# 代数|MATH0006 Algebra 2代写2023

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Assignment-daixieTM为您提供伦敦大学学院 London’s Global University MATH0006 Algebra 2代数代写代考辅导服务！

## Instructions:

1. Check the syllabus: Make sure the book covers the topics that will be taught in the course. Look at the table of contents or the index to see if the book covers the topics you need.
2. Consider your level of experience: Choose a book that matches your level of expertise in the subject. If you are new to the subject, choose a book that is more elementary and easy to understand. If you are more experienced, you may want a book that is more advanced and covers more material.
4. Consider your learning style: Everyone learns differently, so choose a book that matches your learning style. Some books have lots of examples and exercises, while others have more theory and proofs.
5. Check the price: Textbooks can be expensive, so make sure you are comfortable with the price of the book you choose. You may want to consider buying a used copy or renting the book to save money.

Let $G$ be a group and let $H$ be a subgroup. Prove that the following are equivalent. (1) $H$ is normal in $G$.

$(1) \Rightarrow (2)$: Assume that $H$ is a normal subgroup of $G$. We need to show that for every $g \in G, g H g^{-1}=H$. Let $g\in G$ and $h \in H$ be arbitrary. Since $H$ is normal in $G$, we have $ghg^{-1}\in H$. Hence, $gHg^{-1} \subseteq H$. To show the other inclusion, let $x\in H$. Then $g^{-1}xg \in H$ since $H$ is normal, and so $x=gg^{-1}xgg^{-1}=g(g^{-1}xg)g^{-1}\in gHg^{-1}$. Therefore, we have $H\subseteq gHg^{-1}$, which implies that $gHg^{-1}=H$.

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

$(2) \Rightarrow (3)$: Assume that for every $g\in G, g H g^{-1}=H$. We need to show that for every $a \in G, a H=H a$. Let $a \in G$ be arbitrary. We want to show that $aH\subseteq Ha$. Let $h\in H$ be arbitrary. Then, we have $ah=g(g^{-1}ag)h(g^{-1}ag)^{-1}\in gHg^{-1}=H$, where we use that $g^{-1}ag$ is just some element of $G$ that we can conjugate with. Therefore, $ah\in Ha$ for all $h \in H$, which implies that $aH \subseteq Ha$. By a similar argument, we can show that $Ha \subseteq aH$, and so we have $aH=Ha$.

\begin{prob}

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

\end{prob}

\begin{proof}

$(3) \Rightarrow (1)$: Assume that for every $a \in G, a H=H a$. We need to show that $H$ is normal in $G$. Let $g\in G$ and $h\in H$ be arbitrary. We want to show that $ghg^{-1}\in H$. To do this, note that $g^{-1}hg \in H$ since $H$ is closed under conjugation by elements of $G$. Therefore, $ghg^{-1}=g(g^{-1}hg)g^{-1} \in gHg^{-1}$, and so $H$ is normal in $G$.

# 代数|MATH0005 Algebra 1代写2023

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Assignment-daixieTM为您提供伦敦大学学院 London’s Global University MATH0005 Algebra 1代数代写代考辅导服务！

## Instructions:

Linear equations are fundamental to many areas of mathematics, science, and engineering, so it’s important to have a strong understanding of them. And it’s great that the course also covers other important topics in modern mathematics, such as logic, set theory, and functions.

By studying these topics in depth, students can develop important problem-solving skills and critical thinking abilities that will serve them well in future courses and in their careers. Additionally, understanding these foundational concepts will allow students to approach more advanced topics with greater ease and confidence.

Overall, it sounds like MATH0005 is an excellent course for anyone looking to build a strong foundation in modern mathematics.

(i) Exhibit a proper normal subgroup $V$ of $A_4$. To which group is $V$ isomorphic to?

(i) The subgroup $V = {(), (1,2)(3,4), (1,3)(2,4), (1,4)(2,3)}$ is a proper normal subgroup of $A_4$. To see this, note that $V$ is a subgroup of $A_4$ since it is closed under multiplication and inverses. Moreover, it is normal since it is the kernel of the sign homomorphism from $A_4$ to $\mathbb{Z}/2\mathbb{Z}$, which maps even permutations to $0$ and odd permutations to $1$. To see that $V$ is proper, note that it has index $2$ in $A_4$ since $A_4$ has order $12$ and $V$ has order $4$, so $A_4/V$ has order $2$. But there is no subgroup of order $2$ in $A_4$.

The group $V$ is isomorphic to the Klein four-group $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$.

(ii) Give the left cosets of $V$ inside $A_4$.

(ii) The left cosets of $V$ inside $A_4$ are $V$ and $\tau V$, where $\tau$ is any transposition not in $V$. For example, if we take $\tau = (1,2)$, then we have

$V={(),(12)(34),(13)(24),(14)(23)}$

and

$\tau V={(12),(12)(34)(12),(13)(24)(12),(14)(23)(12)}$

\begin{prob}

(iii) To which group is $A_4 / V$ isomorphic to?

\end{prob}

\begin{proof}

(iii) The group $A_4/V$ is isomorphic to $\mathbb{Z}/2\mathbb{Z}$, the cyclic group of order $2$. To see this, note that $A_4/V$ has order $2$, so it is either isomorphic to $\mathbb{Z}/2\mathbb{Z}$ or to the trivial group ${1}$. But if $A_4/V \cong {1}$, then $A_4 = V$, which is false. Therefore, $A_4/V \cong \mathbb{Z}/2\mathbb{Z}$.

# 代数代写 Algebra II|MATH 8807 Boston College Assignment

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Assignment-daixieTM为您提供波士顿学院Boston College MATH 8807 Algebra II 代数学代写代考辅导服务！

## Instructions:

Algebra is a branch of mathematics that deals with the study of mathematical symbols and the rules for manipulating these symbols. In algebra, letters and symbols are used to represent unknown quantities, and mathematical operations such as addition, subtraction, multiplication, division, and exponentiation are used to solve equations and find solutions to problems.

Algebra is used extensively in many areas of mathematics, science, engineering, and economics, among other fields. It is essential in solving problems related to geometry, calculus, and statistics, and is also used in everyday life, such as calculating mortgage payments or determining the best price for a product.

Algebraic expressions can take many forms, such as linear equations, quadratic equations, polynomial equations, and systems of equations. These expressions can be solved using various methods, including factoring, completing the square, and the quadratic formula.

Some fundamental concepts in algebra include variables, coefficients, constants, and equations. Variables are letters or symbols that represent unknown quantities, while coefficients are numbers that multiply the variables in an equation. Constants are fixed numbers, and equations are mathematical expressions that use an equal sign to show that two expressions are equal.

Consider the random walk of a man on the integer line $\mathbb{Z}$ such that, at each step, that the probability to go from position $i$ to position $i+1$ is $p$, and the probability to go from $i$ to $i-1$ is $1-p$. The man “falls off the cliff” if he reaches the position 0. Suppose that the man starts at the initial position $i_0 \geq 1$. Find the probability that he falls off the cliff.

Let $P_i$ be the probability that the man falls off the cliff starting from position $i$. We want to find $P_{i_0}$ given that $i_0 \geq 1$. Note that $P_0 = 1$ since the man has already fallen off the cliff when he starts at position $0$. Also note that $P_i = 0$ for all $i \leq 0$ since the man has already fallen off the cliff.

We can now set up a recurrence relation for $P_i$. If the man is currently at position $i$, then he has two options for his next step: he can either move to position $i+1$ with probability $p$, or he can move to position $i-1$ with probability $1-p$. Thus, we have

$$P_i = p P_{i+1} + (1-p) P_{i-1}$$

for all $i \geq 1$. This is a second-order linear recurrence relation with constant coefficients. We can solve it using the characteristic equation:

$$r^2 – (1-p)r – p = 0.$$

The roots of this equation are

$$r_1 = \frac{1-p + \sqrt{(1-p)^2 + 4p}}{2} \quad \text{and} \quad r_2 = \frac{1-p – \sqrt{(1-p)^2 + 4p}}{2}.$$

Note that $r_1 > 1$ and $r_2 < 0$. Therefore, the general solution to the recurrence relation is

$$P_i = A r_1^i + B r_2^i$$

for some constants $A$ and $B$. We can find $A$ and $B$ by using the boundary conditions $P_0 = 1$ and $P_i = 0$ for all $i \leq 0$. This gives us the system of equations

\begin{align*} A + B &= 1, \ Ar_1^i + Br_2^i &= 0 \quad \text{for all } i \leq 0. \end{align*}

Solving for $A$ and $B$, we get

$$A = \frac{r_2}{r_2 – r_1} \quad \text{and} \quad B = -\frac{r_1}{r_2 – r_1}.$$

Therefore, the probability that the man falls off the cliff starting from position $i_0 \geq 1$ is

$$P_{i_0} = \frac{r_2^{i_0}}{r_2^{i_0} – r_1^{i_0}}.$$

Note that this expression makes sense since $r_1 > 1$ and $r_2 < 0$, so $r_1^{i_0}$ grows exponentially while $r_2^{i_0}$ decays exponentially as $i_0$ gets large. This means that $P_{i_0}$ approaches 1 as $i_0$ increases, which makes sense since the man is more likely to fall off the cliff the further he is from it.

For $1 \leq k \leq n / 2$, find a bijection $f$ between $k$-element subsets of $\{1, \ldots, n\}$ and $(n-k)$-element subsets of $\{1, \ldots, n\}$ such that $f(I) \supseteq I$, for any $k$-element subset $I$.

Let $S$ be a set of size $n$ and let $k$ be an integer such that $1 \leq k \leq n/2$. We will define a bijection $f$ between the set of $k$-element subsets of $S$ and the set of $(n-k)$-element subsets of $S$ that satisfies the given property.

Let $I$ be a $k$-element subset of $S$. We will define $f(I)$ as follows: first, let $J = S \setminus I$ be the complement of $I$ in $S$. Since $|I| = k$, we have $|J| = n-k$. Next, let $T$ be the $k$-element subset of $J$ that contains the $k$ smallest elements of $J$ (with respect to some fixed total ordering of $S$). Note that $T$ is well-defined since $|J| = n-k \geq k$.

Finally, we define $f(I) = T \cup I$. We claim that $f(I)$ is an $(n-k)$-element subset of $S$, and that $f$ is a bijection between the set of $k$-element subsets of $S$ and the set of $(n-k)$-element subsets of $S$.

To see that $f(I)$ is an $(n-k)$-element subset of $S$, note that $|f(I)| = |T \cup I| = |T| + |I| – |T \cap I|$, and $|T| = k$, $|I| = k$, and $|T \cap I| = 0$ (since $T$ and $I$ are disjoint). Therefore, $|f(I)| = 2k \leq 2(n/2) = n$, so $f(I)$ is a subset of $S$ with at most $n$ elements. Moreover, $|f(I)| = n-k$ since $|T| = k$ and $|J| = n-k$, so $f(I)$ is indeed an $(n-k)$-element subset of $S$.