On 27.08.2012 22:52, David Joyner wrote:
On Mon, Aug 27, 2012 at 2:26 PM, Tom Bachmann<e_mc...@web.de> wrote:
I'm not sure if I'm helping, but I'm also not sure if I understand what you
are saying.
Let us fix a set X we are considering the permutation group of, below I will
take X = {1, 2, 3, 4, 5}. A permutation of X is by definition a bijective
function f:X->X. It is specified uniquely by providing the image of every
element. We can write this in the short form
[f(1), f(2), f(3), f(4), f(5)]. In this way, every permutation is
represented by an array of constant size.
Now let's talk about cycles. By definition, a cycle is an ordered subset of
X. In general, the cycle (a_1 ... a_n) represents the unique permutation f
with f(a_1) = a_2, ..., f(a_n) = a_1, and f(x) = x for all x not in {a_1,
.., a_n}. For example, the cycle (1 2 3) denotes the permutation [2, 3, 1,
4, 5].
We can identify the set of cycles with a subset of the set of permutations.
Now let's consider composition. There are two schools of thought. Let me
write * for "ordinary" (where I come from) composition, and . for "weird"
composition. By definition, if f, g are permutations, then the permutation
f*g is the unique mapping such that (f*g)(x) = f(g(x)) ["apply right to
left"], whereas (f.g)(x) = g(f(x)) ["apply left to right"]. [*]
This is correct *as functions*.
If you do this *as permutations* (which *act* on a set and are not
just functions),
then you want the permutations to form an "action" on the set.
The way you want to do it, they aren't an action:
if g1 = (1,2) and g2 = (2,4,5) then (g1*g2)(2) \not= g1(g2(2)):
I'm not sure I get this. Since you are much more knowledgable than me
maybe I should just shut up, but as far as I know, a (left) action of a
group G on a set S is a map phi: GxS -> S, where I shall write g(s) for
phi(g, s) if g in G and s in S. Then the axioms are:
- e(s) = s for all s
- g(h(s)) = (gh)(s) for all s in S, g, h in G
If we let Sym(S) denote the symmetric group on S, by which I mean the
set of all bijections S->S with the "ordinary" (left, or *) composition,
then I believe it is easy to check that an action is the same as a
homomorphism from G to Sym(S). Anyway, we *obviously* want (g1*g2)(2) =
g1(g2(2)).
Wikipedia agrees with this definition of (left) action, btw.
We can similarly define a right action. For example, if Sym'(S) denotes
Sym(S) with the other composition law, then a right action is a
homomorphism G->Sym'(S). More explicitly, it is a map phi:SxG -> S, such
that
- (s)e = s for all s in S
- ((s)g)h = (s)(gh) for all s, g, h.
Where am I wrong?
I think this all just boils down to the fact that for actual
computations with permutations, it is more convenient to work with the
"." operation (i.e. composition on the left) than with the "*"
operation, essentially because we read from left to right. Much of the
confusion probably stems from the fact that these conventions are
completely equivalent (Taking opposite groups interchanges left and
right actions, and the opposite group is naturally isomorphis to the
group we started with).
Thanks,
Tom
sage: G = SymmetricGroup(5)
sage: g1 = G([(1,2)])
sage: g2 = G([(2,4,5)])
sage: g2(2)
4
sage: g3 = g1*g2
sage: a = g2(2); a
4
sage: b = g1(a); b
4
sage: g3(2)
1
It is very important for many people that the group
of permutations yields a group action on the set.
Observe now that any permutation can be written uniquely (up to ordering) as
...
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