Confused with a proof about the simplicity of $A_n$. I am currently self-studying group theory and I am trying to understand why $A_n$ is simple for $n \geq 5$ and I have not made any progress for hours. I have searched for proofs everywhere (including this website) and every proof I've found has confused me. At the moment I am reading a proof found on this website here: https://groupprops.subwiki.org/wiki/Alternating_group_is_simple.
The proof uses induction and the website has a separate article when $n=5$. Luckily, I understand the proof of the base case. The rest of the proof goes as follows:
Let $N$ be a normal subgroup of $A_{n+1}$. Let $H_i$ denote the subgroup of $A_{n+1}$ that stabilizes the letter $i$. Then, each $H_i$ consists of the even permutations on $n$ letters (excluding $i$) and is hence isomorphic to $A_n$. Thus, each $H_i$ is simple. Now, since normality satisfies transfer condition, $N \cap H_i$ is normal in $H_i$ for every $i$. By simplicity of $H_i$, either $N$ contains $H_i$, or $N$ intersects $H_i$ trivially.
Suppose there exists $i$ for which $N$ contains $H_i$. Then, by fact (3) (which says that $A_n$ is a contranormal subgroup inside $A_{n+1}$), $H_i$ is contranormal inside $A_{n+1}$, i.e., its normal closure is $A_{n+1}$. Since $N$ is normal, this forces $N = A_{n+1}$ and we are done.
Otherwise, $N \cap H_i$ is trivial for every $i$. Thus, no nontrivial element of $N$ fixes any letter. Let's use this to show that $N$ can have no nontrivial elements. Suppose $\sigma \in N$ is nontrivial. Then, first observe that in the cycle decomposition of $\sigma$, every element must be in a cycle of the same length $k$ (otherwise, som power of $\sigma$ would fix a letter)(+). Thus, $\sigma$ has $r$ cycles each of size $k$, where $kr = n+1$ (+). Now, if $n \geq 5$, then $n + 1 \geq 6$. Choose an $i$ and a double trransposition $\tau \in A_{n+1}$ such that $\tau$ fixes both $i$ and $\sigma(i)$, but such that $\tau$ does not commute with $\sigma$ (this is possible because $n+1 \geq 6$)(+). Then $\sigma^{-1}\tau\sigma\tau^{-1} \in N$ is a nontrivial element of $N$ fixing both $i$ and $\sigma(i)$, giving the required contradiction. $\square$
The parts that I do not understand in this proof are marked by (+). Why does every element in the cycle decomposition of $\sigma$ have the same length? If that's not the case then how does it come that some power of $\sigma$ would fix a letter? If $\sigma$ has $r$ cycles then why is $kr = n+1$? Lastly, why is it possible to find such a $\tau$ when $n + 1 \geq 6$? It feels like the proof goes way too fast for me and I simply can not understand everything. I am begging for someone to explain this to me. I do not have any teacher or friend that can help me with this so I would seriously appreciate any answer or a link to an easier to understand proof of this theorem!!
 A: $\newcommand{\supp}{\text{supp}}$

*

*Every element in the cycle decomposition of $\sigma$ must have the same length.

Suppose that this is not the case. For simplicity, suppose $\sigma = \sigma_1 \sigma_2$, where $\sigma_1, \sigma_2$ are disjoint cycles of respective length $n_1 < n_2$ (and we have $n_1 +n_2 = n+1$ because $\sigma$ has no fixed point).
Then, $\sigma^{n_1} = \sigma_1^{n_1}\sigma_2^{n_1} = \sigma_2^{n_1} \neq \text{id}_{N}$ because $n_1 < n_2$ and $\sigma_2$ is of order $n_2$. Because $\sigma^{n_1} \in N$, this is a contradiction.
And it is easy to see that in the case where $\sigma = \sigma_1\sigma_2...\sigma_r$ disjoint cycles of the same length $p$, then $\sigma^m$ is either the identity (if $m \equiv 0 \mod r$), or a full-support permutation.

*

*If \sigma has $r$ cycles, then $kr = n+1$
This simply comes from the previous point, and the fact that $\sigma$ has no fixed point, i.e. $|\supp(\sigma)| = n+1$. If we write $\sigma = \sigma_1\sigma_2...\sigma_r$ the decomposition in disjoint cycles, we just proved that they all the same length, and:
$\begin{align*}
\underbrace{|\supp(\sigma)|}_{= n+1} &= \sum_{i=1}^{r} |\supp(\sigma_i)| &\text{because they are disjoint cycles}\\
&= r |\supp(\sigma_1)| &\text{because all the cycles have the same length}\\
&= kr
\end{align*}$

*

*Existence of $\tau$
This is a bit harder to see immediately. You need to find $4$ elements (not $i$ or $\sigma(i)$) $a, b, c, d \in \{1, \dots, n+1\}$ so that $\tau = (ab)(cd)$ does not commute with $\sigma = \sigma_1...\sigma_r$, i.e. we don't want $\t^{-1}\sigma\tau = \sigma$
I don't have the time to write a proper proof now, I will try to come back later - a "nice" way to see this is to understand what the decomposition in cycles of $\tau^{-1}\sigma\tau$ looks like.
