Posted by: yanzhang | June 19, 2008

## Algebra 101 – Why, and Nontrivial Automorphisms

I regretted not learning the basics well while I was in school. Blaming Math 55’s evolution is a cowardly excuse to not blame myself: it surely wasn’t purely the professor’s fault when the students – including myself – were very good as a unit at pretending to understand the material. This is of course why I always tell junior students to learn the basics well and not to skip too many classes, an extension of a popular phenomenon that I will coin as the “those who can’t do and those who f***ed up in their past, teach” principle. While (unlike seemingly everyone else) I did take my share of undergraduate courses, I am still sad that I never had a solid treatment of basic group theory. I knew the theorems and understood the definitions – something easy to do by osmosis – but I never got my hands dirty with problems with real, concrete groups, especially finite ones. The analogy is one between a reader and a writer – I can understand any paper that uses these concepts (and look up more obscure items as I need them), but I have very limited craft in dealing with these objects, to the degree that I am more comfortable in manipulating some of the more “difficult” objects like matrices, fields, algebraic varieties (not schemes. Those things still WTF me ;P) and CW-complexes. This is a gaping hole in my swiss cheese of mathematical knowledge.

This is not a criticism of my peers (most of whom will surely become better mathematicians than I will). Their frequent objection to my view on basics is the following: with the obvious exception of representation theory, we can honestly do a lot of “higher” algebra at a concrete level without really getting dirty with groups, because we are usually using deeper facts about more sophisticated structures. For example, when was the last time you actually used Sylow’s theorem? My objection with my flimsiness with groups is more of a meta-learning concern – I think every branch of mathematics (or general skillset in life, for that matter) can teach a way of thinking and a way of solving problems, both invaluable even if I never use the knowledge in that branch. While I will probably never use Sylow’s Theorems in future research, having used them to solve certain problems will have trained me to think about certain types of general problem-solving situations, and I never think that is a waste of time.

The point of that verbose preface was… to keep myself honest, I’ve been collecting some (easy) algebra problems and working on them. They’re not “deep,” they’re fun and easy, and they teach many little facts that I never knew (and probably will never use). I might post more if I find them, but for now, something cute:

Every finite group of order >=3 has a nontrivial automorphism.

Proof (yeah yeah, I know all of you think this is trivial. I am a beginner remember):

Suppose the group G is not abelian. Then there is some nontrivial element a. Conjugation by a is a group automorphism, and is nontrivial on any element b which doesn’t commute with a, so we are done. In the case that G is abelian, we know G is isomorphic to a direct product of cyclic groups, which we can think of as Z_k under addition. If any one of these, say Z_t, has order >= 3, mapping the generator 1 in that group to an integer in Z_t relatively prime with t and imposing the identity on all the other cyclic groups is a nontrivial automorphism. The only time when this doesn’t work is if all of our cyclic groups has order 2… but the moment we have two of these, we can just define an automorphism to send the generator 1 in one of the groups to the generator 1 in the other (and vice-versa), again applying the identity to everything else.

Feel free to give me more exercises like this!

-Y

Source: Berkeley Problems in Mathematics (de Souza, Silva).

## Responses

1. Define the automorphism by f(x)=x^-1. This is nontrivial for all groups of size >2. Here two problems that I found useful when studying for algebra comps (which I passed at the beginning of the summer).
Determine all homomorphisms S_5->Z_5 and Z_5->S_5.
Determine the size of GL_2(5). How many conjugacy classes does this group have?

I know that there were many others that were useful, but I’d have to look at my notes for examples.

I definitely know what you mean about not getting a good understanding of concrete examples. I now force myself to look at a simple example of any theorem that’s not completely obvious. It’s a habit I highly recommend.

2. thanks man.

Btw, I stumbled upon that solution after posting (inversion). There’s a slight caveat – if every element has order two then this is the trivial automorphism, and you’d have to do the second part anyway. But it is faster than the first part for sure.

I’ll totally work on your problems, maybe even post when I get them =)

As for your technique of looking at a “simple example,” that’s like one of my favorite studying metatechniques. Totally gold – except I forget to do it sometimes =/

3. Whoops. I meant to write GL_2(Z/5Z) (the general linear group over the field with five elements)

4. Nice! Me and a few friends keep a problem page here at grupofundamental.wordpress.com. It’s in spanish, but I think you can understand most of the problems. Could I post this one (it’s about cuteness, not dificulty, I guess)? In exchange, I offer the following problem, that was posted a few days ago:

We know that every group can be embedded inside $S_n$, the group of permutations of n elements. However, clearly $S_n$ is sometimes larger than it is actually needed: Por example, all groups of prime order can be imbedded in… any group of prime order, since they are all isomorphic. So, is there a nicer bound than $|G|!$ for the order of a group that contains isomorphic copies of all groups of order $|G|$? We haven’t solved it yet, but we have explicit numbers for $|G| < 100$… we got them by using Sylow’s Theorems :P.

5. A minor bit of pedantry: in the non-abelian case, you should say that the non-trivial element a lies outside the center of G, otherwise conjugation by a is still the identity.

A nice similar question is: is there a (finite) group with full automorphism group of size three?

6. Was randomly browsing and Jan’s problem caught my eye.

I believe the answer is no. For if G were such a group then G/Z(G), being isomorphic to the subgroup Inn(G) of Aut(G) =~ Z_3, would be cyclic. This implies that G must be abelian. In fact, G must be a product of Z_2’s. This is because the automorphism x -> x^{-1} mentioned above must have order 1 (it can’t have order 2 because 2 doesn’t divide 3!). Now we can finish it off: Aut(Z_1)=Aut(Z_2)=Z_1, while Aut(Z_2 x … Z_2) (at least 2 factors) definitely has more than 3 automorphisms.

7. Er… inversion isn’t a group automorphism in a nonabelian group, is it? It switches the order of multiplication.