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# 数学代写|抽象代数代写Abstract Algebra代考|Homomorphisms and kernel

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## 数学代写|抽象代数代写Abstract Algebra代考|Homomorphisms and kernel

To begin our second part of our survey of group theory, consider the following two functions: $\phi: n \mathbb{Z} \rightarrow \mathbb{Z}$ given by $\phi(x)=x$, and the function $\psi: \mathbb{Z} \rightarrow \mathbb{Z}_n$ given by $\psi(x)=r$, where $r$ is the remainder of $x$ divided by $n$. Neither map is an isomorphism, since $\phi$ is not surjective and $\psi$ is not injective. Yet despite this fact, the maps do preserve the group structure of the domain; that is, $\phi(x+y)=\phi(x)+\phi(y)$ and $\psi(x+y)=\psi(x)+\psi(y)$ (this latter fact is tedious to check, but true nonetheless). Hence, while neither map matches the two groups up perfectly, they at least relate the group structures faithfully. Consequently, we’ll define such maps precisely and turn our attention to their properties.

Definition 6.1. Let $G$ and $G^{\prime}$ be groups. A function $\phi: G \rightarrow G^{\prime}$ is a (group) homomorphism from $G$ to $G^{\prime}$ if and only if $\phi(a b)=\phi(a) \phi(b)$ for all $a, b \in G$.

Notice then that an isomorphism is simply a bijective homomorphism. It’s worthwhile to see what properties of isomorphisms still hold even when the homomorphism isn’t bijective.
Theorem 6.2. Let $\phi: G \rightarrow G^{\prime}$ be a group homomorphism and let $H<G$.
(1) The image of the identity of $G$ under $\phi$ is the identity of $G^{\prime}$.
(2) $\phi\left(g^{-1}\right)=\phi(g)^{-1}$ for all $g \in G$.
(3) $\phi\left(g^n\right)=(\phi(g))^n$ for all $g \in G$ and integers $n$.
(4) If $g \in G$ has finite order, then $\phi(g)$ has finite order and is a divisor of the order of $g$.
(5) $\phi(H)<G^{\prime}$
(6) If $H$ is abelian, then $\phi(H)$ is abelian.
(7) If $A \subset H$ generates $H$, then $\phi(A)$ generates $\phi(H)$.

## 数学代写|抽象代数代写Abstract Algebra代考|Normal subgroups

Theorem 6.10 gives us the opportunity to do something truly innovative. Recall that an isomorphism matched elements from two groups in such a way that the operations on the two groups also matched. A homomorphism still matches the operation, but the function need not be bijective, so individual elements aren’t always matched up perfectly. However, we just saw that it’s not the individual elements that are matched up: it’s the cosets of the kernel that are matched with elements of $G^{\prime}$. Does that mean that, somehow, we can put a group operation on cosets? What would that even mean?
Let’s first play with the kernel $K$ of a group homomorphism $\phi: G \rightarrow G^{\prime}$. If we pick two elements $a^{\prime}, b^{\prime} \in G^{\prime}$ and take their inverse images, we’ll get two cosets $a K$ and $b K$, where $\phi(a)=a^{\prime}$ and $\phi(b)=b^{\prime}$. On the other hand, if we take the inverse image of the product $a^{\prime} b^{\prime}$, we’ll get some coset $c K$, where $\phi(c)=a^{\prime} b^{\prime}$. But wait: since $\phi$ is a homomorphism, we know that $\phi(a b)=\phi(a) \phi(b)=a^{\prime} b^{\prime}$. Thus, we can choose $c=a b$. It therefore makes sense that we would want to say something like, “Define the product of cosets by $a K \cdot b K=(a b) K$.” Will this always work?

Let’s begin with a group $G$ and an arbitrary subgroup $H<G$. What we’re going to attempt is to define an operation on the set $G / H$ of left cosets of $H$ in $G$. The previous paragraph gives us what looks like the natural binary operation to use: given two left cosets $a H, b H \in G / H$, define
$$(a H)(b H)=(a b) H$$
But any time we define a function on cosets – and a binary operation is a function – we have to prove that the function is well-defined. This is now our first crucial theorem.
Theorem 6.11. Let $G$ be a group and $H<G$. Then the binary operation on $G / H$ given by $(a H)(b H)=(a b) H$ is well defined if and only if $g H=H g$ for all $g \in G$.

In other words, this operation makes sense – and only makes sense – when the left and right cosets of $H$ in $G$ are the same. These subgroups form the backbone of much of group theory.

Definition 6.12. Let $G$ be a group and $H<G$. The subgroup $H$ is a normal subgroup of $G$ if and only if $g H=H g$ for all $g \in G$. If $H$ is a normal subgroup of $G$, we write $H<G$.

Using this definition, the operation $(a H)(b H)=(a b) H$ is well-defined if and only if $H$ is a normal subgroup of $G$. Let’s now verify that $G / H$ is a group under this operation when $H$ is a normal subgroup of $G$.

## 数学代写|抽象代数代写Abstract Algebra代考|Homomorphisms and kernel

(一)身份形象 $G$ 在下面 $\phi$ 是的身份 $G^{\prime}$.
(2) $\phi\left(g^{-1}\right)=\phi(g)^{-1}$ 对全部 $g \in G$.
(3) $\phi\left(g^n\right)=(\phi(g))^n$ 对全部 $g \in G$ 和整数 $n$.
(4) 如果 $g \in G$ 有有限阶，那么 $\phi(g)$ 具有有限阶并且是阶的除数 $g$.
(5) $\phi(H)<G^{\prime}$
(6) 如果 $H$ 是交换矩阵，那么 $\phi(H)$ 是阿贝尔的。
(7) 如果 $A \subset H$ 产生 $H$ ，然后 $\phi(A)$ 产生 $\phi(H)$.

## 数学代写|抽象代数代写Abstract Algebra代考|Normal subgroups

$$(a H)(b H)=(a b) H$$

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