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抽象代数作业代写| Units in Z_n. Laws of composition.

1. Units in $\mathbb{Z}_{n}$

Recall that $\mathbb{Z}_{n}$ is the set of congruence classes modulo $n$.

2. Remark 1: Convention on notation

Given a fixed integer $n$ we will often identity an integer number $a \in \mathbb{Z}$ with the corresponding congruence class in $\mathbb{Z}_{n}$. For example we will write

$$
5 \in \mathbb{Z}_{7}
$$

assuming the congruence class of integers [5] modulo 7. In particular we can write

$$
5=19 \text { in } \mathbb{Z}_{7}
$$

since 5 and 19 belong to the same congruence class.

This allows us, by abuse of notation, write

$$
\mathbb{Z}_{n}={0,1,2 \ldots, n-1}
$$

instead of the formally correct

$$
\mathbb{Z}_{n}={[0],[1],[2], \ldots,[n-1]}
$$

An element $k \in \mathbb{Z}{n}$ is called a $u n i t$, if there exists an element $l \in \mathbb{Z}{n}$ such that

$$
k \cdot l=1 \text { in } \mathbb{Z}_{n}
$$

The following theorem provide a complete characterization of units in $\mathbb{Z}_{n}$.

3. Theorem 1: Characterization of units in $\mathbb{Z}_{n}$

Let $n \geqslant 2$ be an integer. The set of units in $\mathbb{Z}{n}\left(\right.$ denoted $\left.\mathbb{Z}{n}^{\times}\right)$consists of all congruence class of integers coprime to $n$ :

$$
\mathbb{Z}{n}^{\times}=\left{k \in \mathbb{Z}{n} \mid \operatorname{gcd}(n, k)=1\right}
$$

Proof. Let $k \in{0, \ldots, n-1}$.

  1. If $\operatorname{gcd}(n, k)=1$, then by the theorem from the previous week, we can find integers $u, v \in \mathbb{Z}$ such that

$$
n u+k v=1 .
$$

Therefore, in $\mathbb{Z}_{n}$ we have

$$
k \cdot v \equiv 1 \quad \bmod n
$$

so that $[v]$ is the multiplicative inverse of $[k]$ in $\mathbb{Z}_{n} .$

  1. Conversely, if we can find a multiplicative inverse to $[k] \in \mathbb{Z}_{n}$ :

$$
k \cdot l \equiv 1 \quad \bmod n
$$

then any common divisor of $k$ and $n$ would also divide 1, so necessarily $\operatorname{gcd}(n, k)=1 .$

4. Example 1

In $\mathbb{Z}_{6}$ the only units are 1 and $5=-1$.

5. Definition 1

Given integer $n \geqslant 2$, define Euler’s function $\varphi(n)$ to be the number of integers in ${1, \ldots, n-1}$ which a coprime with $n$. In other words,

$$
\left|\mathbb{Z}_{n}^{\times}\right|=\varphi(n)
$$

Problem 1: Number $n$ is prime, if and only if $\varphi(n)=n-1 .$

6. Groups

7. Laws of composition

Let $S$ be an arbitrary set.

8. Definition 2: Law of composition

A law of composition (or binary operation) on $S$ is a function

$$
S \times S \rightarrow S
$$

which assigns to any pair $(x, y) \in S \times S$ an element $x * y \in S$. (Instead of $x * y$ often we will write $x \cdot y$, or even $x y$ )

9. Example 2

Addition and multiplication are composition laws on $\mathbb{Z}$.

Division is not a composition law on $\mathbb{R}$, since $x / 0$ is not defined. However, it is a composition law on $\mathbb{R} \backslash{0} .$

10. Example 3: Important example

Let $X$ be an arbitrary set, and consider

$$
S=\mathcal{F}(X, X)
$$

to be the set of all maps $f: X \rightarrow X$. Given a pair of maps $f, g \in \mathcal{F}(X, X)$, we can take their composition and obtain an element $f \circ g \in \mathcal{F}(X, X)$. Hence, the composition of maps defines a law of composition on $\mathcal{F}(X, X)$ (thus the name).

11. Definition 3

Let

$$
\begin{gathered}
S \times S \rightarrow S \
(x, y) \mapsto x * y
\end{gathered}
$$

be a law of composition on $S$. We say that * is

  • associative, if for any $x, y, z \in S$ we have

$$
(x * y) * z=x *(y * z) .
$$

  • commutative, if for any $x, y \in S$

$$
x * y=y * x
$$

If the law of composition is associative, given $x_{1}, \ldots, x_{n} \in S$, it makes sense to write just

$$
x_{1} * x_{2} * \cdots * x_{k}
$$

without any brackets.

Traditionally, we drop the symbol * and denote a law of composition by $(x, y) \mapsto x y($ this is called the multiplicative notation), but if it happens to be commutative, the additive notation $(x, y) \mapsto x+y$ may be used. For the moment, for clarity, we will keep using the symbol *.

12. Definition 4

Let $(S, *)$ be a set with a composition law. We say that $e \in S$ is an identity (or neutral element) if for all $x \in S$ we have

$$
x * e=e * x=x .
$$

Problem 2: Any law of composition has at most one identity element.

Hint: given two identities $e, e^{\prime}$ consider $e * e^{\prime}$.

13. Definition 5

Let $(S, *)$ be a set with a composition law with an identity e. We say that $y \in S$ is invertible if there exists $x \in S$ such that

$$
x * y=y * x=e \text {. }
$$

The inverse of $y$ is denoted by $y^{-1}$.

Clearly, $e$ is invertible with $e^{-1}=e$.

14. Proposition 1

Let $(S, *)$ be a set with an associative composition law and identity. Let $x, y \in S$ be two invertible elements, then $x * y$ is also invertible with $(x * y)^{-1}=y^{-1} * x^{-1}$

Proof. Indeed, we have

$$
(x * y) *\left(y^{-1} * x^{-1}\right)=x *\left(y * y^{-1}\right) * x^{-1}=x * e * x^{-1}=x * x^{-1}=e
$$

where in the first equality we used associativity, in the second the fact that $y^{-1}$ is the inverse of $y$, in the third the property of the identity $e$, and in the last the fact that $x^{-1}$ is the inverse of $x .$

Similarly we check that $\left(y^{-1} * x^{-1}\right) *(x * y)=e$ which proves that $\left(y^{-1} * x^{-1}\right)=(x * y)^{-1}$.

15. Example 4: Key motivating example

Let $\Delta$ be an equilateral triangle. Then it has 6 symmetries:

  • 3 rotations – by $0^{\circ}, 120^{\circ}$ and $240^{\circ}$ (denote them $\tau_{0}, \tau_{1}, \tau_{2}$ )
  • 3 reflections along 3 altitudes (denote them by $\sigma_{1}, \sigma_{2}, \sigma_{3}$ ).

We claim that the composition operation on the set $\left{\tau_{0}, \tau_{1}, \tau_{2}, \sigma_{1}, \sigma_{2}, \sigma_{3}\right}$ defines a law of composition (exercise).

Problem 3: Prove that the above law of composition is associative, has an identity and every element admits an inverse.

16. Definition of a group

17. Definition 6

Let $(G, *)$ be a set with a law of composition. We will say that $(G, *)$ si a group if the following properties hold:

  • * is associative
  • there is an identity $e \in G$
  • every element $x \in G$ is invertible.

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