# 代数代写 Algebra I |MATH 8806 Boston College Assignment

0

Assignment-daixieTM为您提供波士顿学院Boston College MATH 8806 Algebra I 代数学代写代考辅导服务！

## Instructions:

Algebra is a branch of mathematics that deals with symbols and the rules for manipulating these symbols to solve equations and understand relationships between quantities. It involves the use of letters and symbols to represent unknown values, and the use of mathematical operations such as addition, subtraction, multiplication, and division to solve equations and simplify expressions.

Algebra is important in many areas of mathematics, science, engineering, economics, and finance, as it provides a powerful tool for solving problems and understanding relationships between variables. Some of the key concepts in algebra include equations, inequalities, polynomials, functions, and matrices.

Algebraic techniques can be used to solve a wide range of problems, from simple arithmetic calculations to complex systems of equations and models of physical and economic phenomena. Algebra is also used extensively in computer science, cryptography, and other fields that require efficient methods for processing and manipulating large amounts of data.

Let $G$ be a group and let $H$ be a subgroup. Prove that the following are equivalent.

(1) $H$ is normal in $G$.

(2) For every $g \in G, g H^{-1}=H$.

$(1) \Rightarrow (2)$: If $H$ is normal in $G$, then for every $g \in G$ and $h \in H$, we have $ghg^{-1} \in H$. Thus, for any $g \in G$, we have $gHg^{-1} \subseteq H$. On the other hand, $Hg^{-1}g \subseteq g^{-1}Hg$, so $gH^{-1}=Hg^{-1}g \subseteq H$. Hence, $gH^{-1}=H$.

$(2) \Rightarrow (3)$: If $gH^{-1}=H$ for every $g \in G$, then for any $a \in G$ and $h \in H$, we have $ah=ag(g^{-1}h) \in aH$. Therefore, $aH \subseteq Ha$, and the reverse inclusion follows similarly. Thus, $aH=Ha$.

(3) For every $a \in G, a H=H a$.

(4) The set of left cosets is equal to the set of right cosets.

$(3) \Rightarrow (4)$: If $aH=Ha$ for every $a \in G$, then the set of left cosets is ${aH \mid a \in G}$, and the set of right cosets is ${Ha \mid a \in G}$. For any $a,b \in G$, we have $aH=Ha$ and $bH=Hb$, so $aHb=abH=(Ha)b=H(ab)=Hba=bHa$. Hence, the sets of left and right cosets coincide.

$(4) \Rightarrow (1)$: If the set of left cosets is equal to the set of right cosets, then for any $g \in G$ and $h \in H$, there exists $a \in G$ such that $ah=g$. Thus, $g^{-1}ag \in H$ and $g^{-1}hg \in H$, so $ghg^{-1} \in H$ for any $g \in G$ and $h \in H$. Therefore, $H$ is normal in $G$.

$(\Rightarrow)$ Assume that $G/N$ is abelian. Let $x,y\in G$ be arbitrary elements. Then we have $(xN)(yN)=(yN)(xN)$ in $G/N$, which means that $xyN=yxN$. Therefore, we have $xyx^{-1}y^{-1}\in N$. Since $x$ and $y$ were arbitrary, this means that $N$ contains the commutator $[x,y]=xyx^{-1}y^{-1}$ for every pair of elements $x,y\in G$.

$(\Leftarrow)$ Assume that $N$ contains the commutator of every pair of elements of $G$. Let $xN,yN$ be arbitrary elements of $G/N$. We need to show that $(xN)(yN)=(yN)(xN)$, i.e., $xyN=yxN$. Since $N$ is normal in $G$, we have $xNx^{-1}=N$ and $yNy^{-1}=N$. Therefore, we have

\begin{align*} xyN &= xNyN && \text{(since }yN=NyN\text{)} \ &= xNyNy^{-1} && \text{(since }y^{-1}Ny=N\text{)} \ &= xNNy^{-1} && \text{(since }N\text{ is normal)} \ &= xN[y^{-1},x]N && \text{(since }[y^{-1},x]\in N\text{)} \ &= y^{-1}xN[x,y]N && \text{(since }N\text{ is normal)} \ &= y^{-1}NxN[y,x] && \text{(since }[x,y]=[y^{-1},x]^{-1}\in N\text{)} \ &= yNy^{-1}xN && \text{(since }N\text{ is normal)} \ &= yN[x^{-1},y]N && \text{(since }[x^{-1},y]\in N\text{)} \ &= x^{-1}yN[x,y]N && \text{(since }N\text{ is normal)} \ &= x^{-1}NxN[y,x] && \text{(since }[x,y]=[x^{-1},y]^{-1}\in N\text{)} \ &= xNx^{-1}yN && \text{(since }N\text{ is normal)} \ &= yxN && \text{(since }xNx^{-1}=N\text{)} \ &= yNxN && \text{(since }xN=NxN\text{)} \ &= yN(xN) && \text{(since }N\text{ is normal)} \ &= yN(xN), \end{align*}

which shows that $G/N$ is abelian.