Cycle Decomposition/Examples/Permutation in S9/Mistake

Source Work

1996: John F. Humphreys: A Course in Group Theory:

Chapter $9$: Permutations:

Example $9.6$

Mistake

The cycle decomposition for the permutation

$\pi = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ 1 & 4 & 6 & 2 & 8 & 9 & 7 & 5 & 3 \end{pmatrix}$

is $\begin{pmatrix} 1 \end{pmatrix} \begin{pmatrix} 2 & 4 \end{pmatrix} \begin{pmatrix} 3 & 6 & 9 \end{pmatrix} \begin{pmatrix} 5 & 8 \end{pmatrix}$. It is usual to omit the cycles of length $1$, those integers fixed by $\pi$, and so $\pi$ is abbreviated to $\begin{pmatrix} 2 & 4 \end{pmatrix} \begin{pmatrix} 3 & 6 & 9 \end{pmatrix} \begin{pmatrix} 5 & 8 \end{pmatrix}$.

Correction

The cycle decomposition given above includes the $\begin{pmatrix} 1 \end{pmatrix}$, but for some reason omits the $\begin{pmatrix} 7 \end{pmatrix}$.

If you are going to include the one, then you also need to include the other.

Sources

• 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $9$: Permutations: Example $9.6$