Mistake in Schaum's Group Theory?

In summary: Thanks. This is helpful. A 1-1 map between vector spaces is a linear transformation if and only if it preserves the composition of vectors. This is an easy exercise: Every 1-1 map between vector spaces is linear, because composition is associative and commutative.
  • #36
You said earlier that ##AB=I## is given. I think you confused the order now. This is very confusing, as the order is all we talk about. But given ##AB=I## the following doesn't make much sense:
WWGD said:
##BABB^{-1} =I ## implies ##AB=AB^{-1} ## , not necessarily the identity. Maybe we can argue:
##BA=I ## , then ##AB= ABAB=I \rightarrow (AB)^n =I ## , has only identity as solution ( since this is true for all natural ##n## )?
If we allow a left inverse ##CA=I## then we immediately have ##B=IB=(CA)B=C(AB)=CI=C##. The existence of ##C## is the problem. To write it as ##B^{-1}## is cheating.
 
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  • #37
fresh_42 said:
You said earlier that ##AB=I## is given. I think you confused the order now. This is very confusing, as the order is all we talk about. But given ##AB=I## the following doesn't make much sense:

If we allow a left inverse ##CA=I## then we immediately have ##B=IB=(CA)B=C(AB)=CI=C##. The existence of ##C## is the problem. To write it as ##B^{-1}## is cheating.
Yes, my bad, mixed both things up. Let me go for another doppio, a 2-sided doppio. I intended a post to contain left- and right - inverses, but I somehow got logged off a few times and lost track of what I was doing :(.
 
  • #38
Ok, this is the argument I intended:
Assume ##AB=I ##
Then ##BA=B(AB)A=(BA)^2=I ## ( just need associativity) and we can extend to ##(BA)^n =I ## for all n
 
  • #39
WWGD said:
Ok, this is the argument I intended:
Assume ##AB=I ##
Then ##BA=B(AB)A=(BA)^2=I ## ( just need associativity) and we can extend to ##(BA)^n =I ## for all n
That's wrong. If ##BA=C## then all we have is ##C=BA=B(AB)A = (BA)^2=C^2## and per recursion ##C=C^n## for all ##n##. Since there are unipotent matrices which are not the identity, I don't see how ##C=I## should follow.
 
  • #40
fresh_42 said:
That's wrong. If ##BA=C## then all we have is ##C=BA=B(AB)A = (BA)^2=C^2## and per recursion ##C=C^n## for all ##n##. Since there are unipotent matrices which are not the identity, I don't see how ##C=I## should follow.
Yes, use BA or C either way. And, no, not . ##C^n =C ## for _some_ n, but ##C^n =C ## _for all_ n . Do you have a non-ID C with ##C^2=C^3=...=C^n =I ## ? Maybe, I don't know of one. Also DetC =1 here, let's assume Real entries. If not, the determinant is a 2nd, 3rd,...n-th,... root of unity. Is there such number?
 
  • #41
WWGD said:
Yes, use BA or C either way. And, no, not . ##C^n =C ## for _some_ n, but ##C^n =C ## _for all_ n . Do you have a non-ID C with ##C^2=C^3=...=C^n =I ## ?
That's what I said, for all ##n##. And, no, I don't have such a ##C##, but this isn't a formal proof, especially if we do not use linear algebra.
 
  • #42
fresh_42 said:
That's what I said, for all ##n##. And, no, I don't have such a ##C##, but this isn't a formal proof, especially if we do not use linear algebra.
Yes, I know, I never said it is a proof. But this matrix BA satisfies infinitely many polynomials ##C^n-C =0 ##, which is also strange, but, I admit, not a proof (yet?).
 
  • #43
Just experimenting before aiming for a formal proof.
 
  • #44
Well, we have ##C=0## is a solution, too. And theoretically we can have zero divisors from the left and not from the right and similar nasty things. But the question "can we prove ##G=GL(V)## is a group" without using linear algebra, and only ##AB=I## is probably doomed to fail, as @Math_QED mentioned correctly, that it doesn't work in rings. But if it's a group, then we are done. If we do not allow this, we have to use something from the definition of ##G##, and that is linear algebra.
 
  • #45
fresh_42 said:
Well, we have ##C=0## is a solution, too. And theoretically we can have zero divisors from the left and not from the right and similar nasty things. But the question "can we prove ##G=GL(V)## is a group" without using linear algebra, and only ##AB=I## is probably doomed to fail, as @Math_QED mentioned correctly, that it doesn't work in rings. But if it's a group, then we are done. If we do not allow this, we have to use something from the definition of ##G##, and that is linear algebra.
C=0 Is not a solution ##AB=I ## so ##Det(AB)=DetADetB=1 ## , and ##DetC=DetBA =1 ## , but Det0=0. EDIT: ##DetC =1 ##or if you allow Complexes and ##Detc \neq 1##, DetC is a 2nd,3rd,..., nth,... root of unity.
 
  • #46
WWGD said:
C=0 Is not a solution AB=I so Det(AB)=DetADetB=1 , and Det0 =0.
We are still on the necessity part. The damn existence of ##C## is the problem. And ##0## is a solution to ##C=C^n## for all ##n##. If you use the determinant, you have to mention the field! What I say: we will need some argument from LA. In the end we could simply write down the formula for the inverse in terms of ##A## and good is.
 
  • #47
fresh_42 said:
We are still on the necessity part. The damn existence of ##C## is the problem. And ##0## is a solution to ##C=C^n## for all ##n##. If you use the determinant, you have to mention the field! What I say: we will need some argument from LA.
Well, yes, no finite fields, so C^n=C. And I had specified Complexes or Reals. Anyway, I don't see how this relates to Los Angeles, LA ;), but at any rate, I think we may be able to relax condition of linearity. All we need is scaling property and showing the image is not meager, i.e. the image contains a ball about the origin , then if we consider a line segment y=cx from the origin, we consider the multiples of that segment and every point is hit in that way.
 
  • #48
We still have the difficulty that zero divisors are possible as long as we haven't a group. The (unique) solvability of ##x A = B## is equivalent to the group axioms, so we circle around the main topic. Guess we have to visit La La Land.
 
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  • #49
the integers are a group in which all non zero subgroups are normal and isomorphic, but the quotients can be any finite cyclic group.
 
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  • #50
fresh_42 said:
##(0,1) \stackrel{\operatorname{id}}{\hookrightarrow} \mathbb{R}-\{\,0\,\}\stackrel{\varphi}{\rightarrowtail} (-\frac{1}{4},\frac{1}{4}) -\{\,0\,\} \stackrel{+\frac{1}{3}}{\hookrightarrow} (0,1)##
where I only need a bijection ##\varphi## between the real line and an open interval, both without zero to make it easier.
If I understand what your notation conveys, the last bit should be ## \stackrel{+\frac{1}{4}}{\hookrightarrow} (0,1)##, since you're adding 1/4 to each element of the punctured interval (-1/4, 1/4).
 
  • #51
Mark44 said:
If I understand what your notation conveys, the last bit should be ## \stackrel{+\frac{1}{4}}{\hookrightarrow} (0,1)##, since you're adding 1/4 to each element of the punctured interval (-1/4, 1/4).
No, I just wanted to shift it into the interval injectively. The amount didn't matter. With ##\frac{1}{4}## I would have had to deal with the zero, which I avoided by taking ##\frac{1}{3}##.
 
  • #52
fresh_42 said:
We still have the difficulty that zero divisors are possible as long as we haven't a group. The (unique) solvability of ##x A = B## is equivalent to the group axioms, so we circle around the main topic. Guess we have to visit La La Land.
Where are you getting your scalars? We're talking coefficients in a field of characteristic zero now, aren't we?
 
  • #53
WWGD said:
Where are you getting your scalars? We're talking coefficients in a field of characteristic zero now, aren't we?
O.k., but matrices can still multiply to zero. However, since @mathwonk #49 I consider our riddle solved.
 

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