# 数学代写|抽象代数作业代写abstract algebra代考|MATH417

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

ElGamal encryption is a cryptographic protocol based on Diffie-Hellman. We describe now with the assumption that Alice wants to send an encrypted message to Bob.
(1) Alice and Bob settle on a group $G$ and on the base, a group element $g$. The plaintext space $\mathcal{M}$ ean be any set, the ciphertext space $\mathcal{C}$ will be sequences of elements in $G$, and the keyspace is $\mathcal{K}=G$.
(2) Alice and Bob also choose a method of encoding the message into a sequence of elements in $G$; i.e., they choose an injective function $h: \mathcal{M} \rightarrow$ $\operatorname{Fun}\left(\mathbb{N}^*, G\right)$.
(3) They run Diffie-Hellman to obtain their public key $k=g^{a b}$.
(4) Alice encodes her message $m$ with a sequence of group elements $h(m)=$ $\left(m_1, m_2, \ldots, m_n\right)$

(5) Alice sends to Bob the group elements $E_k(m)=\left(k m_1, k m_2, \ldots, k m_n\right)=$ $\left(c_1, c_2, \ldots, c_n\right)$. Note that for each $i$, we have $c_i=g^{a b} m_i$.
(6) To decipher the ciphertext, we have $k^{\prime}=k=g^{a b}$. Bob calculates the $m_i$ from $m_i=c_i k^{-1}=c_i\left(y^{n h}\right)^{-1}$.
[We point out that with a group element of large order, it is not always obvious how to determine the inverse of a group element. We use Corollary $2.1 .12$, which says that $g^{|G|}=1$. Hence, to calculate $g^{-a b}$, without knowing $a$ but only knowing $g^a$, using Fast Exponentiation, Bob calculates
$$\left.\left(g^a\right)^{|G|-b}=g^{a|G|-a b}=g^{-a b} .\right]$$
(7) Since $h$ is injective, Bob can find Alice’s plaintext message $\left(m_1, m_2, \ldots, m_n\right)$.
Example 1.12.3. We use the group $G=U(3001)$ and choose for a base the group element $g=\overline{2}$. (It turns out that $|g|=1500$. In general, one does not have to know the order of $g$, but merely hope that it is high.) The message space is the set of sequences $\mathcal{M}=\operatorname{Fun}\left(\mathbb{N}^*, \mathcal{A}\right)$ where $\mathcal{A}$ is the set consisting of the 26 English letters and the space character.

Alice and Bob decide to encode their messages (define $h: \mathcal{M} \rightarrow$ $\left.\operatorname{Fun}\left(\mathbb{N}^*, G\right)\right)$ as follows. Ignore all punctuation, encode a space with the integer 0 and each letter of the alphabet with its corresponding ordinal, so $\mathrm{A}$ is 1 , B is 2, and so on, where we allow for two digits for each letter. Hence, a space is actually 00 and $\mathrm{A}$ is 01 . Then group pairs of letters simply by adjoining them to make (up to) a four-digit number. Thus, “GOOD-BYE CRUEL WORLD” becomes the finite sequence
$$715,1504,225,500,318,2105,1200,2315,1812 \text {, } 400 \text {, }$$
where we completed the last pair with a space. We now view these numbers as elements in $U(3001)$.

## 数学代写|抽象代数作业代写abstract algebra代考|Semigroups and Monoids

Having introduced groups already, we can see that a semigroup resembles a group but with only the associativity axiom. Note that Proposition 1.2.13 holds in any semigroup. Obviously, every group is a semigroup.

In every semigroup $(S, \circ)$, because of associativity, the order in which we group the operations in an expression of the form $a \circ a \circ \cdots \circ a$ does not change the result. Hence, we denote by $a^k$ the (unique) element $\overbrace{a \circ a \circ \cdots \circ a}^{k \text { times }}$.

Example 1.13.2. The set of positive integers equipped with addition $\left(\mathbb{N}^{>0},+\right)$ is a semigroup.

Example 1.13.3. All integers equipped with multiplication $(\mathbb{Z}, x)$ is also a semigroup. That not all elements have inverses prevented $(\mathbb{Z}, \times)$ from being a group but that does not matter for a semigroup.

Example 1.13.4. Let $S=\mathbb{Z}$ and suppose that $a \circ b=\max {a, b}$. It is easy to see that for all integers $a, b, c$,
$$a \circ(b \circ c)=\max {a, \max {b, c}}=\max {a, b, c}=\max {\max {a, b}, c}=(a \circ b) \circ c .$$
Hence, $(\mathbb{Z}, \circ)$ is a semigroup. There is no integer e such that $\forall a \in$ $\mathbb{Z}, \max {e, a}=a$, because $r$ would need to be less than every integer. Hence, $(\mathbb{Z}, \circ)$ does not have an identity and consequently it cannot have inverses to elements.
Example 1.13.5. Consider the following set of rectangles in the plane
$$S=\left{[a, b] \times[c, d] \subseteq \mathbb{R}^2 \mid a, b, c, d \in \mathbb{R}\right}$$

# 抽象代数代考

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

ElGamal 加密是一种基于 Diffie-Hellman 的加密协议。我们现在假设 Alice 想要向 Bob 发送加密消息。
(1) Alice 和 Bob 确定了一个组 $G$ 在基础上，一个组元素 $g$. 明文空间 $\mathcal{M}$ 可以是任何集合，密文空间 $\mathcal{C}$ 将是元 素序列 $G$, 键空间是 $\mathcal{K}=G$.
(2) Alice 和 Bob 也选择了一种将消息编码为元素序列的方法 $G$; 即，他们选择一个单射函数 $h: \mathcal{M} \rightarrow$ $\operatorname{Fun}\left(\mathbb{N}^, G\right)$ (3) 他们运行 Diffie-Hellman 来获取他们的公钥 $k=g^{a b}$. (4) Alice 对她的信息进行编码 $m$ 具有一系列组元嫊 $h(m)=\left(m_1, m_2, \ldots, m_n\right)$ (5) Alice 将组元素发送给 Bob $E_k(m)=\left(k m_1, k m_2, \ldots, k m_n\right)=\left(c_1, c_2, \ldots, c_n\right)$. 请注意，对于每个 ，我们有 $c_i=g^{a b} m_i$. (6) 为了破译密文，我们有 $k^{\prime}=k=g^{a b}$. 鲍勃计算 $m_i$ 从 $m_i=c_i k^{-1}=c_i\left(y^{n h}\right)^{-1}$. [我们指出，对于大阶群元素，如何确定群元素的逆并不总是显而易见的。我们使用推论 $2.1 .12$ ，它说 $g^{|G|}=1$. 因此，要计算 $g^{-a b}$ ，不知道 $a$ 但只知道 $g^a$ ，使用快速幂运算，Bob 计算 $$\left.\left(g^a\right)^{|G|-b}=g^{a|G|-a b}=g^{-a b} .\right]$$ (7) 自 $h$ 是单射的，Bob 可以找到 Alice 的明文消息 $\left(m_1, m_2, \ldots, m_n\right)$. 示例 1.12.3。我们使用组 $G=U(3001)$ 并选择基组元素 $g=\overline{2}$. (事实证明 $|g|=1500$. 一般来说，不必知 道顺序 $g$ ，但只是希望它很高。) 消息空间是序列的集合 $\mathcal{M}=\operatorname{Fun}\left(\mathbb{N}^, \mathcal{A}\right)$ 在哪里 $\mathcal{A}$ 是由 26 个英文字母 和空格字符组成的集合。

Alice 和 Bob 决定对他们的消息进行编码（定义 $h: \mathcal{M} \rightarrow \operatorname{Fun}\left(\mathbb{N}^*, G\right)$ )如下。忽略所有标点符号，用整 数 0 编码一个空格，字母表中的每个字母都有对应的序数，所以 $\mathrm{A}$ 是 $1 ， \mathrm{~B}$ 是 2，依此类推，我们允许每 个字母有两个数字。因此，一个空格实际上是 00 并且 $A$ 是 01。然后简单地将一对字母组合成 (最多) 个四位数的数字。于是，”GOOD-BYE CRUEL WORLD”变成了有限序列
$$715,1504,225,500,318,2105,1200,2315,1812,400 \text {, }$$

## 数学代写|抽象代数作业代写abstract algebra代考|Semigroups and Monoids

$$a \circ(b \circ c)=\max a, \max b, c=\max a, b, c=\max \max a, b, c=(a \circ b) \circ c .$$

$S=\backslash$ left{[a, b] \times $[c, d] \backslash$ subseteq $\backslash$ mathbb ${R}}^{\wedge} 2 \backslash$ mid a, b, c, d \in \mathbb ${R \backslash \backslash$ right $}$

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