TT's Maths Thread

[TrueTears]

ohhh right i get it, it's pretty easy to think about it intuitively.

Since if we change the order of the factors in the direct product, then the 'names' of each element in the direct product is just simply changed by a permutation of the components in each of the n-tuples.

[TrueTears]

[IMG]http://img849.imageshack.us/img849/8864/factorgroups.jpg[/img]

Eh is there a mistake in the theorem?

If we let and

then according to 15.8 Theorem we have

So that means

However look at Example 15.7 It SHOULD be  

In other words shouldn't the theorem state not the other way around?


Also another question:

We know that every cyclic group is abelian.

Also all subgroups of abelian groups are normal.

Does that imply if a group G is cyclic then it is abelian, now since all groups are a subgroup of itself, then G (being an abelian group) must also be a normal group?


[IMG]http://img189.imageshack.us/img189/67/factorcomputation.jpg[/img]

I completely lost on what this example is trying to explain.

First of all, 'computing' a factor group just means finding which direct product of groups is the factor group is isomorphic to.

Now we know that and so

There since this factor group is finitely generated and it is also abelian then it is isomorphic to a direct product of either of 3 forms below:







Now this is where I am lost. What does the example mean by "the first factor of is left alone" ? Why do you leave it "alone"?

Then "The factor ... is collapsed by a subgroup of order 3" What the hell does it mean that the factor is "collapsed by a subgroup"?

And how does "collapsing" yield a "factor group in the second factor of order 2 that must be isomorphic to " ?

Basically can someone please translate what the hell this example is trying to say into english? Or shall i say, less ambiguous words such as "collapsing"?

that would be great since ive been trying to do this question myself, i've listed out the 8 elements in the factor group and reached the same conclusion as the question itself did, however listing out 8 elements is certainly very troublesome and the example seemed to skip listing the out the whole factor group and find what it's isomorphic to. however I just cant seem to understand the logic behind the example at all!

[kamil9876]

As to the theorem: Yeah you're right on how it should be. In the author's defence his statement is all true except for the last sentence (he should've defined differently, but with the way he has defined his second last sentence is true).

Also another question:

We know that every cyclic group is abelian.

Also all subgroups of abelian groups are normal.

Does that imply if a group G is cyclic then it is abelian, now since all groups are a subgroup of itself, then G (being an abelian group) must also be a normal group?

G is always a normal subgroup of G (whole group is always a normal subgroup of itself), even if it is not abelian. The only interesting thing is whether the smaller subgroups are normal. BTW: there is no such thing as a "Normal Group" only "Normal Subgroup in G", indeed there are examples of groups that are normal subgroups of some group G, but not normal subgroups in another group G'.



Hrmm I find it very weird that you already know the classification theorem for finitely generated abelian groups (it's not really for beginners and you probably won't be able to prove it until you learn much more algebra) and it is not very useful for you here. Here is a longer way of doing it but perhaps it will give you insight:

Basically we want to find the H-Cosets of G.



So if is an arbitrary element of G then the Coset that is in is:



So this tells you something nice: that and are in the same Coset IFF both and .
Now how does this help us? Well you can now see that the Coset that (x,y) is in depends only on the value of x modulo 4, and the value of y modulo 2. Hence the isomorphism.

Hope that builds some intuition.

[TrueTears]

Thanks for the help kamil but i still dont quite get the last question.

Don't you mean x = x' ? why mod 4 the 'a' doesn't change in any coset.

Also <- I get that but how does that relate to the isomorphism? can you break down their explanation for me?

Cheers


Actually nvm, after thinking about it over and over it I think I get it.

For the first value in (x,y) we can have anything from the set \{0, 1, 2, 3\} since any single ordered pair (x,y) with any one of the values for x from that set will be in a different coset. As for the second value y in (x,y) we can have anything from the set \{0, 1\}. Thus the isomorphism to Z4 x Z2

[kamil9876]

You're right, in fact i had "x=x'" but then i changed it coz I think it might be confusing.

[TrueTears]

cheers man, your explanations were great, understand it much better now!

[TrueTears]

Hmmm, I've got another Q...

Compute the factor group

Now using your logic kamil, if we have some

then the H cosets of G is

Which means (x,y) and (x', y') are in the same coset iff both AND

So and

Thus

But clearly that's wrong, since when in fact we know that   !!!!

What went wrong :S

[kamil9876]

Which means (x,y) and (x', y') are in the same coset iff both AND

This is what went wrong, it is true that if (x,y) and (x',y') are in the same Coset then AND , but the converse is not true (example look at so clearly you see that (0,0) and (0,3) are not in the same Coset yet they satisfy the congruences).  It is not as nice as the previous example.

There is a way of doing it but it is probably beyond the scope of where you are at. The other way of doing is to do what you noticed, that the order is 12 and then use the fundamental theorem of finite abelian groups and try to see which one it is isomorphic too.


Though a good way of illustrating the reasoning I used in the previous example is with this slight variation, what is:

isomorphic to?

[TrueTears]

ohhhh i get it, true.

now in regards to your question, isomorphic to   haha, however i am just wondering how do u know when the congruence 'technique' fails and when it doesnt?

Is it because and we see that (x,y) and (x',y') is in the same coset IFF the x values differ by STRICTLY 2 and y values differ by STRICTLY 3. So if we wanted (0,0) to be the "representative" of (0,3)H, (2,0)H and (2,3)H, it can only "represent" (2,3)H in the sense that they are the same coset. However for (0,3)H, (2,0)H, these 2 are clearly different from (0,0)H, they are NOT the same coset.

Now so any (x,y) that satisfies the congruences AND must be in the same coset and if they are in the same coset they must satisfy the congruence.

[kamil9876]

Ok good seems that you are understanding, but try this now:


What is isomorphic to?

How about ?

[TrueTears]

[IMG]http://img35.imageshack.us/img35/957/47409814.jpg[/img]

Just wondering, the result of the dimension of the nullspace or rowspace of a matrix does not change regardless of what field it is over correct?

[IMG]http://img97.imageshack.us/img97/6063/linearmap.jpg[/img]

Also how do I do this q?

cheers

[kamil9876]

Just wondering, the result of the dimension of the nullspace or rowspace of a matrix does not change regardless of what field it is over correct?

Technically that doesn't make sense, you should look at the in as a completely different animal to the in it's just (almost) a symbolic accident that they look the same. e.g your question may not make sense if for example there was a in your matrix or say you were comparing matrices over two completely different fields (example: Field of rational functions on compared with just )

So in general your question doesn't make sense, you shouldn't mix fields (unless you've got some morphism or comparing a field to a subfield or something else that makes sense).

However perhaps it is interesting to restrict this question to the special case of 0,1 matrices (perhaps something to think about when you have spare time)

9. This looks like a very standard first year exercise, just use the definitions.

[TrueTears]

no, actually i think i phrased my question badly, what i mean is, is the underlying "theorems/rules" regarding vector spaces all true regardless or watever field.

eg, say we are reducing a matrix to row echelon form over the field Z_3, does that mean we are still trying to make all the pivots as 1, etc etc??

and yeah i know the definition, but how to apply it? ><

[kamil9876]

ahh ok, Yes in fact you can go over each proof and notice that the only property of real numbers that are used in most of them is the the simple algebra like division, multiplication etc. and hence the same proofs work for fields in general. (some though are special like the fact that every linear transformation has an eigenvalue only works in fields like the complex numbers, though you probably won't be worried by this)

well, take an arbitrary element a+bx, now take an arbitrary scalar , what is , what is ? are they the same? if so then scalar multiplication is satisfied, now we need to do the same for addition...

[TrueTears]

thanks kamil!

yes that's what i dont get, what IS