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In this link here it says that a cycle be decomposed a product of disjoint cycles. However say you consider the element of $S_4$ (The symmetric group of order 24) given by

$(12)(13)(14) = (1432)$. How would you write this as a product of disjoint cycles? An element in $S_4$ is either of the form $(1),(12),(123),(12)(43)$ or $(1234)$. So looking at the last cycle type we can never decompose it into a disjoint cycle unless we have something like $(1)(234)$. However if you decompose it into the form $(1)(234)$ or say $(14)(23)$ would this not contradict it being of the last cycle type?

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    @ZevChonoles Thanks, we should keep the question then.2011-10-12

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$(1432)$ can be written as a product of disjoint cycles like so: $(1432)$. It's a product (with one factor) of disjoint cycles (each cycle in the product is disjoint from any other cycle in the product).

The main problem is that you misread the link. It doesn't say every cycle can be decomposed into a product of disjoint cycles, it says every permutation (on a finite set) can be decomposed into a product of disjoint cycles.

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    @D Lim: A permutation on a finite set $X$ is simply a bijection from $X$ to itself. A cycle is a particular kind of permutation, in which you have a subset $Y$ of objects, $Y=\{a_1,\ldots,a_n\}$, $n\geq 1$, and the cycle acts by mapping every object in $X-Y$ to itself, mapping $a_i$ to $a_{i+1}$ for $i=1,\ldots,n-1$, and mapping $a_n$ to $a_1$.2011-08-07
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$(1432)$ is already a cycle, so there is nothing to decompose. In the decomposition of $(1432)$ into disjoint cycles there is only one term: the cycle $(1432)$ itself. The theorem about the decomposition of an arbitrary element of $S_n$ as a product of disjoint cycles allows the possibility that there is only one cycle in the decomposition.

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    Right that's what I needed to know, that the possibility of only one cycle being in the decomposition is allowed.2011-08-07