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Let $H$ be a normal subgroup of the alternating group $A_n$ ($n\geq5$) such that the 3-cycle $(i j k)\in H$. Then, $H=A_n$.

I think this proof is wrong because I can't see where the hypothesis $n\geq5$ is used.

Proof: I will use two facts: first, if $\sigma\in S_n$, then $\sigma(a_1\ a_2\ ...\ a_k)\sigma^{-1} = (\sigma a_1\ \sigma a_2\ ...\ \sigma a_k)$. Second, remember that $A_n=\langle\{(pqr):k\not=l\not=m\not=k\}\rangle$. Let $(pqr)\in S_n$ a 3-cycle.

  1. If $|\{i,j,k\}\cap\{p,q,r\}|=0$, then, $\sigma=(i\ p)(j\ q)(j\ k)(j\ r)\in A_n$ is such that $\sigma(i)=p,\ \sigma(j)=q,\ \sigma(k)=r$. Then, $(pqr)=\sigma(ijk)\sigma^{-1}\in A_n$.

  2. If $|\{i,j,k\}\cap\{p,q,r\}|=1$, we can suppose $p=i$ and $\sigma=(j\ q)(k\ r)\in A_n$ works as in 1.

  3. If $|\{i,j,k\}\cap\{p,q,r\}|=2$, we have two cases: if $p=i$ and $j=q$, then $\sigma=(i\ k\ r)\in A_n$ works. If $p=i$ and $j=r$, we take $\sigma=(j\ q\ k)\in A_n$.

  4. If $|\{i,j,k\}\cap\{p,q,r\}|=3$, the non trivial case is when $p=i,j=r,k=q$, but this implies $(p\ q\ r) = (i\ k\ j)=(i\ j\ k)^{-1}\in H$.

So, $H$ contain every 3-cycle and, therefore, $H=A_n$.

Veridian Dynamics
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    I am pretty sure that the claim is also true when $n=4$, and it is trivially true for $n \leq 3$. The condition $n \geq 5$ is likely only used to conclude from here that $A_n$ is simple for $n \geq 5$. – N. S. Mar 25 '18 at 00:23

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