# Generated by two subgroups?

1. Jun 8, 2008

### tgt

Consider G a group with two subgroups A and B.

What does it mean to say a subgroup of G generated by the subgroups A and B?

2. Jun 9, 2008

### CompuChip

First look at a simpler case: What does it mean to say that it is generated by A only?

3. Jun 9, 2008

### tgt

Good question. What does it mean?

4. Jun 9, 2008

### eastside00_99

Given any subset S of a group G, we can form the subgroup of G by taking all finite products of the elements of S (we define the empty product to be the identity). This is denoted by <S>. An equivalent definition, is that <S> is the smallest subgroup of G containing S. Since if $$x=s_1\cdots s_n  where  s_i \in S$$, then x must be in the smallest subgroup of G containing S. It is then obvious that the two definitions are equivalent.

So, since the smallest subgroup generated by a subgroup A of G must be A, for G to be generated by A means that A=G. Note that this is only when A is a subgroup. So, for G to be generated by A and B means that there is no subgroup of G which contains A and B other than G. We can always form what is know as the free product of groups, usually denoted either A*B or AB. Then another way of saying it is that G=<AB>. In other words, AB might not necessarily be a subgroup, but you do know that the smallest subgroup containing AB is G. This seems like a generalization of the case when AB=G to the case when AB is not necessarily a subgroup. Note that the free product is always a group, but that there may not be an imbedding of AB into G.

Last edited: Jun 9, 2008
5. Jun 9, 2008

### tgt

The second paragraph isn't clear.

6. Jun 9, 2008

### eastside00_99

(1) So, since the smallest subgroup generated by a subgroup A of G must be A, for G to be generated by A means that A=G. Note that this is only when A is a subgroup.

By definition, the group generated by the subgroup A is A if A is a group. Thus if the group generated by A is G, then A=G whenever A is a subgroup of G.

(2) So, for G to be generated by A and B means that there is no subgroup of G which contains A and B other than G.

By definition, the group generated by A and B is the smallest subgroup containing but A and B. Therefore there is no proper subgroup of G which contains both A and B.

(3) We can always form what is know as the free product of groups, usually denoted either A*B or AB. Then another way of saying it is that G=<AB>.

Let $$AB =\{ab \mid a\in A, b\in B\}$$ It is easy to check that the group generated by A and B is equivalent to the group generated by the set AB. Maybe part of the confusion is that the group generated by A and B would be the group generated by the union of the two groups. This is the only possible meaning of the phrase sense A and B are groups. If they where sets, then it wouldn't be so straightforward. In that case it could mean G=<A>=<B> where A =\= B. But as long as they are groups, they couldn't mean this because then you would have G=A=B.

(4) In other words, AB might not necessarily be a subgroup, but you do know that the smallest subgroup containing AB is G. This seems like a generalization of the case when AB=G to the case when AB is not necessarily a subgroup.

7. Jun 9, 2008

### tgt

3 is unclear.

What happens if the intersection of A and B is non empty? Then the group generated by A and B is not the free product of A and B.

Last edited: Jun 9, 2008
8. Jun 10, 2008

### eastside00_99

There are many types of products of groups. Let AxB denote the cartesian product. I.e., AxB ={(a,b) | a in A and b in B}. This is a group under the multiplication (a,b)(a',b')=(aa',bb'). There is a theorem that states if ab=ba for all a in A and all b in B, the only element of A and B is the identity, then AxB can be embedded into G. Furthermore, if AB ={a,b | a in A, and b in B} =G, then AxB is isomorphic to G.

Proof. We define a map from AxB to G by by (a,b) --> ab. A trivial computation shows that it is a homomorphism. Furthermore, it is onto AB by definition. Now, let (a,b) be in the kernel of our map --- i.e., ab = e. Then a=b^-1. So, a is both in A and in B which implies a = e, whence b=e. Thus the map is injective.

What I think you are getting at is that AB will not necessarily be isomorphic to the group AxB without these conditions.

9. Jun 10, 2008

### eastside00_99

To clearify (3), just use the fact that the group generated by A and B is the smallest subgroup of G which contains A and B. Clearly, this is the smallest subgroup containing AuB or because if H contains A and B then it contains AuB and vise verse. Thus, when it says that the smallest subgroup generated by A and B, think of it as the group of all finite products of elements of AuB.

But, just for your reference there was some ambiguity in the statement "suppose G is group generated by A and B." by a stretch of the imagination this could mean G=<A>=<B> (when A and B are not groups). But sense you have stated explicitly that they are groups, use what I have said above.

10. Jun 10, 2008

### tgt

Why does it have to be finite?

11. Jun 10, 2008

### matt grime

Because that is the definition of 'generate' - any element is a word of finite length in the elements of A and B.

12. Jun 10, 2008

### tgt

I guess what I'm really after is:

How would you work out the subgroup of G generated by two subgroups of G (i.e A and B)?

i.e <AUB>=? for A and B are any subgroup of a group G

Is there a formula or method for calculating such things?

13. Jun 10, 2008

### matt grime

In general no. But in any particular example just 'doing it' will probably give you the answer.

14. Jun 11, 2008

### tgt

Can you prove that in general it is not possible?

15. Jun 11, 2008

### matt grime

It is a genuinely hard problem to determine when two given presentations (generators, relations) define the same group, or indeed to determine almost anything at all about a group. See, e.g. the word problem or that whole gamut of things.

However, in nearly any case you're given, attempting to show that, in this case, <AuB> contains some known generators of G will almost surely work.

For finite groups, you of course have exhaustion, and generically, I suspect that that will be the only option.

I'd also like you to think about your own question: when you ask "what is <AuB>?", you're really asking 'can I identify it with something else that I like more'. The subgroup generated is whatever it is, it is really only interesting to ask if this subgroup is isomorphic to some other group, or is indeed the whole of G. Note that for infinite groups it is entirely possible for <AuB> to be isomorphic to G, and not equal to it.

In finite cases you have other useful things at hand. For instance, take G=S_3, the permutation group of 3 elements. Let A be {e, (12)} and B be {e,(123), (132)}. Then it is obvious that <AuB>=S_3, since <AuB> contains at least 4 elements, and the order must divide 6.

Last edited: Jun 11, 2008
16. Jul 4, 2008

### tgt

What does doing it mean? Is there some procedure?

17. Jul 4, 2008

### CompuChip

In my experience, 'doing it' usually means writing down all the elements you can think of, applying the relations to find all unique ones among them, check if you can construct any new ones, possibly use some group theoretical theorems to check if that might be all (e.g. do all the orders of the elements divide the total number I've found, does it contain the generating group, etc) and hope that's it

I don't think there is really any method that will always definitely give you the entire group, unless you have a computer program or some "physical" meaning (for example, you know that $D_6$, the symmetry group of a triangle, has 6 elements because each symmetry can be identified with a permutation of the corners of which there are 3! = 6).
Indeed I wonder how they proved facts about, for example, the monster group

18. Jul 4, 2008

### matt grime

It means precisely what it states: just do it., get your hands dirty, make some computations, bloody well think about it, for God's sake; it's not hard, it just requires you to put in some work.

19. Jul 4, 2008

### tgt

What happens if we have two infinite groups? At the moment, it seems the free product approach or free product with amagalmation seems best but it dosen't work for all of them. For example, what happens if you are in a large group F where you want to know <A,B> in F. But two or more pairs of subgroups in A and B are identified (for example, considered equal). Then what do you do?

20. Jul 5, 2008

### matt grime

Whatever you can. There are many papers and books on this stuff, try looking up algorithms. Try, for instance, googling for 'the word problem finitely presented groups' and see what comes up.