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**Deveno** the type of function you are talking about is called a 1-cycle. all 1-cycles are equal to the identity:

e = (1) = (1)(2) = (1)(2)(3), etc.

for example, if f(k) = k, and f(x) = x for all other x except k, then f(x) = x for ALL x.

since 1-cycles are trivial, they are omitted in cycle decompositions (just as we typically omit e in products, or 0 in sums).

transpositions are 2-cycles: they are of the form (i j) where i ≠ j. in fact, it is traditional to write them in the form (i j), where i < j (this is handy for making sure lists of transpositions do not contain duplicates).

this rule is helpful to keep in mind:

for a k-cycle, if k is odd, the k-cycle is an EVEN permutation: 1 cycles, 3-cycles, 5-cycles, etc. are even permutations.

if k is even, a k-cycles is ODD. so 2-cycles, 4-cycles, 6-cycles, etc. are all odd permutations.

quick way to keep it straight:

e = (1 2)(1 2) <---even (2 transpositions)

(1 2) <---odd (1 transposition)

if S is a set with 1 element, there is only ONE possible bijection S-->S, the identity. there ARE no 2-cycles on S, and in this one special case, we have that the even permutations are ALL the permutations, so every permutation can only be written as a product of 0 transpositions, because there AREN'T any transpositions. some of the "general theorems" for permutations groups "break down" at the lower ends:

S_{1} has only even permutations, instead of half-and-half.

S_{1} and S_{2} are abelian (and cyclic!). A_{2} is not generated by 3-cycles (it has none). the automorphism group of S_{2} is trivial.

S_{3}: A_{3} is cyclic (and thus abelian). all of the automorphisms of A_{3} are outer (the inner automorphisms are trivial, since A_{3} is abelian)

S_{4}: A_{4} is not simple.

S_{6}: has an outer automorphism (an automorphism that is not of the form x-->gxg^{-1}) <--this is a strange thing.

these anomalies disappear as long as n is large enough:

S_{n},A_{n} are non-abelian

|A_{n}| = (1/2)|S_{n}|

A_{n} is generated by 3-cycles

A_{n} is simple

Aut(S_{n}) = Aut(A_{n}) = S_{n}.

so when considering symmetric (permutation) groups, it pays to make sure that "n is big enough" (you might have an interesting counter-example).