Are a and b in G if ab is in G?

[daveyinaz]

If ab in G, then a and b in G??

I'm not sure I've convinced myself here.

Take the L be the set of all linear transformations from V -> V, where V is a finite dimensional vector space. We know that G contained within L of all invertible linear transformations is a group. Say you have S, T within L and ST = I [ST is functional composition S(T(v))], where I is the identity mapping. Then it must be that ST is invertible (since we know that 1-1 <=> onto <=> invertible in L) since ST is a 1-1 and onto mapping. Which means ST in G and the inverse of (ST)^-1 = T^-1S^-1...but this also implies that T and S are invertible and thus in G?

 

[morphism]

I don't understand what you're saying. You're starting with the assumption that S and T are invertible (otherwise we wouldn't have ST=I). Then you're concluding that S and T are invertible?

 

[daveyinaz]

The only reason you say that I'm starting with S and T are invertible is because V is finite dimensional right? Since DM = I where D is the differentiation and M is the linear transformation such that for a polynomial p, then I(p) = a_0x + (a_1/2)x^2 + ... + (a_k/k+1)x^k+1 on the set of all polynomials with real coefficients yet D nor M is bijective.

 

[morphism]

Yup - in finite dimensions a linear map is invertible iff it's surjective iff it's injective.

 

[daveyinaz]

The thing that bothers me that S and T are automatically invertible to begin with is that it is not known that TS = I.

 

[morphism]

I'm still confused. What exactly are you trying to do?

 

[daveyinaz]

For now show TS = I. Merely to show that S and T are invertible and in G. Since we all know for groups if a and b are in G, then ab in G. In this case, do we have if ab in G, then a and b in G. Since I start with S and T in L (not known whether it is in G or not) and not every S and T in L are invertible.

 

[morphism]

Let me see if I understood that correctly. You're starting with S and T in L such that ST is invertible. And you want to determine whether or not this implies that S and T are invertible. Correct?

 

[daveyinaz]

I'm starting with S and T in L such that ST = I.

The only reason I say ST is invertible is because if ST is the identity mapping on a finite dimensional vector space to itself, then it must be bijective.

 

[morphism]

Yes of course in that case ST is invertible. And it's also equally obvious that S and T are invertible (in fact they're inverses of each other).

 

[daveyinaz]

I don't think it's as obvious as you claim it is. That is to show that TS = I, thereby making T and S invertible and further making them inverses of each other.

 

[morphism]

If ST=I, then this means that T is injective and S is surjective, which in turn implies that T and S are invertible as we've already discussed.

 

[daveyinaz]

Damn it! I knew there was something like that I was not remembering.

Not that it matters now, there is another way to show that TS = I by using bases and image under the bases and blah blah blah...this way just seems more straightforward...

Thanks.

 

[morphism]

No problem.

 

[Pere Callahan]

In general, "If ab in G, then a and b in G" does not hold.
Take G = Z (the integers) and a=b=1/2.

 

[daveyinaz]

Clearly you meant you meant ab to mean addition and not multiplication right?

 

[Pere Callahan]

Yes, I wrote the group multiplication multiplicatively. In the case of Z this means of course addition.