Exercise 4.6.2 (Normal subgroups of $S_n$)

Proof. Let $H$ be a proper nontrivial normal subgroup of $S_n$, where $n\ge 5$. We must prove that $H=A_n$.

Since $H$ is a normal subgroup of $S_n$, then $H\cap A_n$ is a normal subgroup of $A_n$, which is a simple group since $n\ge 5$. Therefore

In the latter case, we have $A_n\le H$, where $|S_n:A_n|=2$, so $A_n$ is a maximal subgroup of $S_n$. Therefore $H=A_n$ since $H$ is a proper subgroup.

Consider now the case $H\cap A_n=\{e\}$. Since $H⊴S_n$, $H{A}_n$ is a subgroup of $S_n$. Moreover $A_n\le H{A}_n$, where $A_n$ is a maximal subgroup of $S_n$, thus $H{A}_n=A_n$ or $H{A}_n=S_n$. If $H{A}_n=A_n$, then $H\le A_n$, so $H⊴A_n$, where $H\ne \{1\}$. Therefore $H=A_n$. If $H{A}_n=S_n$, then

so $|H|=2$:

Then $\sigma$ has order $2$, so $\sigma =(ij)(\cdots )\cdots$ is a product of disjoint transpositions. Consider the transposition $\tau =(jk)$, where $k\ne i,k\ne j$. Then

is also a product of disjoint transpositions. therefore $\tau \ne 1$, and $\mathit{𝜏𝜎}{\tau }^{-1}\ne \sigma$, since $\sigma (i)\ne \mathit{𝜏𝜎}{\tau }^{-1}(i)$, hence $\mathit{𝜏𝜎}{\tau }^{-1}\notin H$. This contradicts the hypothesis $H⊴S_n$. This proves $H=A_n$.

In conclusion, the normal subgroups of $S_n$ are

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