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Below is taken from my linear algebra course lecture notes:

Some facts about permutations of $\{1,2,\dots,n\}$:

Every permutation is a product of disjoint cycles which commute. For example in. $S_{11}$ we have the permutation

$$\sigma = (1 \space{} 3 \space{}7 \space{}11\space{} 8)(2)(4\space{} 6\space{} 5\space{} 10\space{} 9)$$

Such a factorisation $\sigma=\alpha_1\dots\alpha_t$ (where $\alpha$ are some disjoint k-cycles) is unique except for the order in which it is written.

Using this link, I understood this theorem. But what I am wondering is the following:

Since each of the k-cycles are also permutations themselves. Then I could further decompose them into the following:

$$\sigma=(1\space 3 \space 7)(11\space 8)(2)(4 \space 6 \space 5)(10 \space 9)$$

Since the k-cycles involved are still disjoint, does this violates the uniqueness or did I misunderstood something subtle about "factorisation into disjoint cycles"?

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    $\begingroup$ Your mistake is that, for example, $(1\, 3\, 7\, 11\, 8) = (1\, 3\, 7)(7\, 11\, 8)$, not $(1\, 3\, 7)(11\, 8)$ $\endgroup$ – Dylan Jun 7 '15 at 12:32
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Two cycles are said to be disjoint if the sets of elements they act on are disjoint. This is equivalent to saying that $(a_1\ \dotsc\ a_k)$ and $(b_1\ \dotsc\ b_h)$ are disjoint if and only if $a_i \neq b_j$ for every $1 \leq i \leq k$ and $1 \leq j \leq h$.

What you misunderstood is cycle decomposition: $$(1\ 3\ 7\ 11\ 8) \neq (1\ 3\ 7) (11\ 8)$$ For example, the former sends $7 \mapsto 11$ while the latter sends $7 \mapsto 1$. On the other hand, $$(1\ 3\ 7\ 11\ 8) = (1\ 3\ 7)(1\ 11\ 8)$$ but those two $3$-cycles are not disjoint.

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