2
$\begingroup$

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?

$\endgroup$
3
  • $\begingroup$ For $n\ge 5$ the only non-trivial proper normal subgroup of $S_n$ is $A_n$. $\endgroup$ – Hongyi Huang Sep 16 '20 at 12:50
  • $\begingroup$ @Hongyi Huang, in the spirit of the linked post, I assume that parity of a permutation is not an available notion here, let alone $A_n$. $\endgroup$ – user810157 Sep 16 '20 at 13:01
  • $\begingroup$ 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. $\endgroup$ – Hongyi Huang Sep 16 '20 at 13:09
1
$\begingroup$

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 $A_n$.

$\endgroup$
3
  • $\begingroup$ Just for the record (and the interested readers), on "contraposition" vs. "contradiction" see e.g. here: math.stackexchange.com/q/262828/810157 $\endgroup$ – user810157 Sep 16 '20 at 14:59
  • $\begingroup$ For $n=4$ the statement seems to be false: math.stackexchange.com/q/1804908/810157 $\endgroup$ – user810157 Sep 18 '20 at 2:30
  • 1
    $\begingroup$ @user750041 thanks, I’ll fix $\endgroup$ – egreg Sep 18 '20 at 7:29
1
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy