Combinatorics Through Guided Discovery, 2004a

6.1. Permutation Groups 113
Problem 274. Find the cycles of the permutation
( )
1 2 3 4 5 6 7 8 9
.
3 4 6 2 9 7 1 5 8
Problem 275. If two cycles of σ have an element in common, what can we
say about them?
Problem 275 leads almost immediately to the following theorem.
Theorem 6.1.6. for each permutation σ of a set S, there is a unique partition of S each of
whose blocks is the set of elements of a cycle of σ.
More informally, we may say that every permutation partitions its domain into
disjoint cycles. We call the set of cycles of a permutation the cycle decomposition
of the permutation. Since the cycles of a permutation σ tell us σ(x) for every x in
the domain of σ, the cycle decomposition of a permutation completely determines
the permutation. Using our informal language, we can express this idea in the
following corollary to Theorem 6.1.6.
Corollary 6.1.7. Every partition of a set S into cycles determins a unique permutation of
S.
Problem 276. Prove Theorem 6.1.6.
In Problems 273 and Problem 274 you found the cycle decomposition of typical
elements of the group D 4 and of the permutation
( 1 2 3 4 5 6 7 8
) 9
3 4 6 2 9 7 1 5 8
The group of all rotations of the square is simply the set of the four powers
of the cycle ρ =(1234). for this reason it is called a cyclic group3 and is often
denoted by C 4 . Similarly, the rotation group of an n-gon is usually denoted C n .
⇒
Problem 277. Write a recurrence for the number c(k, n) for the number of
permutations of [k] that have exactly n cycles, including 1-cycles. Use it
to write a table of c(k, n) for k between 1 and 7 inclusive. Can you find a
relationship between c(k, n) and any of the other families of special numbers
such as binomial coefficients, Stirling numbers, Lah numbers, etc. we have
studied? If you find such a relationship, prove you are right. (h)
3The phrace cyclic group applies in a more general (but similar) situation. Namely the set of all
powers of any member of a group is called a cyclic group.