# 数学代写|抽象代数作业代写abstract algebra代考|MATH355

## 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 clements in $C_1$ 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 $\left(x^k y^l\right)\left(x^m y^n\right)=x^{k+m} y^{l+n}$.

## 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 $\left\langle g_1, g_2, \ldots, g_k\right| \begin{array}{llll}R_1 & R_2 & \cdots & \left.R_s\right\rangle \text {. Every }\end{array}$ 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.

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.

# 抽象代数代考

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

\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}

$$x^k y^{\ell}=x^m y^n \Longleftrightarrow x^{k-m}=y^{n-\ell} .$$

## 数学代写|抽象代数作业代写abstract algebra代考| presentation and homoomorphisms

$$\varphi(w)=\varphi\left(u_1\right)^{\alpha_1} \varphi\left(u_2\right)^{\alpha_2} \cdots \varphi\left(u_{\ell}\right)^{\alpha_{\ell}} .$$

$$\varphi(g) \stackrel{\text { def }}{=} \varphi\left(u_1\right)^{\alpha_1} \varphi\left(u_2\right)^{\alpha_2} \cdots \varphi\left(u_{\ell}\right)^{\alpha_{\ell}} .$$

$$\varphi(v)=\varphi\left(w_1\right)=\varphi\left(w_2\right)=\cdots=\varphi\left(w_n\right)=\varphi(w) .$$

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