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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 $n\neq4$

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  • $\begingroup$ 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
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    $\begingroup$ 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
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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.

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  • $\begingroup$ 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
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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!

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  • $\begingroup$ 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

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