I want to prove that for $n \geq 3$, the alternating group $A_n$ is generated by the set of all 3-cycles.

Here is my attempt:

Let $\mathcal{S}$ be the set of all 3-cycles in $S_n$, which is a conjugacy class, and let $H=\langle \mathcal{S} \rangle$. For any $\sigma \in S_n$ we have \begin{equation} \sigma H \sigma^{-1}=\sigma \langle \mathcal{S} \rangle \sigma^{-1}=\langle \sigma \mathcal{S} \sigma^{-1} \rangle=\langle \mathcal{S} \rangle=H, \end{equation} so that $H \trianglelefteq S_n$. By a previous exercise (listing the normal subgroups of $S_n$ for $n \geq 5$), $H \in \{1,A_n,S_n\}$. $H$ is obviously non-trivial, and since all elements of its generating set are even permutations it cannot be $S_n$. We conclude that $\langle \mathcal{S} \rangle=A_n$.

The above argument only works for $n \geq 5$. We are left with handling the cases $n \in \{3,4\}$ separately. Both are pretty easy to verify: If $n=3$, $|A_3|=3$ is cyclic and generated by the 3-cycle $(1 \;2 \;3)$. If $n=4$, we have $8$ 3-cycles in $S_4$. Lagrange's Theorem then forces $\langle \mathcal{S} \rangle=A_4$.

Is my proof correct? If not, please help me fix it.


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    $\begingroup$ Sorry, I don't have the attention span to read your proof. But all you have to do, to show $A_n$ is generated by the $3$-cycles, is show that a product of two transpositions can be expressed as a product of $3$-cycles. I.e., just show that $(a\,b)(a\,c)$ and $(a\,b)(c\,d)$ can be expressed as products of $3$-cycles. Seeing as $(a\,b)(a\,c)$ is a $3$-cycle, that just leaves $(a\,b)(c\,d)$. So the identity $(a\,b)(c\,d)=(a\,c\,b)(a\,c\,d)$ does the job, doesn't it? $\endgroup$ – bof Dec 21 '14 at 22:19
  • $\begingroup$ Correct as far as I can see, although I'm not completely sure how did you use Lagrange Theorem, but this special case is easy ayway. $\endgroup$ – Peter Franek Dec 21 '14 at 22:21
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    $\begingroup$ @PeterFranek There are 12 elements of $A_4$, so any subgroup must, by Lagrange, have size a divisor of $12$. Since the subgroup generated by $S$ has at least 8 elements, that subgroup must have 12 helepemts. $\endgroup$ – Thomas Andrews Dec 21 '14 at 22:28

This proof works.

As @bof noted in comments, there is a more direct approach, not using the more advanced result about the normal subgroups of $S_n$ for $n>4$. @hof's answer works for all $n$.

  • $\begingroup$ The OP didn't actually use the fact that $A_n$ is simple, just that $A_n$ is the only proper nontrivial normal subgroup of $S_n$ for $n\geq 5$. It a priori could be that $A_n$ has a normal subgroup that wasn't normal in $S_n$. $\endgroup$ – Matt Samuel Dec 21 '14 at 23:29
  • $\begingroup$ That's true, I guess, but it's essentially the same argument, and it's still a big hammer to pull out compared to @hof's argument. $\endgroup$ – Thomas Andrews Dec 21 '14 at 23:31
  • $\begingroup$ +1 and my apologies for posting an answer in the form of a comment. Perhaps you should quote my comment in your answer, since comments have a way of getting deleted. (Oh, now I remember why I posted a comment. Because I didn't really answer the OP's question, which asked for help in checking his proof.) $\endgroup$ – bof Dec 22 '14 at 17:16

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