# Prove/disprove: $p>3$ prime. There doesn't exist any $H\lhd S_p$, $|H|=p$, such that $S_p/H\cong S_{p-1}$.

While giving my answer here, I thought to the following generalization of the claim therein stated:

Claim. Let $$p>3$$ be a prime. There doesn't exist any $$H\lhd S_p$$, $$|H|=p$$, such that $$S_p/H\cong S_{p-1}$$.

Proof sketch. By contradiction, suppose that such a $$H$$ does exist. Then, there is a surjective homomorphism $$\varphi$$ from $$S_p$$ to $$S_{p-1}$$ with kernel $$H$$. Such a $$\varphi$$ sends conjugacy classes to conjugacy classes. $$H$$ is made up of $$p$$ $$p$$-cycles; the number of $$p$$-cycles in $$S_p$$ is $$(p-1)!>p$$ (for $$p>3$$, as assumed), and thence $$H\setminus\{Id\}\subsetneq \operatorname{Cl}((1...p))$$. Therefore, any element of $$H\setminus \{Id\}$$ is sent into $$\varphi(\operatorname{Cl}((1...p)))$$, which does not contain the identity of $$S_{p-1}$$; but any such element is sent to $$Id$$ by definition of kernel. Contradiction.

(As "minimal corollary", take $$p=5$$ to get the case addressed in the opening link.)

Is this all correct?

• For $n\ge 5$ the only non-trivial proper normal subgroup of $S_n$ is $A_n$. – Hongyi Huang Sep 16 '20 at 12:50
• @Hongyi Huang, in the spirit of the linked post, I assume that parity of a permutation is not an available notion here, let alone $A_n$. – user810157 Sep 16 '20 at 13:01
• Then any element of order $p$ in $S_p$ is a $p$-cycle. With this in mind, it is very straightforward to see that $H$ can never be normal. – Hongyi Huang Sep 16 '20 at 13:09

The generalization is good, but you're proving by contradiction, not by contraposition.

If such a normal $$H$$ subgroup exists, it must have $$p$$ elements by the homomorphism theorems. Since $$p$$ is prime, the group is cyclic and generated by a $$p$$-cycle that, upon relabeling of the elements we permute on, it can be assumed to be $$(123\dots p)$$.

Since $$(12)(1234\dots p)(12)=(2134\dots p)$$ is not a power of $$(1234\dots p)$$, $$H$$ is not normal. Contradiction.

The assumption $$p>3$$ is necessary: indeed, the same argument applied to $$(123)$$ yields $$(213)=(123)^2$$ and, indeed, $$S_3/A_3\cong S_2$$.

The statement is however also true for every $$n>4$$ and follows from the simplicity of $$A_n$$.

• Just for the record (and the interested readers), on "contraposition" vs. "contradiction" see e.g. here: math.stackexchange.com/q/262828/810157 – user810157 Sep 16 '20 at 14:59
• For $n=4$ the statement seems to be false: math.stackexchange.com/q/1804908/810157 – user810157 Sep 18 '20 at 2:30
• @user750041 thanks, I’ll fix – egreg Sep 18 '20 at 7:29

If $$p>3$$ is prime then $$S_p$$ contains exactly three normal subgroups: $$1$$, $$A_p$$, $$S_p$$.

Two approaches:

$$\bullet$$ For $$n\gt4$$, the only nontrivial normal subgroup of $$S_n$$ is $$A_n$$.

$$\bullet$$ The subgroup would have to be generated by a $$p$$-cycle. But, by a theorem, all $$p$$ cycles are conjugate in $$S_p$$. However, there are $$(p-1)!\gt p$$ of them.