I have a group-theoretic lemma that I am having difficulty with:

$\mathbb{S}_n$ has a cyclic quotient group of prime order $p$ iff $p=2$ and $n\geq2$

Proof: $n=2$ is trivial so suppose $n\geq3$.

Let $N\trianglelefteq G$ and suppose $\mathbb{S}_n/N\cong \mathbb{Z}_p$ which is Abelian. By nature of $N$, $ghg^{-1}h^{-1}\in N$ for $g,h\in\mathbb{S}_n$.

jump in logic

Let $g,h$ be 2-cycles i.e. $g=(ab)$ and $h=(ac)$ for a,b,c distinct. Thus, $$ghg^{-1}h^{-1}=(bca)$$ is a 3-cycle and all possible 3-cycles can be obtained this way. Therefore $N$ contains all 3-cycles. Since 3-cycles generate $\mathbb{A}_n$, $N \subseteq \mathbb{A}_n$

jump in logic

$\therefore p=2$ since $|\mathbb{S}_n/\mathbb{A}_n|=2$

Question 1: Why are we picking 2-cycles in particular? Are they picked because they construct the 3-cycles to be used later in the proof?

Question 2: How are concluding that $\mathbb{A}_n\subseteq N$? If we are not using this, how is the jump in logic explained?

P.S. I cannot use the property that $\mathbb{A}_n$ is simple for $n\neq4$

  • $\begingroup$ The facts $N$ contains all $3$-cycles and that $3$-cycles generate $A_n$ imply that $A_n \le N$, not that $N \le A_n$. $\endgroup$ – Derek Holt Feb 20 at 21:03
  • $\begingroup$ That is quite counter-intuitive... especially since the book I'm following says otherwise. $\endgroup$ – Malcolm Feb 20 at 21:13
  • $\begingroup$ @Malcolm Let's begin from the fact that if you assume $N$ contains a $2$-cycle then there is no chance for $N\leq A_n$. So for sure you did something wrong here. Next, if $N$ contains all $3$ cycles then it must contain $A_n$ because $A_n$ is generated by $3$-cycles. $\endgroup$ – Mark Feb 20 at 21:14
  • $\begingroup$ @Mark Okay that makes sense. But then, how does the last 'jump in logic' work if we are not assuming $N=A_n$ (as I thought it continued)? $\endgroup$ – Malcolm Feb 20 at 21:20
  • 1
    $\begingroup$ If $A_n\leq N$ then $N=A_n$ or $N=S_n$, there are no other options. But your assumption is that $S_n/N$ has prime order, hence $N$ can't be equal to $S_n$. Anyway, the proof looks strange to me. Some parts are very unclear. $\endgroup$ – Mark Feb 20 at 21:22

This is trivial because of the following. So assume there exists a surjection $G\rightarrow \mathbb Z_p$ where $p\neq2$. Then since the order of any transposition is $2$ it goes to $0$ in the image. But transpositions generate $S_n;n\geq2$ and hence $Im$ $G=0$ which contradicts surjectivity.

  • $\begingroup$ Very good answer! $\endgroup$ – Mark Feb 20 at 21:07
  • $\begingroup$ Thank you for your enlightful answer. @Mark, thank you for explaining his answer for me as well. I have written my own version of both your comments and referenced your well thought out answers :) $\endgroup$ – Malcolm Feb 20 at 22:12

Thanks to the insight provided by Soumik Ghosh's answer and the extremely clear explanation by Mark, I have managed to translate the quite clever proof into something that I can understand quite simply.

By our hypothesis, $\pi:\mathbb{S}_n\to \mathbb{S}_n/N \cong \mathbb{Z}_p$, for $p \neq 2$.

Since $\mathbb{S}_n$ is generated by transpositions, then any permutation $s=t_1t_2\cdots t_n$ where $t_i$ are transpositions.

However, for $N \trianglelefteq \mathbb{S}_n$, $\pi:\mathbb{S}_n\to \mathbb{S}_n/N$ is a surjective homomorphism so $\forall g\in\mathbb{Z}_p$, $\exists s\in\mathbb{S}_n$ s.t. $\pi(s)=g$.

This implies that $g^2=(\pi(s))^2=\pi(t_1)\cdots\pi(t_n)\pi(t_1)\cdots\pi(t_n)$.

But $\pi$ is abelian so $g^2=\pi(t_1^2)\cdots\pi(t_n^2)=\pi(1)=1$. But every element g is of order $p\neq2$. Contradiction!

  • $\begingroup$ I can also prove along the same lines that if $A_n$ has a cyclic quotient group of prime order, then that prime must be 3 :)! $\endgroup$ – Malcolm Feb 20 at 22:26
  • $\begingroup$ Your proof is almost perfect. Only in the end where you wrote "every element $g$ is of order $p$" you should add "except the identity". So by picking any $g$ which is not the identity we have $g^2\ne 1$ and this is a contradiction. $\endgroup$ – Mark Feb 21 at 0:57

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.