This is probably a dumb question, but I thought I would ask it.

I was reading about bullet proofs, and why they are needed. my understanding is that bullet proofs are needed to prove that all numbers in the equation are positive. It does hat by proving all the number are in a certain range (which is why it is called a range proof), and that range is 'all positive numbers'.

So, Lagrange's Four Square Theorem says all positive numbers are the sum of one to four square numbers.

So I was thinking, would it be sufficient (for the range proof part of the equation) to divide the amounts into four square numbers, transmit only the roots, and use that as the range proof?

For example, Alice has 1024, Bob has 1024, Alice wants to send bob 10.

Alice's balance is stored as 32, Bob's balance is stored as 32, and Alice sends Bob 3+1.

So the blockchain just validates that Alice has 32^2, bob has 32^2, and Alice sent Bob (3^2)+(1^2).

This is just the range proof part. Masking the amount would still be done via RingCT.

The reason I was thinking was that the range proof is still a large part of the blockchain size, and if it could be reduced to just four numbers, it would be much smaller.

Once again, if it's a dumb question, I apologize.

It seemed to work when I wrote some sample code in javascript, but I don't know everything.

There's also performance problems, like there's no good algorithm for converting an integer into a sum of squares. (Best case seems O(N^3)).

This actually has a lot more to do with the Hurwitz Quaterions than the proof as a whole. But the proof posted on wikipedia:

https://en.wikipedia.org/wiki/Lagrange%27s_four-square_theorem

I don't see how it follows that for a given alpha and beta we can find a gamma such that N(beta - alpha gamma) < N(alpha).

I see very clearly how we can find a beta for a given alpha such that N(alpha-beta)<1. And from this I can see that we can find a gamma for a given beta such that N(beta-gamma)<1 And that if we multiply by N(alpha) on both sides we get N(alpha)N(beta-gamma)<N(alpha) and that the left side can be expressed as N(alpha beta - alpha gamma) < N(alpha). I'm just not sure how we can isolate the dependence of (alpha beta) on alpha.

EDIT: New question: How does the result that given Hurtwitz quaternions α and β, we can find a Hurtwitz quaternion γ such that

N(β-αγ)<N(α)

let us know that every right ideal is principal? Also, how do we get the very next equation in the proof:

That there exists a Hurwitz quaternion α such that

αH = pH + (1-li-mj)H

I'm not very well versed in ring theory.

Lagrange's original theorem states that for any fixed values of x_1, ..., x_n and a polynomial p(x_1, ..., x_n) that takes on k values under the permutations of x_1 through x_n, we have that k divides n!. This theorem was used to try to find a quintic formula. When Galois showed that a quintic formula doesn't exist using group theory, he also generalised this theorem to groups, giving it a form that we would recognise today as Lagrange's theorem.

What other theorems have an interesting origin and story behind them?

Let f : [0; 1] -> R

f(0) = 0

f is derivable on [0; 1]

Let c be a value in (0; 1)

Prove that there exists a c so that f'(c) = -f(c)/(c-1)

I first thought of Lagrange's mean value theorem. We know f is derivable so there exists f'(c) = [f(1)-f(0)]/(1-0) <=> f'(c) = f(1).

No idea what to do with this further.

Then I realized that the exercise comes after Rolle's theorem in the book, so I should probably use that.

Well, Rolle's theorem tells us that if f is derivable on [0; 1] then there could be two values a, b on that interval such that f(a) = f(b) and the function changes sign between a and b. But it's not a must. And even if f(0) = f(1), where do I go from here?

Hey yall, I'm having trouble grasping the concept of a coset and how it relates to the order of a group, Lagrange's theorem and the index of a subgroup.

If anyone could help me by explaining what it is, how it's helpful in understanding groups and subgroups and the things I listed above, I would be super thankful.

Prove that a group having finitely many subgroups is finite.

Hey guys,

I'm taking an abstract algebra course and missed the day we covered cosets and lagranges theorem. So far I've studied on my own and understand that cosets help narrow down the possibilities for subgroups (being that the order of the subgroup must divide the order of the mother group) but I feel there is something more fundamental that I am missing. We usually go very quickly in class and it didn't take me very long at all to understand this concept at home. What further is needed to understand and utilize lagranges theorm and cosets?

Thanks!

Title says it all. I need to prove that Z/5Z is the only valid group of order five without using Lagrange's theorem. Do I have to write out all 125 possible permutations of this, and show that each one isn't a group??? That seems weird for just a regular homework problem. Am I being dumb and there's less valid permutations than I think?

Hi, I've been working through a script on group theory by myself, and there's a problem I'm having difficulty solving. Basically, the only thing I know so far is the definition of a group, some examples of groups as well as some basics like orders, left multiplication being bijective and so on. Now, I'm being hit with the following:

1. Prove x^|G| = 1_G for all finite, abelian groups.

I am aware that this is true for any group, but the question only asks for a proof for abelian ones.

Further, it asks me to show that Z_p is the only group of order p (with p prime). I've been able to solve this by assuming the truth of 1., but I'm really not getting how to prove 1. with the limited knowledge I have! It's obviously equivalent to ord x / |G| but I can't seem to find any connection between a groups order and its elements.

I'm not necessarily looking for a solution. Just a header in the right direction would be great.

Thanks in advance.

The Wikipedia article for Lagrange theorem says the infinite version,

for all groups G and subgroups H, card G = card H x | G : H |

is equivalent to the axiom of choice.

It seems plausible, but I don't have a proof.

UPDATE

Call the statement above (LT) for Lagrange theorem. According to this question in mathoverflow, https://mathoverflow.net/questions/256352/is-lagranges-theorem-equivalent-to-ac a statement stronger than Lagrange theorem, LT+, implies choice, which still leaves open whether LT implies LT+.

Find a subgroup of each possible subgroup order of S4, giving brief justification for your answers.

Now, the first part is straightforward enough - the order's being 1, 2, 3, 4, 6, 8, 12, 24. And then subgroup of order 1 is the identity, 24 is S4 itself and I believe order 12 is A4. But as far as the other orders go, how do I work out example subgroups?

Hi! While reviewing for my undergraduate algebra exam, I read, while reviewing, that extending LaGrange's Theorem for finite groups into the infinite gets you the Axiom of Choice? Can someone explain this to me? Wikipedia is surprisingly sparse on this. Also, I'm working with a relatively primitive understanding of both LaGrange's Theorem and AC. For LaGrange's Theorem, I understand it in terms of distinct left cosets of the subgroup (using the Judson Abstract Algebra text if that helps).

The other night my buddy and I were hanging out in an odd group. I tried to interact with them, but it wasn't going anywhere. At one point I even said to my buddy "I wish it was just us. No homo."

question

https://i.imgur.com/Atp2eZ3.png

a

Assume that H < G and that H is neither G nor {e}

then we can choose some element g1 \in G such that g1 is not in H. Using this we can create a left coset g1*H.

• P1. the intersection of this coset and H is the empty set
• P2. The size of this coset is equal to H.

P1 : We have chosen that g1 is not in H.

Suppose that there's some hi, hj in H such that hi != hj

at that

g1 * hi = hj

then

g1 = hj * hi^-1

But g1 was chosen such that it's not in H, this is a contradiction.

P2 : Suppose that there are duplicate elements in some coset g*H

choose h1 != h2 Then

  g * h1   = g * h2

g^-1 g * h1 = g^-1 g * h2

h1 = h2

But these were chosen such that they weren't equal, therefore there are no duplicate elements.

So there's some element in g*H for every element in H, they're the same size.

We've constructed one coset with g1, we could construct another using some g2 and show that the elements of g1H and g2H are disjoint using a similar approach.

We can show that for any element of G that's not already in a coset, it will create a coset sized |H| , using the above approach.

So say that |G| = a

and that |H| = b

And that G is covered with c cosets

Then

a = b * c

And this shows that

|G| = |H| * c

|G| / |H| = c

Which is what was needed ( that |H| divides |G|)

b)

If |G| = p^2 where p is prime, then the factors of |G| are 1, p, p^2.

If H is a proper subgroup of G , and we don't consider H = {e}, then by Lagranges theorem |H| must divide |G|, and the only available size of |H| is then p.

c

The proper subgroups of Z/(p^2) are {e} and Z/(p)

d

I've no idea about this

As the title says, I'm trying to show that for a finite Abelian group G of order n, G has a subgroup of order m for every m that divides n. I can do it with the Sylow theorems, but I have an algebra book that requests a proof well before such machinery is developed and I can't find a good way to do so.

My reasoning: the theorem certainly holds for groups of small order, so let's assume it holds for orders < n and try induction. If G has an element of order n, we're done, so let's say it has an element x of order m. Then <x>, being normal in G, is the kernel of a homomorphism h. Since G/<x> ≅ h(G) we know h(G) has order n/m, and by the induction hypothesis has subgroups of all dividing orders, all of whose pullbacks under h are subgroups of G. So this gets us lots of subgroups of G, but not necessarily of all dividing orders. How do I get the rest? Thanks!

Hello! Trying to figure this theorem out a couple steps along the way and I'm already stuck at one of the first. Just in case, here's the theorem: > Let p be a prime number, and let f(x) be a polynomial of degree n >= 1, not all of whose coefficients are divisible by p. Then the congruence f(x) ≡ (mod p) has at most n solutions in a complete residue system modulo p.

The proof is one of induction and the first step is proving that n = 1 satisfies the theorem. The book I'm using says: > If f is of degree 1, then f(x) = ax + b for some a and b. If p ∤ a, then ax + b ≡ 0 (mod p) has a unique solution. If p | a and p ∤ b, then there is no solution.

This might be dumb, but how do they conclude that?

My thinking (I'm not too sure) but: > you can move b to the other side so that ax + b ≡ 0 (mod p ) -> ax ≡ -b (mod p) and if p doesn't divide a, then there must exist an x such that the statement is true? I'm not too sure why this is the case.

> As for the other I'm a little more confident if p divides a, then ax ≡ 0 (mod p) so it turns to ax + b ≡ b ≡ 0 (mod p) and if p doesn't divide b then the statment doesn't make sense.

Anyways, please feel free to point out what I'm missing. Thanks a ton!