I will not give a proof of this. However, the intuition is that since each
product term βi is
disjoint with all others, they “
act independently” and so the identity will be reached on common multiples.
Things I don’t have:
A good explanation for this
Complete confidence that this is correct
Things I do have:
Relative confidence that this is correcy
Let us
fix σ to be the identity 12-
cycle. So
n=12 and
σ looks like:
Now consider
σ3. We can characterize
σ as taking
x to
x+1 (bar
cyclic behaviour). As such,
σ3 will take
x to
x+3 (bar
cyclic behaviour). This looks like:
Note that
σ is composed of 3 distinct 4-
cycles. Convince yourself this makes sense, and naturally gives rise to the conjecture that if
k∣12 then
σ12 will be comprised of
(12/k)-
cycles.
Now consider
σ8:
Note that
σ8 is comprised of 4 distinct 3-
cycles. Convince yourself this makes sense, and naturally gives rise to the conjecture that the cycle decomposition of
σk consists of
cycles of length equal to the
additive order of
k modulo 12.
Generalizing, we have that if
σ is an
n-
cycle, then
σk comprises
ℓ-
cycles, where
ℓ=LCM(n,k)/k is the
additive order of
k modulo n.
Now consider . We can characterize as taking to (bar cyclic behaviour). As such, will take to (bar cyclic behaviour). This looks like:
Note that is composed of 3 distinct 4-cycles. Convince yourself this makes sense, and naturally gives rise to the conjecture that if then will be comprised of -cycles.
Now consider :
Note that is comprised of 4 distinct 3-cycles. Convince yourself this makes sense, and naturally gives rise to the conjecture that the cycle decomposition of consists of cycles of length equal to the additive order of modulo .
Generalizing, we have that if is an -cycle, then comprises -cycles, where is the additive order of modulo .