I'm reading through an abstract algebra textbook, and one of the exercises is as follows:
Prove that every permutation in $S_n$ is the product of disjoint cycles.
I tried to figure it out on my own, but I got stuck on the special case in which a permutation is a single cycle (e.g. $(1 2 3)$ in $S_3$), since it cannot be disjoint with any other cycle in that permutation group, and you need two elements to perform multiplication. I thought maybe it would be disjoint from the empty set, but then multiplication is not well defined.
I ended up finding a proof here, but it doesn't seem to consider the special case. I also tried looking this question up online, but I can't find any examples of it being asked before. The closest thing I can find is this wiki page, which just says that a cycle is a permutation.
So, my question is as follows:
Is a permutation composed of one cycle a product of disjoint cycles even if there is no multiplication? If so, how?
Best Answer
You're on the right track conceptually when you say it is "disjoint from the empty set." This isn't exactly how to phrase it, but the sentiment is there: it is disjoint from all the other cycles, because there are no other cycles.
This is how I would phrase it: a permutation $p$ is the product of disjoint cycles if, for some $n$, $p$ can be written as $p=\prod_{i=1}^nc_i$ where each $c_i$ is a cycle, and $c_i$ and $c_j$ are disjoint when $i\ne j$.
So, is $(123)$ a product of disjoint cycles? Let's test: is there an $n$ and a set of cycles $c_i$ so that $(123)=\prod_{i=1}^nc_i$? Sure: $n=1$ and $c_1=(123)$.