Therefore, when $n=3$, $G$ does not have a normal subgroup of order $2$.

Now, suppose $G=S_k$ for some $k\in\mathbb{N}$ holds true. We want to prove that $G=S_{k+1}$ is true...

I'm not sure where to go from here, or if induction is even the correct idea here.

user2318081237123
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    Hint: $S_n$ has trivial center when $n\ge3$. See for instance https://math.stackexchange.com/questions/719915/find-the-center-of-the-symmetry-group-s-n – lhf Oct 07 '21 at 01:23

2 Answers2

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Hint, and formulating a bit more general, use the following:

(a) If $N \unlhd G$ and $|N|=2$ then $N \subseteq Z(G)$.

(b) If $n \geq 3$, then $Z(S_n)=\{(1)\}$.

Nicky Hekster
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The only elements of period $2$ are products of transpositions. If $H=\{e, \sigma\}$ is normal then for every $\tau \in S_n$, $$\tau\sigma\tau^{-1}=\sigma$$

This is impossible. Without loss we can assume that either $\sigma$ itself is a transposition $=(12)$ or contains two transpositions, $=(12)(34)\cdots$ in both cases take $\tau=(13)$ for a counterexample.

Rene Schipperus
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