# I Mistake in Schaum's Group Theory?

#### WWGD

Gold Member
Well the 4-,5- and 9-Lemmata should work in $\mathbf{Top}$ as well, but I'm still not convinced.
?? I never heard of those names. Can you ref. please?
Do you agree with my argument using matrices?

#### fresh_42

Mentor
2018 Award
?? I never heard of those names. Can you ref. please?
Do you agree with my argument using matrices?
There is no problem with a finite basis and linearity. I only said, that the element count doesn't work for infinite sets. Interesting would be the case of uncountable dimensions: Can we embed such a vector space in itself without being surjective? Or will Hamel bases, i.e. AC save the day?

#### fresh_42

Mentor
2018 Award
?? I never heard of those names. Can you ref. please?
Do you agree with my argument using matrices?

#### WWGD

Gold Member
There is no problem with a finite basis and linearity. I only said, that the element count doesn't work for infinite sets. Interesting would be the case of uncountable dimensions: Can we embed such a vector space in itself without being surjective? Or will Hamel bases, i.e. AC save the day?
I never thought cardinality/count was enough; that wasn't part of my argument. I thought a key issue was that we are mapping a space to itself. And that matrices are not one-side invertible, unlike some functions. EDIT: I guess re the quotients X/S_1, X/S_2 , we need some condition on maps passing to the quotient. Thanks for the 5-lemma ref. I will see how to fit it.

#### Math_QED

Homework Helper
Why that? What is a counterexample? A quick view on the 4 Lemma seems as if there are inclusions $G_1/H_1 \rightarrowtail G_2/H_2 \rightarrowtail G_1/H_1$. Not sure whether there is also an epimorphism, but the two inclusions are a strong condition.
Exactly the point I was trying to make! It seems so true! Take $G = Z_4 \times Z_2$ and consider the cyclic subgroups generated by $(2,0)$ and $(0,1)$. They are both cyclic of order 2 and hence isomorphic. Quotienting them out gives the two different groups of order $4$.

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#### WWGD

Gold Member
Maybe one way of doing the proof for (fin. dim) V.S is to use the fact that, given an ordered basis $\{ v_1, v_2,...,v_n \}$ there is an isomorphism between $\text L(V,V)$ and $\text Mat_{(n,n)}$. And then you can show a matrix can only have 2-sided inverses, meaning it has to describe a map that is 1-1 and onto.

$AB=I B=A^{-1} \rightarrow BA =A^{-1}A=I$

So B is both right- and left- inverse and we have two-sided inverses.
BAH. My argument here is incorrect; it does not automatically follow. It is an interesting exercise: Show that for A,B same size, AB=I , then BA=I. Not trivial that matrix inverses are necessarily two-sided. EDIT: I don't know if/how this can be generalized to non-commutative rings with identity: if rr'=1 , when does it follow that r'r=1?

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#### StoneTemplePython

Gold Member
BAH. My argument here is incorrect; it does not automatically follow. It is an interesting exercise: Show that for A,B same size, AB=I , then BA=I. Not trivial that matrix inverses are necessarily two-sided.
an easy way to do this for $n$ x $n$ matrices is write the left inverse of $B$ as a linear combination of powers of $B$ (i.e. a polynomial in B). Now right multiply by $B$ and each side is thus equal to the identity matrix, but your polynomial commutes with $B$ (evaluate term by term) which means your left inverse commutes with $B$ as well and hence is an actual inverse.

- - - -
The "easy" polynomial for the above uses Cayley Hamilton. The much more basic result that is actually useful here is that these matrices live in a vectors space with dimension $n^2$ and hence there must be some monic polynomial of degree (at most) $n^2$ that annhilates them... once use you have your polynomial in B equal to zero, multiply on the left by the left inverse of $B$ a suitable number of times and move the lowest order term (the left inverse of B itself), to the right hand side, and rescale as needed. (This is where the first paragraph starts.)

#### Math_QED

Homework Helper
BAH. My argument here is incorrect; it does not automatically follow. It is an interesting exercise: Show that for A,B same size, AB=I , then BA=I. Not trivial that matrix inverses are necessarily two-sided. EDIT: I don't know if/how this can be generalized to non-commutative rings with identity: if rr'=1 , when does it follow that r'r=1?
$AB= I\implies BA = B(AB)B^{-1} = BB^{-1}= I$
An inverse exists by a determinant argument: $AB=I$ means that $\det B \neq 0 \neq \det A$

In general in non-commutative rings $rr'=1$ does not imply $r'r=1$. See e.g. here https://math.stackexchange.com/q/70777/661543

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#### fresh_42

Mentor
2018 Award
$AB= I\implies BA = B(AB)B^{-1} = BB^{-1}= I$
An inverse exists by a determinant argument: $AB=I$ means that $\det B \neq 0 \neq \det A$

In general in non-commutative rings $rr'=1$ does not imply $r'r=1$. See e.g. here https://math.stackexchange.com/q/70777/661543
Yes, but you cheated a bit: the multiplicity of the determinant, the existence of $B^{-1}$ as left and right inverse, which is what we wanted to show! The condition does not hold in arbitrary rings, but in groups. So the essential question behind the task was: Can we show that $GL(V)$ is a group and thus $AB=I \Longrightarrow BA=I$ without using linear algebra? It's easy if there exists a left inverse, but the existence will probably require facts from the definition of the group. Can this be proven, or can we conclude the existence of $CA=I$ only from $AB=I\,$?

#### WWGD

Gold Member
$BABB^{-1} =I$ 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 $n$ )?

#### fresh_42

Mentor
2018 Award
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:
$BABB^{-1} =I$ 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.

#### WWGD

Gold Member
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 :(.

#### WWGD

Gold Member
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

#### fresh_42

Mentor
2018 Award
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.

#### WWGD

Gold Member
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?

#### fresh_42

Mentor
2018 Award
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.

#### WWGD

Gold Member
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?).

#### WWGD

Gold Member
Just experimenting before aiming for a formal proof.

#### fresh_42

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2018 Award
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.

#### WWGD

Gold Member
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.

#### fresh_42

Mentor
2018 Award
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.

#### WWGD

Gold Member
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.

#### fresh_42

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2018 Award
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.

#### mathwonk

Homework Helper
the integers are a group in which all non zero subgroups are normal and isomorphic, but the quotients can be any finite cyclic group.

#### Mark44

Mentor
$(0,1) \stackrel{\operatorname{id}}{\hookrightarrow} \mathbb{R}-\{\,0\,\}\stackrel{\varphi}{\rightarrowtail} (-\frac{1}{4},\frac{1}{4}) -\{\,0\,\} \stackrel{+\frac{1}{3}}{\hookrightarrow} (0,1)$
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).

"Mistake in Schaum's Group Theory?"

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