Tag: algebra

abstract algebra: how many non-isomorphic finite groups with exactly conjugation classes $ n $ are there?

How many non-isomorphic finite groups with exactly $ n $ are there conjugation classes?

All I could prove was that their number is finite for everyone $ n in mathbb {N} $:

Suppose $ G $ has $ n $ conjugation classes $ C_1, … C_n $and suppose $ forall i <n $ $ g_i in C_i $. Then $ | C_i | = (G: C (g_i)) $. From that we can conclude, that $ 1 = | G | ^ {- 1} ( sum_ {i = 1} ^ n | C_i |) = | G | ^ {- 1} ( sum_ {i = 1} ^ n (G: C (g_i))) = sum_ {i = 1} ^ n frac {1} {C (g_i)} $. And it is known that there are only many fine ways to partition $ 1 $ in sum of $ n $ Egyptian fractions

abstract algebra: identity element in a commutative rng.

Let me start with a construction. Take a commutative rng $ A $ and for every finite generated $ A $-module $ M $ take the endomorphism ring $ mathrm {End} _A (M) $ which is itself a commutative ring with identity. Let rng homomorphism $ h_M: A to mathrm {End} _A (M) $ defined by mapping $ a in A $ to the map $ m mapsto am $. Now for the fact that $ M $ finite is generated over $ A $ we must have something $ a in A $, $ h_M (a) $ maps to the identity map in $ mathrm {End} _A (M) $(To see this, apply Cayley-Hamilton to the identity map in $ mathrm {End} _A (M) $ to get an item $ a in A $ such that $ am = m $ for all $ m in M ​​$.) So the image $ h_M (A) $ It is a commutative ring with identity. Note that $ h_M $ it is injective if and only if $ M $ is faithful In other words,

A commutative rng $ A $ it is a ring if and only if there is a finite generated faithful module on $ A $.

I had many questions when trying to get an idea of ​​this construction. So I guess the best answer should be a reference book / article on this topic, but let me write some of my questions:

  1. How canonical is $ h_M $ for each finitely generated module $ M $? Given another finitely generated module $ N $ above $ A $ Is there a unique automorphism $ f in mathrm {Aut} _A (M) $ such that $ h_N = fh_Mf ^ – $ 1?

  2. How does the information for the map work? $ h_M $ for each finitely generated module $ M $ above $ A $ Tell us about the "obstruction" to add an identity to the ring $ A $? This question may be a bit vague, but you should be looking for a "homological" answer.

  3. What are the maps? $ h_M in mathrm {End} _A {M} $ for finely generated module $ M $ above $ A $ tell us about the ring $ A $?

Again, the best answer I would like to get should be a reference that includes this topic to study. Thanks in advance.

complexity theory: Oracle for idempotent algebra

What type of oracle would you need to find the lowest homogeneity term in the expansion of $ phi = (a + b – ab) otimes (b-bc) otimes (a + c-2ac) $ where $ otimes $ It is an idempotent product, that is, $ a otimes a = a $

Second, what type of oracle would I need if I wanted a vector of the sum of coefficients for homogeneous terms in the expansion of $ phi $?

Thanks in advance, Ben

Algebraic equations: problem to follow the steps of algebra

They asked me a question with the solution. However, I have trouble following the mathematical steps that have been taken in the solution. I'm not sure how they went from step 1 to step 2, as shown below:

$ Step 1 … 7 = Vgs + 5 (Vgs – Vt) ^ 2 $

$ Step 2 … 7 = Vgs + 5 (Vgs ^ 2 – 2VgsVt + Vt ^ 2) $

Thanks for any help.

abstract algebra: is the image of a cyclic group $ G $ a normal subgroup?

It is the image of a cyclic group under homomorphism. $ phi: G → G & # 39; $ always
a normal subgroup of $ G & # 39; $?

I think the answer is yes like $ G $ is abelian (since it is isomorphic to $ Z $ or $ Z_n $) So $ phi (ab) = phi (a) phi (b) = phi (ba) = phi (b) phi (a) $. Therefore, the image is also Abelian. However, I am not sure if this shows that it is normal as $ G & # 39; $ It is not necessarily Abelian. Still, we choose $ j in phi (G) $ and $ g in G & # 39; $. We want to show that $ g ^ – 1 jg in phi (G) $. I am stuck in this step since I feel that I lack some key property for everything to work. Any help is appreciated, thanks

reference request – Artin Algebra v / s Herstein Algebra Topics

I am going to take Abstract Algebra next semester and I would like to start now. The book assigned to the class is Herstein's Topics in Algebra, but I have heard that Artin's Algebra is superior.

Now, just that information alone is enough to convince me to choose Artin, but I'm at a crucial moment here. I work full time and, as a result, I can never get to class. So I have to study everything on my own and, due to full time work, I have to manage my time very well.

Given time management and efficiency, should I abandon Herstein for Artin, or should I continue with the assigned textbook?

I ask this because one can be more challenging or have more material than the other, so it may take longer to address the same topic as the other book.

PD: I know the topics that will be discussed. And I also have Hungerford Algebra and D&F Abstract Algebra in addition to Herstein and Artin.

abstract algebra: determination of the ratio of the symmetric group with the group of even permutations

Determine the group of $ S_n / A_n $ until isomorphism

Intuitively, this is the group of strange permutations. However, I am not sure what is isomorphic. I may also be misunderstanding precisely what this problem is asking. Any help is appreciated, thanks.

linear algebra – Nonlinear pendulum solution

I was looking for a solution in the nonlinear pendulum equation and I was wondering why we multiply the equation by dθ / dt when we can integrate directly. I didn't find an answer on the internet except

& # 39; & # 39; Equation (1), although simple in appearance, is quite difficult to solve due to the nonlinearity of the term sinθ. To get the exact solution of the equation. (1), this equation is multiplied by dθ / dt & # 39; & # 39;

Can anyone explain to me why we are doing this? Am I missing something?

linear algebra – Matrix square root operator standard versus original

If I have a non-symmetric matrix whose operator standard is $ leq 1 $ and the square root, its operator norm remains below $ 1 $?

More formally, I want to know if there is always at least one square root for which it is the case, even if it is not true for all of them and under reasonable assumptions that guarantee that there is a square root.

Specifically, suppose I have a non-symmetric square matrix $ A $ with $ | A | _2 leq 1 $ where $ | cdot | $ 2 denotes the norm of the operator (maximum singular value). Is there always a square matrix $ B $ such that $ B ^ 2 = A $ and $ | B | _2 leq 1 $?

I am willing to assume $ A $ It is diagonalizable, although not unitarily diagonalizable. I am also willing to assume $ A $ it's the way $ A & # 39; + Delta $ by some small random disturbance $ Delta $ and matrix $ A & # 39; $. Without at least one of these assumptions, the square root may not exist at all.

I would be interested in a test / counterexample under any non-empty subset of these assumptions. For example, a counterexample that shows that the statement can be false when we assume $ A $ It is diagonalizable would be useful to me.

Homological Algebra – Bimodule Resolutions

I have asked this question about the exchange of Mathematics Stack but I still haven't got an answer. So, I'm asking here.

Let Be an algebra of finite dimension. Let M be a left A module and
N be a correct module A. Choose an injective resolution $ E_1 ^ * $ from $ M $ in $ A $-mod and an injective resolution $ E_2 ^ * $ from $ N $ in mod-$ A $. So, we have $ E_1 ^ * otimes E_2 ^ * $ it is an injective resolution of $ M otimes N $ in the category of $ A $–$ A $-bimodules? I could show that $ E_1 ^ * otimes E_2 ^ * $ are injectable objects in the category of $ A $–$ A $-bimodules. But how to show that it gives a resolution?

Is this also true in case of projective resolution?

Thanks in advance!