The course hasn't officially started yet, but it feels like it. The course website opened last Tuesday and I was finally able to get my hands on the first TMA. Since then, I feel like things have stepped up a gear and I am already having to devote more hours to my study. For starters, there is TMA01 part 1 to do and it was good revision. I did get myself in a bit of a twist to start with but things worked out ok in the end. It is amazing how one tiny slip can lead you down a blind alley. Then there is the sudden burst of activity on the forums to follow; a bit more revision of I1 to do; talking to my tutor about one of the problems I found in one of the exercises (that was far more interesting than I realised) and trying to keep up with my work with the books at the end of the course. Phew! I won't be able to keep this up at this rate. I may have to scale back to just doing the TMA's and the end of course work.
Other than that I have finished the chapter on Homomorphisms (GTB2) and have begun the chapter on Group Actions (GTB3). This is my final group theory book. I don't mind group theory too much and I find it ok to work with. In the chapter on Homomorphisms (I wish that word wasn't such a mouthful!) we find that they are a bit like Isomorphisms but with the strict one-one and onto condition relaxed. Homomorphisms are basically functions that map one group to another and in the mapping process some of the domain group properties are preserved. For example, the identity in the domain is mapped to the identity in the codomain, as are inverses and powers of elements. Also elements that are conjugate in the domain are also conjugate in the codomain.
The meat of the chapter comes with the discussion of Kernels and Images of a homomorphism. This is a bit like the discussion of Kernels and Images of a linear transformation in the Linear Algebra blocks. The Kernel of a homomorphism is the set of elements of the domain that map to the identity element in the codomain. The Image is the usual notion of an image of a function. It turns out that the Kernel is a normal subgroup of the domain group and the Image is a subgroup of the codomain group. The fact that the Kernel is a normal subgroup of the domain means, of course, that its cosets partition the domain and this leads on to the Correspondence Theorem, a return to Quotient Groups and finally the Isomorphism Theorem (I ain't going to discuss all that!). It's all a bit of a handful to remember. Reasonably straight forward at the time but instantly forgettable!
Tuesday 24 January 2012
Monday 16 January 2012
Starting to get course fatigue
I was starting to get course fatigue this week. I have been working on M208 since mid February last year and it is hard to sustain the momentum over that length of time. Each day I attempt to get through a page or two, but this week I found myself doing even less than that. The thought crept into mind that I just wanted to just get the last 5 books over with, so I could start consolidating what I have already learnt.
I think part of the problem with studying is that you are being led by the nose through a lot of material not all of which is going to be exciting. The tendency after a while is to think, "oh no, not another complex idea to grapple with" or "groan, not another proof to understand". I am a firm believer that you learn the most from exploring ideas on your own and there isn't much room for this on this course. It is all very much spoon feeding.
To entertain myself I was thinking about groups in general and one question that I came up with was what if you had the situation where all elements of a set were self-inverse under composition; could that set form a group or not? I had been thinking about finite groups of low order and imagining Cayley tables where the identity element is always along the leading diagonal.
I posed this question on the OUSA M208 forum but I quickly realised that it was not a question for the novice. It was also not easy to pose the question logically, anyway. A couple of theorems occurred to me. One was a corollary to Lagrange's theorem that if G is a group of prime order p, then G is a cyclic group. This is because the only numbers that can divide p are 1 and p and so the subgroups of G can only have order 1 or order p. These correspond to the subgroups {e} and G (e is the identity element of G). Moreover, as the only element with order 1 is e, the other elements in G must have order n, so the group G is cyclic.
So if the order of a finite group is prime then the group is cyclic. If we now consider these prime order groups where the order is greater than 2, then we can show that the elements of these groups cannot all be self-inverse. Why? Because to be a cyclic group we require that there is g element G of order n but the orders of g are all 2 (because they are self-inverse).
Hence, if a set contains p self-inverse elements and p is a prime number bigger than 2, then this set cannot form a group!
Interestingly, the sets with either one and two self-inverse elements can form groups under composition. {e} is a group and {e, g} can be a cyclic group since g can generate the group.
The other theorem which tackles this problem head on is Theorem 3.2 of GTA4. Let G be a group, with order greater than 2, in which each element except the identity has order 2. Then the order of G is a multiple of 4. So this means that groups of order 3 or more in which all the elements are self-inverse have to have an order of 4, 8, 12, 16...This agrees with what I deduced from the corollary to Lagrange's theorem, but is more stringent.
I think part of the problem with studying is that you are being led by the nose through a lot of material not all of which is going to be exciting. The tendency after a while is to think, "oh no, not another complex idea to grapple with" or "groan, not another proof to understand". I am a firm believer that you learn the most from exploring ideas on your own and there isn't much room for this on this course. It is all very much spoon feeding.
To entertain myself I was thinking about groups in general and one question that I came up with was what if you had the situation where all elements of a set were self-inverse under composition; could that set form a group or not? I had been thinking about finite groups of low order and imagining Cayley tables where the identity element is always along the leading diagonal.
I posed this question on the OUSA M208 forum but I quickly realised that it was not a question for the novice. It was also not easy to pose the question logically, anyway. A couple of theorems occurred to me. One was a corollary to Lagrange's theorem that if G is a group of prime order p, then G is a cyclic group. This is because the only numbers that can divide p are 1 and p and so the subgroups of G can only have order 1 or order p. These correspond to the subgroups {e} and G (e is the identity element of G). Moreover, as the only element with order 1 is e, the other elements in G must have order n, so the group G is cyclic.
So if the order of a finite group is prime then the group is cyclic. If we now consider these prime order groups where the order is greater than 2, then we can show that the elements of these groups cannot all be self-inverse. Why? Because to be a cyclic group we require that there is g element G of order n but the orders of g are all 2 (because they are self-inverse).
Hence, if a set contains p self-inverse elements and p is a prime number bigger than 2, then this set cannot form a group!
Interestingly, the sets with either one and two self-inverse elements can form groups under composition. {e} is a group and {e, g} can be a cyclic group since g can generate the group.
The other theorem which tackles this problem head on is Theorem 3.2 of GTA4. Let G be a group, with order greater than 2, in which each element except the identity has order 2. Then the order of G is a multiple of 4. So this means that groups of order 3 or more in which all the elements are self-inverse have to have an order of 4, 8, 12, 16...This agrees with what I deduced from the corollary to Lagrange's theorem, but is more stringent.
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