• 沒有找到結果。

1 PROBLEM SET 1 DUE: FEB.24 Problem 1 (1) Let G be the set of integers which can be written as the sum of two squares, that is, G = {x ∈ Z|x = a

N/A
N/A
Protected

Academic year: 2021

Share "1 PROBLEM SET 1 DUE: FEB.24 Problem 1 (1) Let G be the set of integers which can be written as the sum of two squares, that is, G = {x ∈ Z|x = a"

Copied!
2
0
0

加載中.... (立即查看全文)

全文

(1)

1

PROBLEM SET 1 DUE: FEB.24 Problem 1

(1) Let G be the set of integers which can be written as the sum of two squares, that is, G = {x ∈ Z|x = a2+ b2, a, b ∈ Z} . Show that G is a monoid under the multiplication of integers.

(2) Let H be the set of integers which can be written as the sum of four squares, that is, H = {x ∈ Z|x = a2+ b2+ c2+ d2, a, b, c, d ∈ Z} . Show that H is also a monoid under the multiplication of integers.

*(3) Let N be the set of integers which can be written as the sum of eight squares, that is, N = {x ∈ Z|x = a21+ a22+ a23+ a42+ a25+ a26+ a27+ a28, a1, ...., a8∈ Z} . Is it true that N is also a monoid under the multipli- cation of integers?

Problem 2

Consider the lattice Z2⊂ R2, the points on the plane which has inte- gral coordinates, and let C be the convex cone formed by two rays starting from the origin (0, 0) (where we assume these two rays do not lie in the same straight line). Then show that S = C ∩ Z2 is a monoid. Try to prove that the group generated by the monoid S is Z2.

Problem 3

Show that every group of order ≤ 5 is abelian.

Problem 4

Show that there are only two non-isomorphic groups of order 4, namely the cyclic one, and the product of two cyclic groups of order 2.

Problem 5

Let G be a group such that Aut(G) is cyclic. Prove that G is abelian.

Problem 6

Prove the following statement:

(2)

2

Let G be a group, S a set of generators for G, and G0 another group.

Let f : S → G0 be a map. If there exists a homomorphism f of G into G0 whose restriction to S is f , then there is only one.

Problem 7

Let G be a group and let H ,H0 be subgroups. By a double coset of H ,H0 one means a subset of G of the form HxH0.

(1) Show that G is a disjoint union of double cosets.

(2) Let {c} be a family of representatives for the double cosets. For each a ∈ G denote by [a]H0 the conjugate aH0a−1 of H0 . For each c we have a decomposition into ordinary cosets

H =[

c

xc(H ∩ [c]H0)

where xc is a family of elements of H ,depending on C. Show that the elements xcc form a family of left coset representatives for H0 in G; that is,

G =[

c

[

xc

xccH0,

and the union is disjoint.

Problem 8

Let G be a group and H a subgroup of finite index. Prove that there is only a finite number of right cosets of H , and that the number of right cosets is equal to the number of the left cosets.

參考文獻

相關文件

The first case occurs when #1 is the second candidate after the sub- ordinate player’s rejection point and the best applicant before the subor- dinate player’s rejection point is

More precisely, it is the problem of partitioning a positive integer m into n positive integers such that any of the numbers is less than the sum of the remaining n − 1

An alternative way to proceed would be to directly estimate that one of the unknown constants C 1 :::C 4 which is appropriate to the context (e.g. C 1 if we were using the excess

好了既然 Z[x] 中的 ideal 不一定是 principle ideal 那麼我們就不能學 Proposition 7.2.11 的方法得到 Z[x] 中的 irreducible element 就是 prime element 了..

Let T ⇤ be the temperature at which the GWs are produced from the cosmological phase transition. Without significant reheating, this temperature can be approximated by the

The observed small neutrino masses strongly suggest the presence of super heavy Majorana neutrinos N. Out-of-thermal equilibrium processes may be easily realized around the

Given a connected graph G together with a coloring f from the edge set of G to a set of colors, where adjacent edges may be colored the same, a u-v path P in G is said to be a

Elsewhere the difference between and this plain wave is, in virtue of equation (A13), of order of .Generally the best choice for x 1 ,x 2 are the points where V(x) has