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数学代写|现代代数代考Modern Algebra代写|MATH612

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数学代写|现代代数代考Modern Algebra代写|Product of Subsets

Let $A$ and $B$ be nonempty subsets of the group $G$. The product $A B$ is defined by
$$A B={x \in G \mid x=a b \text { for some } a \in A, b \in B} .$$
This product is formed by using the group operation in $G$. A more precise formulation would be
$$A * B={x \in G \mid x=a * b \text { for some } a \in A, b \in B},$$
where $*$ is the group operation in $G$.
Several properties of this product are worth mentioning. For nonempty subsets $A, B$, and $C$ of the group $G$,
\begin{aligned} A(B C) & ={a(b c) \mid a \in A, b \in B, c \in C} \ & ={(a b) c \mid a \in A, b \in B, c \in C} \ & =(A B) C . \end{aligned}
It is obvious that
$$B=C \Rightarrow A B=A C \text { and } B A=C A,$$
but we must be careful about the order because $A B$ and $B A$ may be different sets.

数学代写|现代代数代考Modern Algebra代写|Properties of the Product of Subsets

Let $A, B$, and $C$ denote nonempty subsets of the group $G$, and let $g$ denote an element of $G$. Then the following statements hold:
a. $A(B C)=(A B) C$.
b. $B=C$ implies $A B=A C$ and $B A=C A$.
c. The product $A B$ is not commutative.
d. $A B=A C$ does not imply $B=C$.
e. $g A=g B$ implies $A=B$.
Statements $\mathbf{d}$ and $\mathbf{e}$ have obvious duals in which the common factor is on the right side.

We shall be concerned mainly with products of subsets in which one of the factors is a subgroup. The cosets of a subgroup are of special importance.

Let $H$ be a subgroup of the group $G$. For any $a$ in $G$,
$$a H={x \in G \mid x=a h \text { for some } h \in H}$$
is a left coset of $H$ in $G$. Similarly, $H a$ is called a right coset of $H$ in $G$.
The left coset $a H$ and the right coset $H a$ are never disjoint, since $a=a e=e a$ is in both sets. In spite of this, $a H$ and $H a$ may happen to be different sets, as the next example shows.

Example 3 Consider the subgroup
$$K={(1),(1,2)}$$
of
$$G=S_3={(1),(1,2,3),(1,3,2),(1,2),(1,3),(2,3)} .$$
For $a=(1,2,3)$, we have
\begin{aligned} a K & ={(1,2,3),(1,2,3)(1,2)} \ & ={(1,2,3),(1,3)} \end{aligned}
and
\begin{aligned} K a & ={(1,2,3),(1,2)(1,2,3)} \ & ={(1,2,3),(2,3)} . \end{aligned}
In this case, $a K \neq K a$.

现代代数代写

数学代写|现代代数代考Modern Algebra代写|Product of Subsets

$$A B={x \in G \mid x=a b \text { for some } a \in A, b \in B} .$$

$$A * B={x \in G \mid x=a * b \text { for some } a \in A, b \in B},$$

\begin{aligned} A(B C) & ={a(b c) \mid a \in A, b \in B, c \in C} \ & ={(a b) c \mid a \in A, b \in B, c \in C} \ & =(A B) C . \end{aligned}

$$B=C \Rightarrow A B=A C \text { and } B A=C A,$$

数学代写|现代代数代考Modern Algebra代写|Properties of the Product of Subsets

A. $A(B C)=(A B) C$;
B. $B=C$暗示$A B=A C$和$B A=C A$。
c.乘积$A B$不可交换。
D. $A B=A C$并不暗示$B=C$。
E. $g A=g B$暗示$A=B$。

$$a H={x \in G \mid x=a h \text { for some } h \in H}$$

$$K={(1),(1,2)}$$

$$G=S_3={(1),(1,2,3),(1,3,2),(1,2),(1,3),(2,3)} .$$

\begin{aligned} a K & ={(1,2,3),(1,2,3)(1,2)} \ & ={(1,2,3),(1,3)} \end{aligned}

\begin{aligned} K a & ={(1,2,3),(1,2)(1,2,3)} \ & ={(1,2,3),(2,3)} . \end{aligned}

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