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?
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Is this in $S_6$ ? – lhf Jul 26 '16 at 13:17
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yes if aϵS6 in two line notation – W. G. Jul 26 '16 at 14:31
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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 $a(6) = 6$.
Mees de Vries
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And one-line notation, which is how OP is interpreting (1432), would have all of the numbers listed, so every permutation of {1,2,3,4,5,6} in one-line notation would have all 6 numbers 1,2,3,4,5,6 listed in some order. Since (1432) has only four numbers listed and yet denotes a permutation of six numbers, it can't be intended as one-line notation. – anon Jul 26 '16 at 13:24
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So if I wrote aϵS6 in two line notation as shown in the link https://en.wikipedia.org/wiki/Permutation the corresponding two line notation would for a be a(1)=1, a(2)=4, a(3)=3, a(4)=1, a(5)=5 and a(6)=6 where S=(1,2,3,4,5,6)? – W. G. Jul 26 '16 at 14:28
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Rendering the $a$ with $a(1)=1, a(2)=4, a(3)=3, a(4)=2, a(5)=5$ and $a(6)=6$ in two-line notation would go as follows: $$ a = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6\ 1 & 4 & 3 & 2 & 5 & 6 \end{pmatrix}. $$ In cycle notation, this would just be $a = (2;4)$. – Mees de Vries Jul 26 '16 at 14:37