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Let $S_n$ denote the symmetric group on $n$ letters.
Let $\pi, \rho \in S_n$ both be permutations on $S_n$.
Then $\pi$ and $\rho$ are disjoint if and only if:
- $(1): \quad i \notin \Fix \pi \implies i \in \Fix \rho$
- $(2): \quad i \notin \Fix \rho \implies i \in \Fix \pi$
That is, each element moved by $\pi$ is fixed by $\rho$ and (equivalently) each element moved by $\rho$ is fixed by $\pi$.
That is, if and only if their supports are disjoint sets.
We may say that:
- $\pi$ is disjoint from $\rho$
- $\rho$ is disjoint from $\pi$
- $\pi$ and $\rho$ are (mutually) disjoint.
Note of course that it is perfectly possible for $i \in \Fix \pi$ and also $i \in \Fix \rho$, that is, there may well be elements fixed by more than one of a pair of disjoint permutations.
- 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $2$: The Symmetric Groups: $\S 79$
- 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $9$: Permutations: Definition $9.4$