Talk:Abstract Algebra/Older Version

Automorphisms are conceptually tricky? Says who?


 * I mean that they probably won't "click" as easily with someone learning about groups as an experienced group theorist would. We talk about homomorphisms between two groups and then we move to homomorphisms between one group - it might be a little confusing for some Dysprosia 15:54, 18 Oct 2003 (UTC)

Hi. I hope you don't mind my changing and adding to your page. You seem to be the only author, but I've also seen suggestions to "be bold" in editing.

I've a suggestion. What about moving cosets into a subsection of subgroups? To me, they don't need their own section, and it would make introducing normal subgroups in the subgroups section easier. For now, I'll throw something in on normal subgroups under the subgroups section. Mjeff 15:38, 21 Feb 2004 (UTC)


 * I don't mind at all. Just keep in mind not to introduce concepts that haven't been dealt with, but that's not such a big problem... In any case, I particularly like where you put the stuff on kernels :)
 * Re to cosets, I'm thinking that when we introduce factor groups in the "Special groups" section I might slot them in there.... Dysprosia 23:30, 21 Feb 2004 (UTC)

Hey,

I just finished taking some classes on abstract algebra and I thought it would be fun to do a little bit more over the summer. One of the other comments said not to introduce any new concepts, but I noticed a couple things that would be nice to have, like a bit about cyclic groups and subgroups. Would anybody mind if I added some stuff like that? I also thought it would be nice to have more proofs, especially some easier ones at the beginning so people can get the hang of it.

stdarg 8:00, 10 May 2004

new here
I just found this project and it looks like worth helping. I can fill out proofs here and there. Is the plan to put all basic group theory under this one book?

I was scanning through this page and noticed that all the while, the notation a*b was used and then in one of the problems, the * operator was suddenly dropped and the notation ab was used instead. Perhaps this change in notation needs a word or two to avoid confusion to people who is reading the page for the very first time.

Also, I would think that it would be extremely useful to introduce the following theorem on subgroups:
 * Let G be a group and let H be a subset of G. Then H is a subgroup of G if and only if for all a, b in H, ab-1 is also in H.

The part about a*b and ab is now in section, 1.1. I agree about the Subgroup Criterion and will add it when I have some time, if no one else does. MikeBorkowski

I hope there are templates for formulating theorems, lemmas, propositions, corollaries, exercises and remarks. Can anyonw help me with that? Pm5 1:38, December 7, 2005.

So I was looking at section 3.3 Problems without answers, and after attempting these problems, it seems that they should be expanded into the book text. They are not elementary problems. At the very least, a proof outline should be provided.

Here are my issues:

Problem one: This proof lays the foundation for the definition of the order of an element, but does not define it. In a finite group, after all, any element taken to the group order is the identity, but that isn't the order of every element.

Problem two: But then in the very next problem, the order of the element is used as part of a very deep theorem about the structure of all finite abelian groups. This hardly seems appropriate after the first section that introduces subgroups.

The other problems are typical "proof left to the reader" types, but their consequence should be expanded in the text

beej 14:13, 9 December 2005 (UTC)

A question to all editors here:

The use here - starting in the homomorphim section - of 'f(a)' is perhaps a little non standard for abstract groups. Might it be better to use the form 'aφ' (the function identifier after the variable) instead which is a little more usual?

Subgroups
Is the empty set and the operator a subgroup?Shiney (talk) 22:29, 19 April 2008 (UTC)

Goodness! No-one has answered you?

The empty set is not a group (and so not a subgroup) because there is no identity element. The set containing the identity element is a subgroup. --beej (talk) 16:06, 7 September 2008 (UTC)