# Cyclic quotient group of permutation group $\mathbb{S}_n$

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 $$n\neq4$$

• The facts $N$ contains all $3$-cycles and that $3$-cycles generate $A_n$ imply that $A_n \le N$, not that $N \le A_n$. – Derek Holt Feb 20 at 21:03
• That is quite counter-intuitive... especially since the book I'm following says otherwise. – Malcolm Feb 20 at 21:13
• @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. – Mark Feb 20 at 21:14
• @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)? – Malcolm Feb 20 at 21:20
• 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. – Mark Feb 20 at 21:22

## 2 Answers

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.

• Very good answer! – Mark Feb 20 at 21:07
• 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 :) – 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!

• 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 :)! – Malcolm Feb 20 at 22:26
• 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. – Mark Feb 21 at 0:57