数学代写|抽象代数作业代写abstract algebra代考|MATH GU4041

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数学代写|抽象代数作业代写abstract algebra代考|Defining Groups from a Presentation

A group defined by a presentation is similar to a free group in that it is first understood through its symbols, rather than the symbols representing some function, matrix, or number. In a group defined by a presentation, the elements are simply reduced words in the generators but the relations impose additional simplifications beyond just the power rules that hold in any group.
Example 1.10.6. To illustrate similarities and differences in various sets of relations, consider the following three groups.
\begin{aligned} &G_{1}=\left\langle x, y \mid x^{3}=y^{7}=1, x y=y x\right\rangle \ &G_{2}=\left\langle a, b \mid a^{3}=b^{7}=1, a b=b^{2} a\right\rangle \ &G_{3}=\left\langle u, v \mid u^{3}=v^{7}=1, u v=v^{2} u^{2}\right\rangle \end{aligned}
In $G_{1}$, since $x y=y x$, in any word in the generators $x$ and $y$, all the $x$ symbols can be moved to the left. Thus, all elements in $G_{1}$ can be written as $x^{k} y^{\ell}$. Furthermore, since $x^{3}=y^{7}=1$, then $x^{i} y^{j}$ with $0 \leq i \leq 2$ and $0 \leq j \leq 6$ give all the elements of $G_{1}$. We claim that all 21 of these elements are distinct. To prove this, we must show that $x^{i} y^{j}$ are distinct for $0 \leq i \leq 2$ and $0 \leq j \leq 6$. If $x^{k} y^{\ell}=x^{m} y^{n}$, we have
$$x^{k} y^{\ell}=x^{m} y^{n} \Longleftrightarrow x^{k-m}=y^{n-\ell}$$
By Corollary $1.3 .7$, since $x^{3}=1$, the order of $x^{k-m}$ divides 3 and, since $y^{7}=1$, the order of $y^{n-\ell}$ divides 7 . Since $x^{k-m}=y^{n-\ell}$, then the order of this element must divide $\operatorname{gcd}(3,7)=1$. Hence, $x^{k-m}=y^{n-\ell}=1$. Thus, 3 divides $k-m$and 7 divides $n-\ell$, but if we assume that $0 \leq k, m \leq 2$ and $0 \leq \ell, n \leq 6$, then we conclude that $k=m$ and $n=\ell$. This proves the claim. Hence, $G_{1}$ is a group of order 21 in which the elements operate as $\left(x^{k} y^{l}\right)\left(x^{m} y^{n}\right)=x^{k+m} y^{l+n}$. It is easy to see that $G_{1} \cong Z_{3} \oplus Z_{7}$ and by Exercise 1.9.18, we deduce that $G_{1} \cong Z_{21}$

In $G_{2}$, from the relation $a b=b^{2} a$, we see that all the $a$ symbols may be moved to the right of any $b$ symbols, though possibly changing the power on b. In particular,
$$a^{n} b=a^{n-1} b^{2} a=a^{n-2}(a b) b a=a^{n-2} b^{2} a b a=a^{n-2} b^{4} a^{2}=\cdots=b^{2^{n}} a^{n}$$
and also
$$a^{n} b^{k}=b^{2^{n}} a^{n} b^{k-1}=b^{2^{n}} b^{2^{n}} a^{n} b^{k-2}=\cdots=b^{k 2^{n}} a^{n}$$

数学代写|抽象代数作业代写abstract algebra代考|Presentations and Homomorphisms

Suppose that $G$ has a presentation $\left\langle g_{1}, g_{2}, \ldots, g_{k} \mid R_{1} \quad R_{2} \quad \cdots, R_{s}\right\rangle$. Every element $w \in G$ is a word in the generators, $w=u_{1}^{\alpha_{1}} u_{2}^{\alpha_{2}} \cdots u_{\ell}^{\alpha_{\ell}}$, with $u_{i} \in$ $\left{g_{1}, g_{2}, \ldots, g_{k}\right}$, so for a homomorphism $\varphi: G \rightarrow H$ we have
$$\varphi(w)=\varphi\left(u_{1}\right)^{\alpha_{1}} \varphi\left(u_{2}\right)^{\alpha_{2}} \cdots \varphi\left(u_{\ell}\right)^{\alpha_{\ell}}$$
Hence, $\varphi$ is entirely determined by the values of $\varphi\left(g_{1}\right), \varphi\left(g_{2}\right), \ldots, \varphi\left(g_{k}\right)$.

When trying to construct a homomorphism from $G$ to a group $H$, it is not possible to associate arbitrary elements in $H$ to the generators of $G$ and always obtain a homomorphism. The following theorem makes this precise.
Theorem 1.10.10 (Extension Theorem on Generators)
Let $G=\left\langle g_{1}, g_{2}, \ldots, g_{k} \mid R_{1} \quad R_{2} \quad \cdots, R_{s}\right\rangle$ and let $h_{1}, h_{2}, \ldots, h_{k} \in H$ be elements that satisfy the relations $R_{i}$ as the generators of $G$ when replacing $g_{i}$ with $h_{i}$ for $i=1,2, \ldots, k$. Then the function $\left{g_{1}, g_{2}, \ldots, g_{k}\right} \rightarrow H$ that maps $g_{i} \mapsto h_{i}$ for $i=1,2, \ldots, k$ can be extended to a unique homomorphism $\varphi: G \rightarrow H$ that has $\varphi\left(g_{i}\right)=h_{i}$.
Proof. We define the function $\varphi: G \rightarrow H$ by $\varphi\left(g_{i}\right)=h_{i}$ for $i=1,2, \ldots, k$ and for each element $g \in G$, if $g=u_{1}^{\alpha_{1}} u_{2}^{\alpha_{2}} \cdots u_{\ell}^{\alpha_{\ell}}$ with $u_{j} \in\left{g_{1}, g_{2}, \ldots, g_{k}\right}$, then
$$\varphi(g) \stackrel{\text { def }}{=} \varphi\left(u_{1}\right)^{\alpha_{1}} \varphi\left(u_{2}\right)^{\alpha_{2}} \cdots \varphi\left(u_{\ell}\right)^{\alpha_{\ell}} .$$
By construction, $\varphi$ satisfies the homomorphism property $\varphi(x y)=\varphi(x) \varphi(y)$ for all $x, y \in G$. However, since different words can be equal, we have not yet determined if $\varphi$ is a well-defined function.

Two words $v$ and $w$ in the generators $g_{1}, g_{2}, \ldots, g_{k}$ are equal if and only if there is a finite sequence of words $w_{1}, w_{2}, \ldots, w_{n}$ such that $v=w_{1}, w=w_{n}$, and $w_{i}$ to $w_{i+1}$ are related to each other by either one application of a power rule (as given in Proposition 1.2.12) or one application of a relation $R_{j}$. Since the elements $h_{1}, h_{2}, \ldots, h_{k} \in H$ satisfy the same relations $R_{1}, R_{2}, \ldots, R_{s}$ as $g_{1}, g_{2}, \ldots, g_{k}$, then the same equalities apply between the words $\varphi\left(w_{i}\right)$ and $\varphi\left(w_{i+1}\right)$ as between $w_{i}$ and $w_{i+1}$. This establishes the chain of equalities
$$\varphi(v)=\varphi\left(w_{1}\right)=\varphi\left(w_{2}\right)=\cdots=\varphi\left(w_{n}\right)=\varphi(w) .$$
Hence, if $v=w$ are words in $G$, then $\varphi(v)=\varphi(w)$. Thus, $\varphi$ is a well-defined function and hence is a homomorphism.

数学代写|抽象代数作业代写abstract algebra代考|Defining Groups from a Presentation

$$G_{1}=\left\langle x, y \mid x^{3}=y^{7}=1, x y=y x\right\rangle \quad G_{2}=\left\langle a, b \mid a^{3}=b^{7}=1, a b=b^{2} a\right\rangle G_{3}$$

$0 \leq j \leq 6$ 给出所有元溸 $G_{1}$. 我们声称所有 21 个元䮖都是不同的。为了证明这一点，我们 必须证明 $x^{i} y^{j}$ 是不同的 $0 \leq i \leq 2$ 和 $0 \leq j \leq 6$. 如果 $x^{k} y^{\ell}=x^{m} y^{n}$ ，我们有
$$x^{k} y^{\ell}=x^{m} y^{n} \Longleftrightarrow x^{k-m}=y^{n-\ell}$$

$x^{k-m}=y^{n-\ell}=1$. 因此，3个除法 $k-m$ 和 7 分 $n-\ell$ ，但如果我们假设 $0 \leq k, m \leq 2$ 和 $0 \leq \ell, n \leq 6$ ，那么我们得出结论 $k=m$ 和 $n=\ell$. 这证明了这一说法。因此， $G_{1}$ 是一 组 21 阶的元蜮，其中元拜的操作为 $\left(x^{k} y^{l}\right)\left(x^{m} y^{n}\right)=x^{k+m} y^{l+n}$. 很容易看出 $G_{1} \cong Z_{3} \oplus Z_{7}$ 通过练习 $1.9 .18$ ，我们推断出 $G_{1} \cong Z_{21}$

$$a^{n} b=a^{n-1} b^{2} a=a^{n-2}(a b) b a=a^{n-2} b^{2} a b a=a^{n-2} b^{4} a^{2}=\cdots=b^{2^{n}} a^{n}$$

$$a^{n} b^{k}=b^{2^{n}} a^{n} b^{k-1}=b^{2^{n}} b^{2^{n}} a^{n} b^{k-2}=\cdots=b^{k 2^{n}} a^{n}$$

数学代写|抽象代数作业代写abstract algebra代考|Presentations and Homomorphisms

$$\varphi(w)=\varphi\left(u_{1}\right)^{\alpha_{1}} \varphi\left(u_{2}\right)^{\alpha_{2}} \cdots \varphi\left(u_{\ell}\right)^{\alpha_{\ell}}$$

Uleft{g_{1}, g_{2}, Vdots, g_{k}}right } \rightarrowH H那映射 $g_{i} \mapsto h_{i}$ 为了 $i=1,2, \ldots, k$ 可以扩 展为唯一的同态 $\varphi: G \rightarrow H$ 具有 $\varphi\left(g_{i}\right)=h_{i}$.

$$\varphi(g) \stackrel{\text { def }}{=} \varphi\left(u_{1}\right)^{\alpha_{1}} \varphi\left(u_{2}\right)^{\alpha_{2}} \cdots \varphi\left(u_{\ell}\right)^{\alpha_{\ell}}$$

$$\varphi(v)=\varphi\left(w_{1}\right)=\varphi\left(w_{2}\right)=\cdots=\varphi\left(w_{n}\right)=\varphi(w) .$$

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