Permutation Cycle Question

Question

If a=(1,4,3,2), then a(1)=1, a(2)=4, a(3)=3, and a(4)=2. Does a(5)=5 and a(6)=6 where these one cycles are neglected? Or does the permutation not exist for these values?

Answer 1

Cycle notation is usually interpreted as follows: a cycle maps a number which occurs in the cycle to the next element of that cycle. Thus, if $a = (1\;4\;3\;2)$, then $a(1) = 4$, because $4$ occurs immediately after $1$. Similarly, $a(4) = 3$ and $a(3) = 2$, and finally $a(2) = 1$ (at the end of the cycle you "jump back").

If $a \in S_6$, then the "full notation" for $a$ would be $a = (1\;4\;3\;2)(5)(6)$, but indeed cycles of length 1 are usually omitted. So $a(5) = 5$ and $a(6) = 6$.