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.]
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