# 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)
Yes, exactly: given an arbitrary 3-cycle $(i,j,k)$, we want to show that there exists an even permutation $\sigma$ with $\sigma(1) = i$, $\sigma(2) = j$, and $\sigma(3) = k$. (Of course, if we remove the requirement that $\sigma$ be even, then it is straightforward to see that such a permutation always exists.)
I more or less have a proof, but it is much more bare-bones than the direction you propose. The problem is that it is a very cumbersome (and I'm not entirely sure if valid) proof by cases. Here's a draft.
Let $\rho \in S_n$ be an arbitrary 3-cycle.
Let $i, j, k \in \mathbb{N}_n$ be such that $\rho = (i, j, k)$.
Our goal is to show that there exists an even permutation $\sigma \in A_n$ such that $\sigma(1) = i$, $\sigma(2) = j$ and $\sigma(3) = k$.
We will proceed by cases.
We have that $|\{1,2,3\} \cap \{i, j, k\}|$ is either $0$, $1$, $2$ or $3$.
Case 1: $|\{1,2,3\} \cap \{i, j, k\}| = 0$. (This case is only possible when $n \ge 6$.)
The permutation $\sigma = (1, i, 2, j)(3, k)$ is even and has the desired property.
Case 2: $|\{1,2,3\} \cap \{i, j, k\}| = 1$.
Without loss of generality, suppose that $\{1,2,3\} \cap \{i, j, k\} = \{1\}$.
We have two further sub-cases: $i = 1$ or $i \ne 1$.
If $i = 1$, then the permutation $\sigma = (2, j)(3, k)$ is even and has the desired property.
Now suppose $i \ne 1$.
Assume, without loss of generality, that $j = 1$.
Then, the permutation $\sigma = (1, i, 3, k, 2)$ is even and has the desired property.
Case 3: $|\{1,2,3\} \cap \{i, j, k\}| = 2$.
Assume, without loss of generality, that $\{1,2,3\} \cap \{i, j, k\} = \{1, 2\}$.
We have three further sub-cases: either $i = 1 \land j = 2$, or $(i = 1 \land j \ne 2) \lor (i \ne 1 \land j = 2)$, or $i \ne 1 \land j \ne 2$.
(In other words, if $\sigma$ exists, it either fixes 2, 1, or 0 letters of those in $\{1, 2, 3\}$.)
Suppose $i = 1 \land j = 2$.
Since $n \ge 5$, there exists an element $l \in \mathbb{N}_n \setminus \{1, 2, 3, k\}$.
Then, the permutation $\sigma = (3, k, l)$ is even and has the desired property.
Now suppose $(i = 1 \land j \ne 2) \lor (i \ne 1 \land j = 2)$.
Without loss of generality, assume $(i = 1 \land j \ne 2)$.
This implies $k = 2$.
Then, the permutation $\sigma = (2, j, 3)$ is even and has the desired property.
Now suppose $i \ne 1 \land j \ne 2$.
If $i = 2$ and $j = 1$, then the permutation $\sigma = (1, 2)(3, k)$ is even and has the desired property.
Otherwise, without loss of generality, assume $i = 2$ and $k = 1$.
Since $n \ge 5$, there exists an element $l \in \mathbb{N}_n \setminus \{1, 2, 3, k\}$.
Then, the permutation $\sigma = (1, 2, j, l, 3)$ is even and has the desired property.
Case 4: $|\{1,2,3\} \cap \{i, j, k\}| = 3$.
That is, $\{i, j, k\} = \{1, 2, 3\}$.
We have that any (not necessarily even) permutation $\sigma$ has the desired property if and only if it can be written as a disjoint product $\sigma = \alpha\beta$, where $\alpha$ is the permutation: $\begin{pmatrix} 1 & 2 & 3 \\ i & j & k \\ \end{pmatrix}$
If $\alpha$ fixes all letters (is the identity), take $\beta$ to be the identity; then $\sigma$ will be even.
If $\alpha$ fixes one letter, $\alpha$ will be a transposition, so take $\beta$ to be another (disjoint) transposition (we can do this because $n \ge 5$); then $\sigma$ will be even.
If $\alpha$ fixes no letters, $\alpha$ will be a 3-cycle, so take $\beta$ to be the identity; then $\sigma$ will be even.
------------
In all cases, we found an even permutation $\sigma$ with the desired property.
Also, $\rho$ was arbitrary.
Hence a $\sigma$ always exists for any 3-cycle.
Let me know what you think. --plammens (talk) 14:56, 10 October 2021 (UTC)
Feel free to post it up. Once it's done it is probably best extracted and made into a lemma, then we can have multiple proofs of that one aspect of it.
You are strongly encouraged to make full use of the custom $\LaTeX$ constructs designed for $\mathsf{Pr} \infty \mathsf{fWiki}$. --prime mover (talk) 21:02, 10 October 2021 (UTC)