Gromov's Theorem On Groups Of Polynomial Growth Puns

So, the Chinese remainder theorem for a polynomial ring R = F[X*1, X2, ..., Xn] (let field F = ℝ) states that if I1, I2, ..., Im* are pairwise coprime ideals of R and I = I*1* ∩ I*2* ∩ ... ∩ I*m, then R/I is isomorphic to R/I1* × R/I*2* × ... × R/I*m. Let us denote this isomorphism by f. For polynomial p in R, one can compute f(p + I): first, construct Gröbner bases for ideals Ij* by using Buchberger's algorithm, and then use general polynomial division to get the remainders r*j* of p when divided by Gröbner bases of I*j, 1≤j≤m, and then f(p + I) = (r1, r2, ..., rm*). So, it goes in one way.

Now my question is: Is there a constructive way to compute p + I ∈ R/I if remainders r*j* are known? In other words, does there exist an algorithm for inverse of f?

For the following polynomials use the Bounds Theorem to determine an upper bound for the modulus of their roots.

6x^2 + 3x^3 + 2x + 12

10.5x^3 +12x + 3

15x^4 + (8-6i)x^3 + 3x + x^2 +1

I 'think' I know Bounds theorem but this is unlike any other problem we have done so far.

Graphing x^3 -4x^2 +4x +8, you only get one root, approx. (-0.931, 0). What is the explanation for this?

I missed a number theory class but am familiar with lagrange interpolation. The problems on my pset lend themselves to lagrange interpolation and i read that the process of lagrange interpolation is related to the CRT. Can anyone explain how or direct me to a good resource?

So the issue I'm having is in a part of the proof where we suppose that G has no elements of order p^2 and conclude that all (non-identity) elements of G have order p. From there Gallian wants to show that for any 'a' in G, the subgroup <a> is normal by a proof by contradiction. Here's where I'm having trouble. So we suppose that <a> isn't normal, and Gallian claims there has to exist some 'b' in G such that bab^-1 is not an element of <a>, therefore <a> and <bab^-1 > are distinct subgroups.

I don't know why <a> not being normal implies that bab^-1 isn't an element of <a>, when the coset rules state that a subgroup, H, of G is normal if and only if xHx^-1 is contained in H for all x in G. So if Gallian is trying to show that <a> isn't normal, shouldn't he be trying to show that b<a>b^-1 isn't contained in <a>? I guess the question I'm asking is, how does he know specifically that 'a' itself doesn't commute with 'b'? Couldn't 'a' commute with 'b' and it's some other element of <a> that doesn't commute with 'b'? Like, wouldn't it be more rigorous for Gallian to argue something like, "Then there exists some 'b' in G such that b(a^i )b^-1 is not an element of <a>, where 0<i<p"? It just bothers me that he assumes it has to be 'a' without demonstrating it.

Edit: A secondary question: From the fact (which I'm still unsure about) that <a> being not normal implies bab^-1 not in <a>, and <a> and <bab^-1 > are both distinct subgroups of order p, Gallian states that the intersection of <a> and <bab^-1 > is {e}. I don't know exactly how he comes to that conclusion, but here is my reasoning, I'd like somebody to tell me if this is valid:

Since bab^-1 isn't an element of <a>, I can say that 'a' =/= bab^-1 . Therefore a^2 =/= (bab^-1 )^2 . Likewise a^3 =/= (bab^-1 )^3 .... and so on up to p-1: a^(p-1) =/= (bab^-1 )^(p-1). So the only time they are equal is when 'p' is the exponent and they both equal 'e'. I'm just not sure I'm allowed to treat the not-equals sign (=/=) like that.

Edit 2: Third question. After showing that the intersection of <a> and <bab^-1 > is just {e}, Gallian says that the left cosets of <bab^-1 > in <a> are given by a<bab^-1 >, a^2 <bab^-1 >, a^3 <bab^-1 >,...,a^(p-1) <bab^-1 >. Then he says that b^-1 has to be contained in one of these cosets.

*Thm 1: Let G be a group of order 2p, where p is a prime greater than 2. Then G is isomorphic to Z_2p or D_p*

In the part of the proof where Gallian tries to prove that G~D_p (I am using ~ to represent "is isomorphic to"), he assumes G is non-cyclic so that its elements have either order 2 or order p, shows that it is impossible for all non-identity elements to be order 2, so he calls an element of order p "a". He then picks an element of G, "b", that is not an element of <a>. This is how he shows that |b|=2, and where I started having trouble following:

&gt;...we have that |b| = 2 or p. Because |<a>INTR<b>| divides |<a>|=p and <a>=/=<b> we have that |<a>INTR<b>|=1.

**I understand that |<a><b>|=|<a>||<b>|/|<a>INTR <b>| but I don't know how he got |<a>INTR<b>| divides |<a>| from that.**

Later in the proof, after showing |b|=2, he shows that |ab|=2, and therefore that ab=ba^-1 , he states:

&gt;...this relation completely determines the multiplication table for G... Since the multiplication table for all noncyclic groups of order 2p is uniquely determined by the relation ab=ba^-1 , all noncyclic groups of order 2p must be isomorphic to each other.

*I get that the relation allows any member of G (or any non-cyclic group of order 2p) to be written in the form a^i b^j or b^j a^-i but I don't get how that means all such groups are isomorphic.* With things like subgroups, I'm used to applying tests to determine that they are subgroups, and my instinct is to do the same when proving that something is an isomorphism. But Gallian seems to assume that it is trivial that an isomorphism exists, and I couldn't find one on my own.

As for the second theorem: *The group of rotations of a cube is isomorphic to S_4.*

Gallian starts out the proof by saying:

&gt;Since the group of rotations of a cube has the same order as S_4, we need only prove that the group of rotations is isomorphic to a subgroup of S_4

I don't understand why that is the case at all.

Gallian gives the designations 1,2,3,4 to four diagonals connecting the 8 corners of the cube (i.e. connecting the back right corner of the top face to the front left corner of the bottom face), and states that the rotation group induces a permutation group on the diagonals. He then goes on to try to show that there are 24 such permutations induced by the rotations. In d

This was one of the questions on my final exam in algebra 2; I’ve got most of a solution if anyone attempts it 😄

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?

As the title says, I'd like to know if there is a simple proof for the triviality of the nth homotopy group of S^k if n<k without talking about CW-complexes and the cellular approximation theorem.

Does anyone know any way to do this? Maybe some way to show that every map from S^n to S^k is homotopic to a map which is not surjective?

EDIT: I am also interested if someone knows a way to skip "some" of the details of the cellular approximation theorem, it doesn't have to be a completely different way. I used the triviality of these homotopy groups in a proof and would like to show why these groups are trivial without taking too much time.

I'm sure this is well known, but here is an elementary proof of the Fundamental Theorem of Arithmetic using group theory.

Let k be a natural number greater than 1 and let G be Z_k . It is easily proven k has some prime factorization p_1 ^n_1 ... p_r^n_r . As well, there is a subgroup for every factor of k, let P_i be any subgroup of order p_i ^n_i . The pairwise intersection of different P_i must be trivial by Lagrange's theorem, so we can compute the size of P_1 P_2 ... P_r to be exactly the product of |P_1||P_2|...|P_r|=k.

Let p^s be the highest power of some prime that appears in a prime factorization of k. It corresponds to a subgroup P of G of order p^s . It's pairwise intersection with any P_i is either trivial or of order p^t by Lagrange's theorem. If its intersection is trivial I can compute P_1 P_2 ... P_r P to have k p^s elements, so s must be 0. Otherwise it is of order k p^s /p^t , so s must equal t meaning it lies entirely in some P_i .

Collecting this information we have that if p^s appears in the prime factorization of k, p must be one of the p_i to a power at most n_i . If they were anything less than n_i we would have something less than k, so they must be n_i . The primes and their powers are unique.

I don't think this proof begs the question at all, but let me know if there is some part that seems circular.

Edit: This proof relies on the only factors of a prime power being that prime to a smaller power. You can prove this by induction. Let G be Z_p^s . Suppose q is a prime not equal to p dividing p^s . If q does not divide p^(s-1) then I can say the following: There is a group P of order p^(s-1) and a group Q of order q. Their intersection is trivial, so |PQ|=p^(s-1) q. Both P and Q are normal, so PQ is a subgroup and its order divides p^s . Doing this division we get p/q is an integer, so no such q exists since p is prime. The base case is showing for all primes p they are not divisible by different primes q which is by definition.

In the theory of Lie groups there's a really nice theorem which states that if G is a compact connected Lie group, and T<G is a maximal torus, then every element of G is conjugate to an element of T.

Does this theorem also hold for finite groups? (Replacing maximal torus with maximal Abelian subgroup). I'm interested in particular in the case of the finite symplectic group over GF(p) where p is a prime. I've looked hard into the literature, but it's pretty dense so if somebody could point me to where to find the answer, it would be super useful!

Okay this seems so obvious that I'm scared to ask, but I have not seen anyone explain things this way:

We can view growth in terms of the number of leaves in a tree where the # branches are limited to n. This tree structure might be a decision tree modeling a computation (for example). Now what's the difference between a polynomial and exponential growth when looking at the structure of these trees?

Well, polynomial growth can be thought of as increasing the number of branches in our tree only. So O(n^2) is a tree of depth two with n branches at each vertex. Notice the tree doesn't growth in length when viewed this way. n^k would also have a fixed depth k.

While exponential growth will cause this same tree to grow in depth.

So polynomial problems result in trees which don't need to grow in length while exponential problems require it. That's visually very clear, but does it capture the difference?

I should mention that these trees are nice because they are both polynomial time verifiable in the sense that checking a solution is equivalent to walking down a single path of the tree. So they are obviously not exponential to verify.

----Update

cost = # leaves of tree which has n branches at each vertex

n^k - As n grows this tree doesn't grow in depth

k^n - As n grows this tree does grow in depth

So the difference I'm focused on is the ability to draw a line in the sand and say that, when viewed using this analogy, certain problems only grow 'wide' while others must grow long. It's a physical analogy you could animate, and the the count of leaves would be your Y axis. That would better demonstrate the reason for the curvature

And this is due to that limitation on vertex degree being n

So if someone does a polynomial time reduction of a problem in NP, what they are actually doing is preventing the tree from growing in depth.

I recently joined and saw it jump from 357 thousand to 358 thousand. I'm curious what it was a month ago, 6 months, 1 year, etc.

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He really made a difference.

Hey guys question from first year engineering undergrad here. We are about halfway through the linear algebra course and it's been pretty fun. I was just wondering, are systems of nonlinear equations something that's looked at and worked on often in math? If so, does it also have deep connections to other fields in math? When I asked my prof what he researched on he replied with "lie algebra," which he described to be similar to linear algebra but much deeper. Is there an analogous for nonlinear systems?