Order of Product of Disjoint Permutations/Examples/Non-Disjoint Permutations in S9

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 $9.9$