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## 数学代写|抽象代数作业代写abstract algebra代考|Subgroups Generated by a Subset

Let $S$ be a subset of a group $G$. Since subgroups of $G$ are closed under the operation and under taking inverses, any subgroup of $G$ that contains $S$ must contain all elements obtained by repeated operations or inverses from elements in $S$.
Definition 1.7.1
Let $S$ be a nonempty subset of a group $G$. We define $\langle S\rangle$ as the subset of words made from elements in $S$, that is to say
$$\langle S\rangle=\left{s_{1}^{\alpha_{1}} s_{2}^{\alpha_{2}} \cdots s_{n}^{\alpha_{n}} \mid n \in \mathbb{N}, s_{i} \in S, \alpha_{i} \in \mathbb{Z}\right}$$
Note that the $s_{i}$ are not necessarily distinct.
Proposition 1.7.2
For any nonempty subset $S$ of a group $G,\langle S\rangle \leq G$.
Proof. First of all $\langle S\rangle$ is nonempty since it contains $S$. For any two elements $x=s_{1}^{\alpha_{1}} s_{2}^{\alpha_{2}} \cdots s_{n}^{\alpha_{n}}$ and $y=t_{1}^{\beta_{1}} t_{2}^{\beta_{2}} \cdots t_{m}^{\beta_{m}}$ in $\langle S\rangle$, we have
$$x y^{-1}=s_{1}^{\alpha_{1}} s_{2}^{\alpha_{2}} \cdots s_{n}^{\alpha_{n}} t_{m}^{-\beta_{m}} \cdots t_{2}^{-\beta_{2}} t_{1}^{-\beta_{1}} .$$
This product is again an element of $\langle S\rangle$ so by the One-Step Subgroup Criterion $\langle S\rangle \leq G$

By virtue of Proposition $1.7 .2,\langle S\rangle$ is called the subgroup generated by $S$. (In the analogy with vector spaces, a subgroup generated by a subset is like the span of a set of elements in a vector space, which is a subspace.)

## 数学代写|抽象代数作业代写abstract algebra代考|Center, Centralizer, Normalizer

Proposition $1.7 .2$ gave us a way to construct subgroups of a group. However, a number of subsets defined in terms of equations also turn out to always be subgroups. Many play central roles in understanding the internal structure of a group so we present a few such subgroups here.

Proof. Note that $1 \in Z(G)$, so $Z(G)$ is nonempty. Let $x, y \in Z(G)$. Then
$$(x y) g=x(y g)=x(g y)=(x g) y=(g x) y=g(x y)$$
so $Z(G)$ is closed under the operation. Let $x \in Z(G)$. By definition $x g=g x$ so $g=x^{-1} g x$ and $g x^{-1}=x^{-1} g$. Thus, $x^{-1} \in Z(G)$ and we conclude that $Z(G)$ is closed under taking inverses.

Note that $Z(G)=G$ if and only if $G$ is abelian. On the other hand, $Z(G)={1}$ means that the identity is the only element that commutes with every other element. Intuitively speaking, $Z(G)$ gives a measure of how far $G$ is from being abelian. The center itself is an abelian subgroup. However, $Z(G)$ is not necessarily the largest abelian subgroup of $G$.

Example 1.7.9. Let $F$ be $\mathbb{Q}, \mathbb{R}, \mathbb{C}$, or $\mathbb{F}{p}$ (where $p$ is prime). In this example, we prove that $$Z\left(G L{n}(F)\right)={a I \mid a \neq 0},$$
where $I$ is the identity matrix in $\mathrm{GL}{n}(F)$. By properties of matrix multiplication, for all matrices $B \in \mathrm{GL}{n}(F)$ we have $B(a I)=a(B I)=a B=(a I) B$. Hence, ${a I \mid a \neq 0} \subseteq Z\left(\mathrm{GL}_{n}(F)\right)$. The difficulty lies is proving the reverse inclusion.

Suppose $1 \leq i, j \leq n$ with $i \neq j$. Let $E_{i j}$ be the $n \times n$ matrix consisting of zeros in all entries except for a 1 in the $(i, j)$ th entry. The matrix $E_{i j}$ is not in $\mathrm{GL}{n}(F)$ but $I+E{i j}$ is, since $\operatorname{det}\left(I+E_{i j}\right)=1$. Since $B I=I B$ for all $B \in \mathrm{GL}{n}(F)$, then $B\left(I+E{i j}\right)=\left(I+E_{i j}\right) B$ if and only if $B E_{i j}=E_{i j} B$. Thus, all $B \in Z\left(\mathrm{GL}{n}(F)\right)$ satisfy the matrix product $B E{i j}=E_{i j} B$.

The matrix product $B E_{1 j}$ is the matrix of zeros everywhere except for its $j$ th column being the first column of $B$. Similarly, $E_{1 j} B$ is the matrix of zeros everywhere except for its first row being the $j$ th row of $B$. (See Exercise 1.7.16.) Thus, for a particular $j \geq 2$, the identity $B E_{1 j}=E_{1 j} B$ implies that
$$b_{j k}= \begin{cases}0 & \text { if } k \neq j \ b_{11} & \text { if } k=j .\end{cases}$$
If $B \in Z\left(\mathrm{GL}{n}(F)\right)$, then $B E{1 j}=E_{1 j} B$ for all pairs $2 \leq j \leq n$. Therefore, all off-diagonal elements of $B$ are zero and $b_{j j}=b_{11}$ for all $j$, i.e., all diagonal elements of $B$ are equal. This establishes $Z\left(\mathrm{GL}_{n}(F)\right) \subseteq{a I \mid a \neq 0}$ and we deduce (1.7).

## 数学代写|抽象代数作业代写abstract algebra代考|Subgroups Generated by a Subset

lleft 的分隔符缺失或无法识别

$$x y^{-1}=s_{1}^{\alpha_{1}} s_{2}^{\alpha_{2}} \cdots s_{n}^{\alpha_{n}} t_{m}^{-\beta_{m}} \cdots t_{2}^{-\beta_{2}} t_{1}^{-\beta_{1}} .$$

## 数学代写|抽象代数作业代写abstract algebra代考|Center, Centralizer, Normalizer

$$(x y) g=x(y g)=x(g y)=(x g) y=(g x) y=g(x y)$$

$$Z(G \operatorname{Ln}(F))=a I \mid a \neq 0,$$

$$b_{j k}=\left{0 \quad \text { if } k \neq j b_{11} \quad \text { if } k=j .\right.$$

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assignmentutor™您的专属作业导师
assignmentutor™您的专属作业导师