# Lazy Group Proofs and Efficiently Using Categories

1. Oct 29, 2013

### Mandelbroth

From Artin's Algebra: "Prove that the set $\operatorname{Aut}(G)$ of automorphisms of a group $G$ forms a group, the law of composition being composition of functions."

Of course, we could go through and prove that the four group axioms in the standard definition of a group hold for $\operatorname{Aut}(G)$. However, I'm lazy and like to do things a little more elegantly.

Suppose we define a group as a category with one object such that all its morphisms are automorphisms. Doesn't the proof follow trivially from this (id est, taking the set of the group as the only object in a category...)?

This is a good lead-in to another, related question: how often is category theory likely to simplify algebraic proofs like this? I was only recently introduced to category theory, and I'm going back through algebra to see if it simplifies things considerably. In other words, how often do situations like the above happen?

2. Oct 29, 2013

### Office_Shredder

Staff Emeritus
I don't see how the proof follows trivially from this. The morphisms of a category do not have to satisfy the group laws.

In particular, what if I try to copy your proof, but instead of the morphisms being automorphisms, they are just any endomorphism of a group.

Last edited: Oct 29, 2013
3. Oct 29, 2013

### Mandelbroth

If we say the morphisms are automorphisms, as I have above, which do they not satisfy?

4. Oct 29, 2013

### Office_Shredder

Staff Emeritus
If you just tell me the morphisms are automorphisms, I will say "why do you have inverses?" And then you'll say because every automorphism is an invertible function. At that point, what was the point of using category theory? To get associativity out of the way? If you want to claim that the automorphisms are the morphisms of a category rigorously you need to prove that anyway, and either way everyone knows that function composition is associative so you don't need to prove it unless you're being an incredible stickler (at which point, again, you would need to show associativity to get that you have a category).

5. Oct 29, 2013

### Robert1986

Well, you could define a group that way, but at some point, you'll have to show that this definition of group and the "normal" definition are the same. And then there's all the stuff that Office_Shredder said.

I'm not very experienced with category theory, but it seems that one of the best uses is not in proving theorems of various branched of math, but in giving seemingly different areas of math a common language. So, for example, when a new mathematical object is being studied, one can guess at what types of theorems should be proven.

6. Oct 29, 2013

### Mandelbroth

We're trying to show that the set of automorphisms on a group form a group under composition. By assumption, all the elements of this set of automorphisms are, as I thought might seem trivial, automorphisms. I have defined a group to be a category with a single object such that all its morphisms are automorphisms (I proved the equivalence of this to the standard definition yesterday).

I don't see what problem you have with this. Would you please rephrase it differently? Maybe I'm missing something, but reading the above sounded slightly incoherent to context, as if you didn't read what I've written or I've written something that wasn't understood. The point of this was to avoid the group axioms, so I don't know what you're trying to say by bringing them back into this.

7. Oct 29, 2013

### Office_Shredder

Staff Emeritus
The overarching problem is that you can't have proven anything because all you've done so far is assert that a group is a category with a single object, such that the automorphisms are the morphisms. This is certainly true, but you have yet to prove anything at all about the automorphisms that relates to proving they are anything other than arrows in a category, which by itself is not enough to show that they are a group (and to prove that they are the morphisms of a category, you have to show they satisfy almost every group axiom anyway, so any savings you might get are really swept under the rug at this point). I don't understand your end goal now; what are you proving about the automorphism set if you're not trying to show that they satisfy the group axioms?

Or are you trying to say screw everything you know about group theory, I'm starting over from scratch with a category definition which makes proving this one lemma easier.

8. Oct 29, 2013

### Mandelbroth

I don't know if this was unclear. I am, in a way, applying the definition twice.

We have a group, which consists of an object (what our object is here is unimportant in this context) and a set of automorphisms on the object (the group elements). Then, we imagine another category, taking the object to be the set of automorphisms from the first category (the group elements) and the morphisms to be all the automorphisms on the original group (the automorphisms of the automorphisms of the original, unimportant object).

In my mind, the second category "trivially" satisfies the definition of a group.

What "morphisms of a category" are you referring to? I feel like one of us (probably me) is missing something important here.

This. I'm just interested in seeing where it goes. All the other exercises I've done using categories are harder (in my opinion) than if I'd just used the group axioms and related concepts.

9. Oct 29, 2013

### Office_Shredder

Staff Emeritus
So what you have proven is that given a group G, there is a category which consists of the following:
A single object, which is G, and morphisms Aut(G).

On the other hand, if you give me just Aut(G) there is a very natural question for me to ask - how do I multiply two elements in G? I'm going to assume you know how to answer that question (I haven't thought about) but clearly if I give you two different sets of morphisms then G is going to be a different group.

So if you take as a category
A single object Aut(G), and Aut(G) as the morphisms

notice that Aut(G) is NOT Aut(Aut(G)). You have shown there exists SOME group structure on Aut(G), but you don't know that it induces the multiplication which is composition of functions, and in most cases I'm guessing it doesn't. Also, you can't just take your morphisms to be Aut(Aut(G)) because you don't know Aut(G) is a group yet.

If you want to prove that a group is a category of a single object whose arrows are the automorphisms, then you have to prove the automorphisms have an identity and are associative - that's all the work you need to do to show that Aut(G) is a group in the first place!

Basically, you shouldn't expect category theory to do math for you. Its main power is in applying results in one category to another by observing that the statement is purely category theoretic - for example, there is probably a way to simultaneously prove Aut(G) is a group for G a group, or G a vector space, or G a ring, but it's not going to be easier than just proving any one of those is a group on its own.

Last edited: Oct 29, 2013
10. Oct 29, 2013

### jgens

Not quite. If you were to go through the work to establish that this second category is, in fact, a category, then the proof reads exactly the same as the proof that Aut(G) is a group (under the usual definition).

Well if you only care about groups, then there is little perspective to be gained by thinking of groups as categories with one objects and all arrows isomorphisms. However if you are interested in generalizations of the group concept, this does lead quite naturally to groupoids. There is some interesting theory about them and right now they have a few applications in algebraic topology, so I suppose look there if you're interested.

11. Oct 29, 2013

### Mandelbroth

This is the second category described. I don't know where $\operatorname{Aut}(\operatorname{Aut}(G))$ comes from, but this section was helpful. Thank you.

Well, again, I'm interested in doing the math myself. I just thought it would be a fun way to practice some new concepts (to me, at least). In all the other exercises I've worked, it makes the problems more difficult.

12. Oct 29, 2013

### Office_Shredder

Staff Emeritus
The Aut(Aut(G)) came from the fact that if your object in the category is H=Aut(G) and you want to realize it the same way you were realizing G, then your morphisms should be Aut(H) = Aut(Aut(G)).

When I say that category theory won't do math for you, I mean that if you want to prove an interesting result about a specific category you usually aren't going to do it without just diving in and doing a bunch of algebra. Category theory is just a way of generalizing your algebra - for example instead of proving the automorphisms of a group form a group, you can prove that the automorphisms of any category (or perhaps ones that are reasonably well behaved only, I'm not terribly familiar with the field) form a group, and then you know the automorphisms of a vector space or a ring or even a set are all groups as well. The proof is going to basically be the same as it is for the automorphisms of a group, and you save yourself the trouble of proving it for every other category you want the result for.

Last edited: Oct 30, 2013