# Group and Ring

## Definition 1

$\cdot: G\times G \to G$ is binary operation where we write $x\cdot y =xy$ for $x,y\in G$. $(G,\cdot)$ is group if it satisfies the followings (a) For all $x,y,z \in G$, $(x\cdot y)\cdot z = x\cdot (y\cdot z)$. (b) $\exists e \in G$, such that $x\cdot e = e\cdot x = x$, for all $x\in G$. (c) For every $x \in G$, $\exists \hat{x} \in G$ such that $x \hat{x} = \hat{x}x = e$.

## Definition 2

Let $(G,\cdot)$ be a group and $H \subset G$ be a subgroup satisfying the three requirements of group as described in Definition 1. We write $(H,\cdot) \leq (G,\cdot)$.

## Definition 3

Let $(G,\cdot)$ is a group and $(H,\cdot) \leq G$. $H$ is normal subgroup if $g^{-1}hg\in H$ for all $g \in G, h \in H$. We write $H \trianglelefteq G$.

## Definition 4

Let $G$ be a group and $H$ be a subgroup of $G$. For $x \in G$, we define a coset as follows: \begin{align} xH := \{xh| h\in H\} \end{align} We call it left coset ($\because xh \neq hx$ if the group is not abelian).

## Definition 5

Let $(G, \cdot)$ be a group and $(H,\cdot)$ be a normal subgroup of $G$. Then we define quotient group as follows: $G/H := \{xH| x \in G \}$ For notational simplicity, we write a coset $\overline{x} := xH$.

## Theorem 6

Let $G$ be a group and $H$ be a subgroup of $G$. For $x,y \in G$, \begin{align} xH = yH \iff y^{-1}x \in H \end{align}

<proof> $\Rightarrow$ Suppose that $xH = yH$. Then $xh_1 = yh_2$ for some $h_1, h_2 \in H$, i.e. $y^{-1}x = h_2 h^{-1}_1 \in H.$

$\Leftarrow$ Suppose that $y^{-1}x \in H$, i.e. $y^{-1}x = h_0$ for some $h_0 \in H$. i.e. $x = yh_0$.

For all $h \in H$, $xh = yh_0h\in yH$. $\therefore xH \subset yH$.

Similarly, $y=xh^{-1}_0$. For all $h \in H$, $yh=xh^{-1}_0h \in xH$.

$\therefore yH \subset xH.$

$\therefore xH = yH$.

$\tag*{\square}$

## Theorem 6

Let $G$ be a group and $H$ is a normal subgroup of $G$. The binary operation $\cdot$ on quotient group $G/H$ is defined as $xH\cdot yH := xyH$. where $x,y \in G$.

Then the binary operation is well defined if and only if $H$ is a normal subgroup.

<proof> $\Leftarrow$ Suppose that $H$ is a normal subgroup of $G$.

We want to show that $xH = x^\prime H, yH = y^\prime H \Rightarrow xyH = x^\prime y^\prime H$. \begin{align} \begin{split} (x^\prime y^\prime)^{-1} &= y^{\prime -1}x^{\prime -1} xy \\ &=(y^{\prime -1}y)(y^{\prime -1}x^{\prime -1}xy) \end{split} \end{align} Since $H \trianglelefteq G$, $y^{\prime -1}x^{\prime -1}xy \in H$.

$\therefore (y^{\prime -1}y)(y^{\prime -1}x^{\prime -1}xy) \in H.$

$\therefore$ The binary operation is well defined.

$\Rightarrow$ Take $y = y^\prime = g, x= h, x^\prime = 1.$

Then $y^{\prime -1}x^{\prime -1}xy = g^{-1}hg \in H.$

$\therefore H \trianglelefteq G.$

$\tag*{\square}$

## Definition 7

$(A, +, \cdot)$ is a ring if

(a) $(A, +)$ is an abelian group

(b) The binary operation $\cdot$ is associative and distributive over $+$

(c) $\cdot$ is commutative

(d) $\exists 1 \in A$ such that $x1=1x = x$ for all $x \in A$.

## Note

1) The identity 1 is unique. $(\because 1^\prime$: another identity in $A \Rightarrow 1= 1^\prime 1 = 1^\prime 1 = 1^\prime$) 2) $1=0 \iff A = 0$ (zero ring) $(\because \Rightarrow ) x\in A, x = x1 = x0 =0. \Leftarrow)$ trivial.$)$

## Definition 8 Let $A,B$ be rings. $f: A\to B$ is a ring homomorphism if

1. $f(x+y) = f(x) + f(y)$
2. $f(xy) = f(x)f(y)$
3. $f(1)=1$

## Definition 9

$S \subset (A, +, \cdot)$ is a subring if $(S, +, \cdot)$ is a ring.

## Definition 10

Let $A$ be a ring. $\mathfrak{a} \subset A$ is an ideal of $A$ if $(\mathfrak{a},+) \leq (A, +)$ and $A\mathfrak{a} \subset \mathfrak{a}$, i.e. $[x\in A, y \in \mathfrak{a} \Rightarrow xy \in \mathfrak{a}]$.

## Remark 11

Since $(A, +)$ is abelian and $(\mathfrak{a}, +) \leq (A,+)$, we can think of the quotient group $(A/\mathfrak{a}, +)$. Moreover, we can give a multiplication on $A/\mathfrak{a}$ by $\overline{x}\cdot\overline{y}:=(x+\mathfrak{a})(y+\mathfrak{a}) := xy + \mathfrak{a}$ where $x,y \in A$.

## Theorem 12

Let $A$ be a ring and $J$ be an additive subgroup. The binary operation $\overline{x}\cdot\overline{y}:= (x+J)(y+J)$ is well defined if and only if $\mathfrak{a}$ is an ideal, where $x,y \in A$.

<proof> $\Rightarrow$ Suppose that $\overline{x}\cdot\overline{y}$ is well defined where $x,y \in A$. We want to show that for all $a \in A, j\in J$, $aj \in J$. Take $x:=a \in A$ and $y:=0 \in A$. Then, $(a+J)J = J$. Since $a\in a+J, aj \in (a+J)J = J$.

$\therefore J$ is an ideal

$\Leftarrow$ Suppose that $J$ is an ideal We want to show that $x_1 - x_2 \in J, y_1 -y_2 \in J \Rightarrow x_1y_1 -x_2y_2 \in J$ where $x_1, x_2, y_1, y_2 \in J$. Since $J$ is an ideal and $x_1 - x_2 \in J, y_1 - y_2 \in J$, $y_1(x_1-x_2) \in J, x_2(y_1-y_2) \in J$. Since $J$ is an additive subgroup, \begin{align} \begin{split} J &\ni y_1(x_1-x_2) + x_2(y_1-y_2) \\ &=y_1x_1 = y_1x_2 + x_2y_1 - x_2y_2 \\ &=x_1y_1 - x_2y_1 + x_2y_1 - x_2y_2 \\ &= x_1y_1 - x_2y_2 \end{split} \end{align} The second equality holds because $A$ is ring so $\cdot$ is commutative.

$\therefore x_1y_1 - x_2y_2 \in J$

$\tag*{\square}$

## Reference

• Lecture note from Hobin Jeong.
• 이인석, 『선형대수와 군』,서울대학교출판문화원

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