Talk:Alternating Group is Simple except on 4 Letters

In step two, the article reads (paraphrased) "let $\sigma$ be an even permutation; define $i = \sigma(1)$, $j = \sigma(2)$, $k = \sigma(3)$; then $\sigma (1\; 2\; 3) \sigma^{-1} = (i\; j\; k)$; we have that $i$, $j$, $k$ are arbitrary". But the last sentence is not immediately clear to me. To be precise, I don't immediately see how given any arbitrary distinct $i,j,k \in \mathbb{N}_n$ we can find an even permutation $\sigma$ such that $\sigma(1) = i$, $\sigma(2) = j$, and $\sigma(3) = k$ (which is what I assume the statement "we have that $i,j,k$ are arbitrary" is actually saying). I think this uses the hypothesis that $n \ge 5$. (Moreover, if one were to be pedantic, one could say that no, $i,j,k$ are not at all arbitrary, since we defined them as $i = \sigma(1)$, $j = \sigma(2)$, $k = \sigma(3)$, where $\sigma$ is the particular even permutation that we happened to pick.) --plammens (talk) 09:11, 9 October 2021 (UTC)


 * Right, let me see if I understand. we have picked an even permutation.


 * Having picked such a permutation, we then define $i$, $j$ and $k$ as the "arbitrary" elements to which the "equally arbitrary" even permutation map from $1$, $2$ and $3$.


 * Yes you're right, that's important, it's not $\tuple {i, j, k}$ that is arbitrary, it's the even permutation we pick that is arbitrary.


 * Having done that, we then invoke a result which says $\map {\paren {\sigma \rho \sigma^{-1} } } x = \map \rho x$. (Which exists, we just need to find the link to it.)


 * Then we say that this result holds for any arbitrary $i$, $j$ and $k$, but we need to prove that a mapping from $\paren {1, 2, 3}$ to $\tuple {i, j, k}$ occurs in some even permutation for every $\tuple {i, j, k}$.


 * Hmm.


 * Okay, so as long as we can prove that $S_3$ is embedded in $A_n$ for all $n \ge 5$ (or $n \ge 4$, we may be able to do that as well) then we are done. Can we prove that $A_{n - 1}$ is embedded in $A_n$? I believe we can. Then all we need to do is show that $S_3$ is embedded in $A_4$, which I think we may already have somewhere.


 * Feel free to fix it up, any way you think makes sense. --prime mover (talk) 10:32, 9 October 2021 (UTC)