数学代写|MATH455 Modern Algebra

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这是加州州立大学洛杉矶分校 现代代数课程的代写成功案例。

MATH455课程简介

Course Description/Objectives: Groups, subgroups, cyclic groups, permutation groups,Lagrange’s theorem, direct products, finitely generated Abelian groups, group homomorphisms,and factor groups (sometimes called quotient groups). If we have time then we will cover rings and fields.

Notes:

Notes on the dihedral group can be found here.
Lemma about the identity element and inverses in subgroups.
Lemma about what cyclic subgroups look like.
The division algorithm statement and proof
Classification of homomorphisms from cyclic groups: lemma and main theorem.
A proof that the set of permutations of a set is a group under composition.

Prerequisites 

Student learning outcomes: Students who successfully complete this course will be able to: 1.Understand the definition of a group and determine whether or not a set with a given binary operation is a group; 2. Do computations in the various groups, such as the symmetric groups,the integers modulo n, dihedral groups, matrix groups, etc. These computations include the following: taking the product of two elements, finding the order of an element, and finding the inverse of an element; 3. Understand direct products of groups and do computations in a direct product; 4. Prove or disprove that a subset of a group is a subgroup of the given group; 5.Compute the subgroup generated by a given element of a group; 6. Understand the classification theorem of cyclic groups and find all of the subgroups of a cyclic group; 7. Understand Lagrange’s theorem and its consequences; 8. Compute the cosets of a subgroup of a group; 9. Use the fundamental theorem of finitely generated groups to determine all the finite Abelian groups of a given order; 10. Understand the definition of a homomorphism; 11. Determine whether a map is a homomorphism; 12. Compute the kernel of a homomorphism; 13. Apply the fundamental (first) homomorphism theorem; 14. Understand the definition of a normal subgroup and determine whether a subgroup is normal; 15. Do computations in a factor group; 16. Prove theorems about all of the above mathematical objects.

MATH455 Modern Algebra HELP（EXAM HELP， ONLINE TUTOR）

By the symmetry between $a$ and $b$, it suffices to prove that $\operatorname{gcd}(a, n) \leq \operatorname{gcd}(b, n)$. Set $d=\operatorname{gcd}(a, n)$ for convenience. Since $a \equiv b(\bmod n), n \mid(a-b)$. Thus there is some integer $k$ with $a-b=k n$. We rearrange to $b=a-k n$. Since $d \mid a$ and $d \mid n$, we can write $a=d a^{\prime}, n=d n^{\prime}$ for some integers $a^{\prime}, n^{\prime}$. We now have $b=d a^{\prime}-k d n^{\prime}=d\left(a^{\prime}-k n^{\prime}\right)$. Hence $d \mid b$. But we also have $d \mid n$, so $d$ is a common factor of $b, n$. Thus, by definition of gcd, we have $d \leq \operatorname{gcd}(b, n)$.

The neutral additive element is 0 and the neutral multiplicative element is [1]. The zero divisors are those elements that have a factor in common with 9 , namely 3 and 6. To justify, note that $3 \odot3=6 \odot3=6 \odot6=0$.

We have $[1] \odot1=2 \odot5=4 \odot7=8 \odot8$, so 1,2,4,5,7,8 are the units.
We define $\mathbb{Z}_3 \times \mathbb{Z}_3=\left{(a, b): a \in \mathbb{Z}_3, b \in \mathbb{Z}_3\right}$, the set of ordered pairs of elements, one from $\mathbb{Z}_3$ and one from another copy of $\mathbb{Z}_3$. We define operations in the natural way, i.e. componentwise:
$(a, b) \oplus\left(a^{\prime}, b^{\prime}\right)=\left(a \oplus_3 a^{\prime}, b \oplus_3 b^{\prime}\right) \quad$ and $\quad(a, b) \odot\left(a^{\prime}, b^{\prime}\right)=\left(a \odot_3 a^{\prime}, b \odot_3 b^{\prime}\right)$.

The neutral additive element is $([0],[0])$ and the neutral multiplicative element is $([1],[1])$. Just half of the nonzero elements are zero divisors: $([0],[1]) \odot([1],[0])=([0],[2]) \odot([2],[0])=([0],[0])$. Note that although $\mathbb{Z}_3 \times \mathbb{Z}_3$ and $\mathbb{Z}_9$ both have nine elements, they have different multiplicative structure.

Just half of the nonzero elements are units (those that are not zero divisors): $([1],[1]) \odot([1],[1])=$ $([2],[1]) \odot([2],[1])=([1],[2]) \odot([1],[2])=([2],[2]) \odot([2],[2])=([1],[1])$. Note that every invertible element happens to be its own inverse, in this ring. This is quite unusual.

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