Well, 6 = 2 * 3. So, if you could write down a permutation that has two commuting parts, one of order 2 and one of order 3, what's the order of their product?

I would guess that IF I could find and write down a permutation with two commuting parts it's product would have an order of six(?)

However, I don't know how to find a permutation that has two commuting parts, one of order 2 and one of order 3. And I don't know why if I found a permutation that has two commuting parts, one of order 2 and one of order 3, why it would have an order of six.

Thanks for your help anyway Hurkyl. I appreciate it. But I don't think I have an aptitude for this course.

Let x and y be commuting elements of a multiplicative group, and x has order 2 and y has order 3.

Let z = xy and write the first 12 powers of z in the form:
z^k = x^i * y^j

And simplify as much as possible.

Exercise 2:

Determine a sufficient condition for two permutations to commute by randomly picking cycles and multiplying them. (You can probably figure it out after a few dozen examples)

Alright. I have pretty much given up on this course (I am thinking about dropping it), but I will give the exercise a shot (since I do not like giving up). I will approach this exercise by giving some elementry definitions to see if I understand the question(s).

Part 1

Multiplicative Group:

A non-empty set in which properties of a group apply and whose binary operation is multiplication.

Commutative Property of the group:

Given x, y which are elements of the Group G, x o y = y o x in which the product of x,y is an element of the group G.

Order of an element:

The order of an element a which is an element of G is the least positive integer n for which a^{n} = u (the identity element).

So from what I know (which is not much) I think that if x is of order 2 then

x^{2} = u

and

y^{3} = u.

Given that z = xy

then the first 12 powers of z are:

z1 = x y
z2 = x2 y2 = u y2 = y2
z3 = x3 y3 = x2 x u = u x u = x
z4 = x4 y4 = (x2)2 y y3 = (u)2 y u = y
z5 = x5 y5 = (x2)2 x y2 y3 = u x y2 u = x y2
z6 = x6 y6 = (x2)3 (y3)2 = (u)3 (u)2 = u u = u
z7 = x7 y7 = (x2)3 x (y3)2 y = (u)3 x (y)2 y = x y
z8 = x8 y8 = (x2)4 (y3)2 y2 = (u)4 (u)2 y2 = u u y2 = y2
z9 = x9 y9 = (x2)4 x (y3)3 = (u)4 x (u)3 = u x u = x
z10 = x10 y10 = (x2)5 (y3)3 y = (u)5 (u)3 y = u u y = y
z11 = x11 y11 = (x2)5 x (y3)3 y2 = (u)5 x (u)3 y2 = u x u y2 = x y2
z12 = x12 y12 = (x2)6 (y3)4 = (u)6 (u)4 = u u = u

Summary

z^{1} = xy
z^{2} = y^{2}
z^{3} = x
z^{4} = y
z^{5} = xy^{2}
z^{6} = u
z^{7} = xy
z^{8} = y^{2}
z^{9} = x
z^{10} = y
z^{11} = xy^{2}
z^{12} = u

This is what I have so far. Part 2 in my next reply.

Yes. I think I got this. Is this because the Group is finite?

I am not sure what you are saying here:

You are referring to my loose definition of the Commutative Property of the group correct?

Could you explain the second point. If you feel you are simplifying it too much you probably are not.

BTW NateTG, as far as I know we have yet to talk about isomorphism. If we had, I wasn't aware of it. But I don't recall a lecture on isomorphism as of yet.

I am not sure of what to take permutations. Do I simply take any group, say S_{3}, work out the permutations, then find a condition in which the permutations of S_{3} commute?

BTW, after doing the first exercise, I noticed that the elements produced looked like the dihedral group D_{6}. I'm not sure what that would mean in the context of the excerises you gave me Hurkyl.

Another note: I am very appreciative of both the efforts of Hurkyl and NateTG. I know it can be frustrating trying to explain certain concepts to me, but if one has the patience, I will try my best to figure things out. Thankyou.

You might try a few different permutation groups, just in case S_{3} isn't representative of the general case. Varying the types of things over which you search is good practice in general.

And, it turns out, that it is important in this case; S_{3} really isn't a very good representative of the general case. While you can find permutations that commute in S_{3}, it probably won't be enough to suggest the general result. I would suggest considering permutations over 4 or more objects.

Almost; it's because z has finite order.

When studying complex numbers, did you ever have to do the fun problem of computing the value of i^2003? The trick is to decompose the exponent as:

The same principle is at work in groups. If x has finite order, that means that x^k = 1 for some nonzero value of k. Then, if we want to know what x^n is for n >= k (or n < 0), we can apply the above trick to reduce the problem to computing x^m for some m in the range [0, k).

And, incidentally, you may or may not have seen the entire group. In particular, you have only touched the cyclic subgroup that is generated by z; in principle, there could be just about anything else in the rest of the group! For example, the group itself could be nonabelian, even though x and y commute. "Nonabelian group" just says that some elements don't commute, it doesn't say all elements don't commute.

I did not intend to point you towards D_{6}. I'm tired and my algebra book isn't handy so I may be wrong, but isn't D_{6} nonabelian? If so, then TMK the (sub)group I had you generate only bears a superficial resemblance to D_{6}.

But it's a good observation nonetheless. It is often useful to express a group in terms of its generators.

I just finished reading your last post Hurkyl. It might take me a little time to absorb what you are saying. But I will address some points at this time.

No I have not. Strangely enough, the last time I remember complex numbers was in high school which was many years ago. However I do see what you are suggesting.

Right. I understand this.

I believe it is. I am not sure but can the permutations of D6 be expressed as the symmetries of an equilateral triangle? If so, I don't think that D6 is abelian.

I started out with S_{5} and picked two permutations of S_{5} of the order 3 and 2:

(1 2 3)(4 5)

I see that

(1 2 3)(4 5) = (4 5)(1 2 3)

Therefore these two cycles commute(?) Is this what you meant?

It seems to me from the two permutations that I picked, the product of two disjoint cycles will always commute. Is this the condition that you were inferring in the second exercise?

IIRC stands for something like "If I recall correctly"

Hurkyl: Disjoint cycles is sufficient not necessary.
For example:
a=(1 2 3)(4 5)
and
b=(1 2 3)

commute (since b=a^{4}), but are not disjoint cycles.

But, thanks to you, I'm going to have to go home, and see if I can categorize all commutative subgroups of S_{i}.

wubie:

D_{6} is not abelian (commutative).

any finite group generated by a single element a is going to be of the form
G = 1 a a^{2} a^{3} a^{4} a^{5}...
and since a^{x}*a^{y}=a^{x+y}=a^{y}*a^{x}
any group generated by a single element will be commutative.

So, if you generated a subgroup using a single element and got D_{6} you made a mistake somewhere.