## 数学代写|抽象代数作业代写abstract algebra代考|MATH2022

2022年10月12日

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## 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.
Definition 1.7.7
The center $Z(G)$ is the subset of $G$ consisting of all elements that commute with every other element in $G$. In other words,
$$Z(G)={x \in G \mid x g=g x \text { for all } g \in G} .$$
Proposition 1.7.8
Let $G$ be any group. The center $Z(G)$ is a subgroup of $G$.

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, ${a I \mid a \neq 0} \subseteq Z\left(\operatorname{GL}_n(F)\right)$. 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 $B E{i j}=E_{i j} B$.

## 数学代写|抽象代数作业代写abstract algebra代考|Lattice of Subgroups

In order to develop an understanding of the internal structure of a group, listing all the subgroups of a group has some value. However, showing how these subgroups are related to each other carries more information. The lattice of subgroups offers a visual representation of containment among subgroups.
Denote by $\operatorname{Sub}(G)$ the set of all subgroups of the group $G$. Obviously, $\operatorname{Sub}(G) \subseteq \mathcal{P}(G)$. For any two subgroups $H, K \in \operatorname{Sub}(G), H \subseteq K$ if and only if $H \leq K$. Consequently, $(\operatorname{Sub}(G), \leq)$ is a poset, namely the subposet of $(\mathcal{P}(G), \subseteq)$ on the subset $\operatorname{Sub}(G)$.

Proof. We know that $(\mathcal{P}(G), \subseteq)$ is a lattice: the least upper bound of any two subsets $A$ and $B$ is $A \cup B$ and the greatest lower bound is $A \cap B$. By Proposition 1.6.10, for any two $H, K \in \operatorname{Sub}(G)$, we also have $H \cap K \in \operatorname{Sub}(G)$, so $H \cap K$ their greatest lower bound in the poset $(\operatorname{Sub}(G), \leq)$.

In contrast, for $H, K \leq G$, the union $H \cup K$ is generally not a subgroup. By Exercise 1.7.13, $\langle H \cup K\rangle$ is the smallest (by inclusion) subgroup of $G$ that contains both $H$ and $K$, and thus the least upper bound of $H$ and $K$ in $(\operatorname{Sub}(G), \leq)$. Since every pair of subgroups of $G$ has a least upper bound and a greatest lower bound in $(\operatorname{Sub}(G), \leq)$, the poset is a lattice.

The construction given in the above proof for a least upper bound of $H$ and $K$, namely $\langle H \cup K\rangle$ is called the join of $H$ and $K$.

Since $(\operatorname{Sub}(G), \leq)$ is a poset, we can create the Hasse diagram for it. By a common abuse of language, we often say “draw the lattice of $G$ ” for “draw the Hasse diagram of the poset $(\operatorname{Sub}(G), \leq)$.” The lattice of a group shows all subgroups and their containment relationships.

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

$$Z(G)={x \in G \mid x g=g x \text { for all } g \in G} .$$

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

$$Z\left(G L_n(F)\right)={a I \mid a \neq 0},$$

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