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

The subgroup of translations of $G$ consists of all translations that are an integer multiple of $2 \overrightarrow{P Q}$. Some other transformations in $G$ include

• reflections through a vertical line $L_1$ through $P$ or any line parallel to $L_1$ displaced by an integer multiple of $\overrightarrow{P Q}$;
• reflection through the horizontal line $L_3=\overleftrightarrow{P Q}$
• rotations by an angle of $\pi$ about $P$, $Q$, or any point translated from $P$ by an integer multiple of $\overrightarrow{P Q}$.

It is possible to describe $G$ with a presentation. Let $s_i$ be the reflection through $L_i$, for $i=1,2,3$. We claim that
$$G=\left\langle s_1, s_2, s_3 \mid s_1^2=s_2^2=s_3^2=1,\left(s_1 s_3\right)^2=1,\left(s_2 s_3\right)^2=1\right\rangle .$$
In order to prove the claim, we first should check that $s_1, s_2, s_3$ do indeed generate all of $G$. By Exercise $1.11 .8, s_1 s_3$ corresponds to rotation by $\pi$ about $P$ and $s_2 s_3$ corresponds to rotation by $\pi$ about $Q$. By Exercise $1.11 .7 s_2 s_1$ corresponds to a translation by $2 \overrightarrow{P Q}$. In order to obtain a reflection through another vertical line besides $L_1$ or $L_2$, or a rotation about another point besides $P$ or $Q$, we can translate the strip to center it on $P$ and $Q$, apply the desired transformation $\left(s_1, s_2, s_1 s_3\right.$, or $\left.s_2 s_3\right)$, and then translate back. For example, the rotation by an angle of $\pi$ about $Q_3$, can be described by $\left(s_2 s_1\right)^2 s_2 s_3\left(s_2 s_1\right)^{-2}$. This shows that our choice of generators is sufficient to generate $G$.

Next, we need to check that we have found all the relations. The relations $s_i^2=1$ are obvious. The last two relations are from the fact that $s_1 s_3$ and $s_2 s_3$ are rotations by $\pi$ so have order 2 . Note that $\left\langle s_1 s_2\right\rangle \cong \mathbb{Z}$ is the infinite subgroup of translation and $\left\langle s_1, s_2\right\rangle \cong D_{\infty}$ (see Example 1.10.7). From the relation, $\left(s_1 s_3\right)^2=1$, we see that
$$s_1 s_3-\left(s_1 s_3\right)^{-1}-s_3^{-1} s_1^{-1}-s_3 s_1$$
because $s_1^2=1$ and $s_3^2=1$. Hence, $s_3$ commutes with $s_1$. Similarly $s_3$ commutes with $s_2$. Hence, all elements in $G$ can be written as an alternating string of $s_1$ and $s_2$ or an alternating string of $s_1$ and $s_2$ followed by $s_3$. For example, rotation by $\pi$ about $Q_3$ is
\begin{aligned} \left(s_2 s_1\right)^2 s_2 s_3\left(s_2 s_1\right)^{-2} &=s_2 s_1 s_2 s_1 s_2 s_3 s_1 s_2 s_1 s_2=s_2 s_1 s_2 s_1 s_2 s_1 s_2 s_1 s_2 s_3 \ &=\left(s_2 s_1\right)^4 s_2 s_3 . \end{aligned}

## 数学代写|抽象代数作业代写abstract algebra代考|A Brief Background on Cryptography

In this section, we will study an application of group theory to cryptography, the science of keeping information secret.

Cryptography has a long history, with one of the first documented uses of cryptography attributed to Caesar. When writing messages he wished to keep in confidence, the Roman emperor would shift each letter by 3 to the right, assuming the alphabet wraps around. In other words, he would substitute a letter of $\mathrm{A}$ with $\mathrm{D}, \mathrm{B}$ with $\mathrm{E}$ and so forth, down to replacing $\mathrm{Z}$ with $\mathrm{C}$. To anyone who intercepted the modified message, it would look like nonsense. This was particularly valuable if Caesar thought there existed a chance that an enemy could intercept orders sent to his military commanders.

After Caesar’s cipher, there came letter wheels in the early Renaissance, letter codes during the American Civil War, the Navajo windtalkers during World War II, the Enigma machine used by the Nazis, and then a whole plethora of techniques since then. Military uses, protection of financial data, and safety of intellectual property have utilized cryptographic techniques for centuries. For a long time, the science of cryptography remained the knowledge of a few experts because both governments and companies held that keeping their cryptographic techniques secret would make it even harder for “an enemy” to learn one’s information security tactics.

Today, electronic data storage, telecommunication, and the Internet require increasingly complex cryptographic algorithms. Activities that are commonplace like conversing on a cellphone, opening a car remotely, purchasing something online, all use cryptography so that a conversation cannot be intercepted, someone else cannot easily unlock your car, or an eavesdropper cannot intercept your credit card information.

Because of the proliferation of applications of cryptography in modern society, no one should assume that the cryptographic algorithm used in any given instance remains secret. In fact, modern cryptographers do not consider an information security algorithm at all secure if part of its effectiveness relies on the algorithm remaining secret. But not everything about a cryptographic algorithm can be known to possible eavesdroppers if parties using the algorithm hope to keep some message secure. Consequently, most, if not all, cryptographic techniques involve an algorithm but also a “key,” which can be a letter, a number, a string of numbers, a string of bits, a matrix or some other mathematical object. The security of the algorithm does not depend on the algorithm staying secret but rather on the key remaining secret. Users can change keys from time to time without changing the algorithm and have confidence that their messages remain secure.

# 抽象代数代考

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

• 通过垂直线的反射 $L_1$ 通过 $P$ 或任何平行于 $L_1$ 位移的整数倍 $\overrightarrow{P Q}$;
• 通过水平线反射 $L_3=\overleftrightarrow{P Q}$
• 旋转一个角度 $\pi$ 关于 $P, Q$, 或任何翻译自的点 $P$ 的整数倍 $\overrightarrow{P Q}$.
可以形容 $G$ 带演示文稿。让 $s_i$ 成为反映 $L_i$ ，为了 $i=1,2,3$. 我们声称
$$G=\left\langle s_1, s_2, s_3 \mid s_1^2=s_2^2=s_3^2=1,\left(s_1 s_3\right)^2=1,\left(s_2 s_3\right)^2=1\right\rangle .$$
为了证明这个说法，我们首先应该检查 $s_1, s_2, s_3$ 确实产生了所有 $G$. 通过运动 $1.11 .8, s_1 s_3$ 对应于旋转 $\pi$ 关 于 $P$ 和 $s_2 s_3$ 对应于旋转 $\pi$ 关于 $Q$. 通过运动 $1.11 .7 s_2 s_1$ 对应于翻译 $2 \overrightarrow{P Q}$. 为了通过另一条垂直线获得反射， 除了 $L_1$ 或者 $L_2$, 或围绕另一点旋转 $P$ 或者 $Q$ ，我们可以平移条带使其居中 $P$ 和 $Q$, 应用所需的转换 $\left(s_1, s_2, s_1 s_3\right.$ ，或者 $\left.s_2 s_3\right)$, 然后翻译回来。例如，旋转一个角度 $\pi$ 关于 $Q_3$, 可以描述为 $\left(s_2 s_1\right)^2 s_2 s_3\left(s_2 s_1\right)^{-2}$. 这表明我们选择的生成器足以生成 $G$.
接下来，我们需要检查是否找到了所有关系。关系 $s_i^2=1$ 很明显。最后两个关系来自以下事实: $s_1 s_3$ 和 $s_2 s_3$ 是旋转 $\pi$ 所以有订单 2 。注意 $\left\langle s_1 s_2\right\rangle \cong \mathbb{Z}$ 是平移的无限子群，并且 $\left\langle s_1, s_2\right\rangle \cong D_{\infty}$ (参见示例 1.10.7) 。从关系来看， $\left(s_1 s_3\right)^2=1$ ，我们看到
$$s_1 s_3-\left(s_1 s_3\right)^{-1}-s_3^{-1} s_1^{-1}-s_3 s_1$$
因为 $s_1^2=1$ 和 $s_3^2=1$. 因此， $s_3$ 通勤 $s_1$. 相似地 $s_3$ 通勤 $s_2$. 因此，所有元素 $G$ 可以写成一个交替的字符串 $s_1$ 和 $s_2$ 或交替的字符串 $s_1$ 和 $s_2$ 其次是 $s_3$. 例如，旋转 $\pi$ 关于 $Q_3$ 是
$$\left(s_2 s_1\right)^2 s_2 s_3\left(s_2 s_1\right)^{-2}=s_2 s_1 s_2 s_1 s_2 s_3 s_1 s_2 s_1 s_2=s_2 s_1 s_2 s_1 s_2 s_1 s_2 s_1 s_2 s_3 \quad=\left(s_2 s_1\right)^4 s_2 s_3 .$$

## 数学代写|抽象代数作业代写abstract algebra代考|A Brief Background on Cryptography

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