Cycle decomposition of an element of prime order

Question

I am reading the proof of the theorem that every alternating group $A_n$ is simple for $n \ge 5$ in Artin's Algebra.

In one step, Artin said

We are given that $N$ is a normal subgroup and that it contains a permutation $x$ different from the identity ... Our first step is to note that a suitable power of $x$ will have prime order, say order $l$. We may replace $x$ by this power, so we may assume that $x$ has order $l$. Then the cycle decomposition of $x$ will consist of $l$-cycles and $1$-cycles.

I think I understand the first statement (correct me if I am wrong) because $N$ is finite and hence $x^L=e$ for some $L$. We can always find a prime factor $l$ of $L$ and write it as $(x^{(L/l)})^l=e$.

But I don't understand how he can arrive at the last statement, that the cycle decomposition of $x$ will consist of $l$-cycles and $1$-cycles.

Moreover, later in his proof, he said

Case 1: $x$ has order $l \ge 5$.

How the indices are numbered is irrelevant, so we may supposed that $x$ contains the $l$-cycle $(12345\cdots l)$, say $x=(12345\cdots l)y$, where $y$ is a permutation of the remaining indices...

But didn't he just stated that the cycle decomposition of $x$ consists only of $l$-cycles and $1$-cycles?

Answer 1

Let $\sigma=\sigma_1\sigma_2\sigma_3 [...] \sigma_n$, where $\sigma_i$ are all disjoint cycles. If the order of $\sigma$ is $l$ for prime $l$ and the order of $\sigma_i$ is $r_i$, then $l$ is the least common multiple of $r_1, r_2, r_3, [...], r_n$ since that's how the order of products of elements works. Thus, $r_i \mid l$, so $r_i$ is $1$ or $l$. Therefore, $\sigma$ is the product of disjoint $l$-cycles or $1$-cycles. This is a pretty common theorem, so Artin must've assumed the reader already knew about it.

Also, in the last statement, he's simply taking out an $l$-cycle from the decomposition of $x$. Since you know the decomposition of $x$ is just $l$-cycles and $1$-cycles, then $y$ is just a bunch of $l$-cycles and $1$-cycles composed together. There is no contradiction with the first statement.

Answer 2

$\newcommand{\Span}[1]{\left\langle #1 \right\rangle}$First of all, the last statement does not contradict the previous ones: $y$ will be a product of $l$-cycles and $1$-cycles.

Then, if the permutation $x$ has prime order $l$, you may use the orbit-stabilizer theorem, which tells you that the orbits of $G = \Span{x}$, a group of prime order $l$, have size dividing $l$, thus either $1$ or $l$. As the orbit of $\alpha$ is $$ \alpha, \alpha^{x}, \dots, \alpha^{x^{l-1}}, $$ you get the required decomposition.