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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?

Alexander Gruber
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Matt Raymond
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1 Answers1

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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 $c_1=(123)$.

Alexander Gruber
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