My last course M208 Pure Mathematics, lightly touched upon complex analysis, only as far as arithmetic and basic manipulation of complex numbers, is concerned. My current plans include completing the Open University course M337, Complex Analysis, in 2014. As such, I am planning a very slow study of the course, prior to starting it, next year.
When I say slow, I mean glacial.
I am in no rush whatever. So, alongside my Groups, and number theory preparation over the next few months; I am planning on studying a page or two from the first couple of units in M337, every few days.
I am hoping that the basic concepts will then slowly simmer in my unconscious mind, over the next year, prior to starting the course, proper, next October.
I don't have any number theory books to pre-study, prior to October this year, other than my trusty copy of 'How to Prove it' by Velleman. It contains some nice set theory along with some logic and practise at working out basic proofs. I would recommend that book to anyone who wants extra practice at creating proofs from first principles, in a systematic, logical way.
Thursday, 4 July 2013
Monday, 1 July 2013
Conjugate Groups, Abelian Groups and Equivalence Relations.
Whilst conducting some revision of my material from the Open University course M208 Pure Mathematics, today; I noticed something quite satisfying, within one of the proofs of that book, (P.16, Unit GTB1, Conjugacy).
When I studied this material, last year ( I think it was around May/June time?), I had glossed over many of the proofs in the book, and leaned some of the material on face value. This strategy served me well, such that I had enough time to practise questions and gain an overview of each topic, to allow me to pass the course.
However, I am now going through some of those longer proofs, in some very fine detail. Why? Well, I want to enrich the material that I learnt last year; I want to gain more experience of reading proofs, so that I can begin to write some of my own; and It is providing good, previously missed connections, between the material that I have already studied.
This leads me onto the chosen proof of today's post:
'In any group G, the relation 'is conjugate to, denoted by ~, is an equivalence relation on the set of elements of G'.
I just want to highlight one small part of that proof. For those interested in seeing the entire proof; let me know, and I will happily reproduce it, once I sort out my LaTex editor, next week. Alternatively, any search on the web, will probably locate it somewhere.
The part that caught my attention, was in the section of the proof ' E2 SYMMETRIC'.
This is where we are trying to prove that for all elements x,y of G; then if x~y , it follows that y~x.
The strategy used is one where the equivalence relation x~y is denoted as y = gxg^-1. And the proof uses some basic manipulations to show that this is equivalent to x = g^-1 y (g^-1)^-1; or simply put, y~x.
The first step seems simple enough:
If y = gxg^-1, then g^-1 y g = g^-1(gxg^-1)g
[All we have done here is multiply both sides of the expression by g^-1 on the left and g on the right, of each monomial part].
It then follows that through the associative property for groups; we can further manipulate the expression to give:
g^-1 y g = g^-1(gxg^-1)g
= (g^-1 g) x (g^-1 g)
= exe [e is the identity element of Group G]
= x.
Now, this particular proof takes one final step, to show that this group is symmetric.
Why? you might ask.
Well, what we have been left with, is x = g^-1 y g. The problem with this expression, is that although it looks very similar to the expression that we need to find; you will see that the g^-1 and the g are actually reversed (they are the wrong way around).
This matters, because the proof actually makes no mention of whether the group G, is an Abelien group, or not.
An Abelian group is one where the group has the additional - commutative property - such that:
For all g1, g2 elements of G; g1 o g2 = g2 o g1.
So, the order of the elements, before they are operated on, does not matter in an Abelian group. In an Abelian group, the expression x = g y g^-1 and x = g^-1 y g, would be equivalent.
Hence, because our proof makes no mention of the group G, being Abelian, we can assume that we need a final stage to this proof, in order to complete it. We need to 'swap the two conjugating elements back around.
This act of 'swapping around' the conjugating elements g and g^-1, makes the whole proof work for a non- Abelian group, as well as an Abelian group.
Thus, we need this final step:
Since
(g^-1)^-1 = g, (this is basic manipulation of indices);
substituting this into our expression, we have
g^-1 y (g^-1)^-1 = x.
Here, all we have done is restated the expression, in a slightly different way, but it is still essentially the same form of expression.
And finally, we have shown that g^-1 conjugates y to x, so y is conjugate to x. It doesn't matter that the element conjugating is actually g^-1, since as long as we have the inverse element, (g^-1)^-1, on the right of the conjugated element, it all still makes sense. All we ever need, is at least one element of G, to conjugate x~y and y~x, in order for the group to be symmetric, in this way. It matters not whether that element is g or g^-1 or or if you want to go crazy, ((g^-1)^-1)^-1; or any other random element of G, for that matter.
Therefore, since y is conjugate to x and x is conjugate to y; the Group G has the symmetric property.
I'm sorry if my lack of LaTex editor made this post difficult to follow. I will endeavour to fix that little problem, by next week.
So, by delving into this particular proof, in detail, I have been able to subtly shift my own understanding of Groups; by naturally combining an earlier concept (Abelian Groups), with the concepts of equivalence relations and conjugacy.
Neat!
[Caveat: I am a student, rather than an authority on Group Theory; so if you notice an error in my reasoning / preceeding proof; please let me know, so that I may correct it.]
When I studied this material, last year ( I think it was around May/June time?), I had glossed over many of the proofs in the book, and leaned some of the material on face value. This strategy served me well, such that I had enough time to practise questions and gain an overview of each topic, to allow me to pass the course.
However, I am now going through some of those longer proofs, in some very fine detail. Why? Well, I want to enrich the material that I learnt last year; I want to gain more experience of reading proofs, so that I can begin to write some of my own; and It is providing good, previously missed connections, between the material that I have already studied.
This leads me onto the chosen proof of today's post:
'In any group G, the relation 'is conjugate to, denoted by ~, is an equivalence relation on the set of elements of G'.
I just want to highlight one small part of that proof. For those interested in seeing the entire proof; let me know, and I will happily reproduce it, once I sort out my LaTex editor, next week. Alternatively, any search on the web, will probably locate it somewhere.
The part that caught my attention, was in the section of the proof ' E2 SYMMETRIC'.
This is where we are trying to prove that for all elements x,y of G; then if x~y , it follows that y~x.
The strategy used is one where the equivalence relation x~y is denoted as y = gxg^-1. And the proof uses some basic manipulations to show that this is equivalent to x = g^-1 y (g^-1)^-1; or simply put, y~x.
The first step seems simple enough:
If y = gxg^-1, then g^-1 y g = g^-1(gxg^-1)g
[All we have done here is multiply both sides of the expression by g^-1 on the left and g on the right, of each monomial part].
It then follows that through the associative property for groups; we can further manipulate the expression to give:
g^-1 y g = g^-1(gxg^-1)g
= (g^-1 g) x (g^-1 g)
= exe [e is the identity element of Group G]
= x.
Now, this particular proof takes one final step, to show that this group is symmetric.
Why? you might ask.
Well, what we have been left with, is x = g^-1 y g. The problem with this expression, is that although it looks very similar to the expression that we need to find; you will see that the g^-1 and the g are actually reversed (they are the wrong way around).
This matters, because the proof actually makes no mention of whether the group G, is an Abelien group, or not.
An Abelian group is one where the group has the additional - commutative property - such that:
For all g1, g2 elements of G; g1 o g2 = g2 o g1.
So, the order of the elements, before they are operated on, does not matter in an Abelian group. In an Abelian group, the expression x = g y g^-1 and x = g^-1 y g, would be equivalent.
Hence, because our proof makes no mention of the group G, being Abelian, we can assume that we need a final stage to this proof, in order to complete it. We need to 'swap the two conjugating elements back around.
This act of 'swapping around' the conjugating elements g and g^-1, makes the whole proof work for a non- Abelian group, as well as an Abelian group.
Thus, we need this final step:
Since
(g^-1)^-1 = g, (this is basic manipulation of indices);
substituting this into our expression, we have
g^-1 y (g^-1)^-1 = x.
Here, all we have done is restated the expression, in a slightly different way, but it is still essentially the same form of expression.
And finally, we have shown that g^-1 conjugates y to x, so y is conjugate to x. It doesn't matter that the element conjugating is actually g^-1, since as long as we have the inverse element, (g^-1)^-1, on the right of the conjugated element, it all still makes sense. All we ever need, is at least one element of G, to conjugate x~y and y~x, in order for the group to be symmetric, in this way. It matters not whether that element is g or g^-1 or or if you want to go crazy, ((g^-1)^-1)^-1; or any other random element of G, for that matter.
Therefore, since y is conjugate to x and x is conjugate to y; the Group G has the symmetric property.
I'm sorry if my lack of LaTex editor made this post difficult to follow. I will endeavour to fix that little problem, by next week.
So, by delving into this particular proof, in detail, I have been able to subtly shift my own understanding of Groups; by naturally combining an earlier concept (Abelian Groups), with the concepts of equivalence relations and conjugacy.
Neat!
[Caveat: I am a student, rather than an authority on Group Theory; so if you notice an error in my reasoning / preceeding proof; please let me know, so that I may correct it.]
Saturday, 8 June 2013
A Group Theory diversion
I was re-reading unit GTB1 tonight, of the Open University course M208 Pure Mathematics.
I just thought I would note down something that I have noticed, but which I have only just realized, on reviewing the material, this year.
Just a bit of background; M208 was the first pure mathematics course that I completed, and part of it contained sections on Group Theory. As any diligent group theorist would do, I began by learning the Axoims that lead to the definition of a 'Group', from a set of any given elements.
Now, one of those axioms is the property of associativity. That is:
For all g1, g2, g3 elements of G,
g1 o (g2 o g3) = (g1 o g2) o g3.
(the numbers are supposed to be subscript, but my Latex editor is currently 'up the Shoot'.)
Whilst I was studying the course, I had assumed that a group probably needed three elements in a set, in order to meet this axiom's criteria.
I say I assumed this, however, to be honest, I had never actually given it any real thought.
So, on revision of this material in 'slow time'; this is the sort of question, that seems to be popping into my head. A promising sign, in my eyes.
So, I glanced at the issue tonight and, of course, I quickly concluded that the axioms do not state that you need three elements in a set, in order to meet their requirements.
For example, if I have a set {1}; this could be a group under multiplication. Why? Well, it meets the axioms of Closure, Identity, Inverses and Associativity; where g1, g2 and g3 are all the set element {1}. There are no axioms that say that the elements in a set must be different, or that the elements must be distinct.
Why? Well, the axioms do not go into numbers of elements, or other topics, for that matter. The way that they are written, means that they do not assume that there is more than one element or even that the elements actually exist.
Having said that the Axioms certainly do not state, that there should be a none existent element, or that there should be a group that exists, which contains less than three elements.
So, the axioms certainly don't prove that there is a set of less than three elements that can form a group.
Logical, but confusing, I think.
I just thought I would note down something that I have noticed, but which I have only just realized, on reviewing the material, this year.
Just a bit of background; M208 was the first pure mathematics course that I completed, and part of it contained sections on Group Theory. As any diligent group theorist would do, I began by learning the Axoims that lead to the definition of a 'Group', from a set of any given elements.
Now, one of those axioms is the property of associativity. That is:
For all g1, g2, g3 elements of G,
g1 o (g2 o g3) = (g1 o g2) o g3.
(the numbers are supposed to be subscript, but my Latex editor is currently 'up the Shoot'.)
Whilst I was studying the course, I had assumed that a group probably needed three elements in a set, in order to meet this axiom's criteria.
I say I assumed this, however, to be honest, I had never actually given it any real thought.
So, on revision of this material in 'slow time'; this is the sort of question, that seems to be popping into my head. A promising sign, in my eyes.
So, I glanced at the issue tonight and, of course, I quickly concluded that the axioms do not state that you need three elements in a set, in order to meet their requirements.
For example, if I have a set {1}; this could be a group under multiplication. Why? Well, it meets the axioms of Closure, Identity, Inverses and Associativity; where g1, g2 and g3 are all the set element {1}. There are no axioms that say that the elements in a set must be different, or that the elements must be distinct.
Why? Well, the axioms do not go into numbers of elements, or other topics, for that matter. The way that they are written, means that they do not assume that there is more than one element or even that the elements actually exist.
Having said that the Axioms certainly do not state, that there should be a none existent element, or that there should be a group that exists, which contains less than three elements.
So, the axioms certainly don't prove that there is a set of less than three elements that can form a group.
Logical, but confusing, I think.
Wednesday, 5 June 2013
Pure Mathematics
Good lord, has it really been 8 weeks since my last post? I've had a bit of a break from all things studying for the last two months, as I had to defer my O.U module, Astrophysics, whilst I recovered from some health issues.
Before I left, I managed a couple of TMA's and also some of the research elements; and whilst I am sad to leave this interesting subject, I can't help but feel that it is probably actually for the best; as I can now, truly concentrate, on my new found love affair with pure mathematics. My degree profile will also probably look a a little more focussed, since it will contain all level 3 maths modules.
Also, in a strange twist of fate; whilst I have just effectively added another year to my level 3 study schedule; I hadn't realized that by doing Astrophysics, I would have been unable to complete Complex analysis, before moving onto the MSc. This would have been quite foolhardy, in my opinion. So, this could be a blessing in disguise.
It now means that I will take two pure maths modules in October, as planned (number theory and logic, with groups and geometry), but because I will now need to study more level 3 modules in 2014, I can now add complex analysis, to my quiver. I had planned to self study the complex analysis books, that were kindly sent to my by Chris F; but I think that Chris was quite right when he suggested, some time ago, that any future sponsor for post-graduate maths research / work, would question why one did not have such an important module.
Anyway, the study break has given me a good opportunity to start slowly re-reading some of my O.U module Units from M208, Pure Mathematics, which I completed last year. I have forgotten a fair bit, and there was some of it that I never entirely understood, the first time around (epsilon-delta, I'm looking at you!)
So, I have a rather lazy summer ahead, with some hopefully enjoyable revision of pure maths. I also plan to dust off my copies of Hardy's - Course in Pure Mathematics; Brannan's - Geometry and Spivak's - Calculus (Which should really be called Real Analysis, judging by the content of the book).
Before I left, I managed a couple of TMA's and also some of the research elements; and whilst I am sad to leave this interesting subject, I can't help but feel that it is probably actually for the best; as I can now, truly concentrate, on my new found love affair with pure mathematics. My degree profile will also probably look a a little more focussed, since it will contain all level 3 maths modules.
Also, in a strange twist of fate; whilst I have just effectively added another year to my level 3 study schedule; I hadn't realized that by doing Astrophysics, I would have been unable to complete Complex analysis, before moving onto the MSc. This would have been quite foolhardy, in my opinion. So, this could be a blessing in disguise.
It now means that I will take two pure maths modules in October, as planned (number theory and logic, with groups and geometry), but because I will now need to study more level 3 modules in 2014, I can now add complex analysis, to my quiver. I had planned to self study the complex analysis books, that were kindly sent to my by Chris F; but I think that Chris was quite right when he suggested, some time ago, that any future sponsor for post-graduate maths research / work, would question why one did not have such an important module.
Anyway, the study break has given me a good opportunity to start slowly re-reading some of my O.U module Units from M208, Pure Mathematics, which I completed last year. I have forgotten a fair bit, and there was some of it that I never entirely understood, the first time around (epsilon-delta, I'm looking at you!)
So, I have a rather lazy summer ahead, with some hopefully enjoyable revision of pure maths. I also plan to dust off my copies of Hardy's - Course in Pure Mathematics; Brannan's - Geometry and Spivak's - Calculus (Which should really be called Real Analysis, judging by the content of the book).
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