Order of Product of Disjoint Permutations/Examples/Non-Disjoint Permutations in S9
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Example of Order of Product of Non-Disjoint Permutations
Consider the permutation given in cycle notation as
- $\rho = \begin{pmatrix} 1 & 2 & 3 & 4 \end{pmatrix} \begin{pmatrix} 2 & 6 & 7 \end{pmatrix} \begin{pmatrix} 3 & 9 \end{pmatrix}$
Its order is given by:
- $\order \rho = 7$
and not $\lcm \set {4, 3, 2} = 12$.
Proof
$\rho$ is the product of $3$ cyclic but not disjoint permutations.
Thus we cannot use Order of Product of Disjoint Permutations to determine its order.
By compositing cycles, we obtain:
- $\rho = \begin{pmatrix} 1 & 2 & 6 & 7 & 3 & 9 & 4 \end{pmatrix}$
which is of order $7$.
$\blacksquare$
Sources
- 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $9$: Permutations: Example $9.9$