I'll use a longer cycle to help describe two techniques for writing disjoint cycles as the product of transpositions:
Let's say $\tau = (1, 3, 4, 6, 7, 9) \in S_9$
Then, note the patterns:
Method 1: $\tau = (1, 3, 4, 6, 7, 9) = (1, 9)(1, 7)(1, 6)(1, 4)(1, 3)$
Method 2: $\tau = (1, 3, 4, 6, 7, 9) = (1, 3)(3, 4)(4, 6)(6, 7)(7, 9)$
Both products of transpositions, method $1$ or method $2$, represent the same permutation, $\tau$. Note that the order of the disjoint cycle $\tau$ is $6$, but in both expressions of $\tau$ as the product of transpositions, $\tau$ has $5$ (odd number of) transpositions. Hence $\tau$ is an odd permutation.
Now, don't forget to multiply the transpositions you obtain for each disjoint cycle so you obtain an expression of the permutation $S_{11}$ as the product of the product of transpositions, and determine whether it is odd or even:
$\sigma = (1, 4, 10)(3, 9, 8, 7, 11)(5, 6)$.
Two permutations are disjoint if any point moved by one permutation is fixed by the other permutation. In other words, in the disjoint cycle decomposition of the two permutations, there is no overlap in the points that are written out. Equivalently, two permutations are disjoint if and only if they have disjoint support (the support of a permutation is the set of points moved by the permutation).
For example, in $S_9$, the permutations $(12)(34)$ and $(569)$ are disjoint permutations because the sets $\{1,2,3,4\}$ and $\{5,6,9\}$ are disjoint. The permutations $(12)(34)$ and $(561)$ are not disjoint.
Best Answer
We write this permutation on its standard form $$\sigma=\left(\begin{array}\\ 1&2&3&4&5\\ 5&1&2&3&4 \end{array}\right)$$ and this a cycle since $$1\overset \sigma\rightarrow5\overset \sigma\rightarrow4\overset \sigma\rightarrow3\overset \sigma\rightarrow2\overset \sigma\rightarrow1$$ so $$\sigma=(1\ 5\ 4\ 3\ 2)$$