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# 数学代写|抽象代数代写Abstract Algebra代考|MATH411 Turning a set of cosets into a group

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## 数学代写|抽象代数代写Abstract Algebra代考|Turning a set of cosets into a group

We begin by revisiting Example 19.1. Let $H={1,3,9}$ be a subgroup of $U_{13}$. The distinct cosets of $H$ are the following:

$1 H={1,3,9}=3 H=9 H$.

$2 H={2,6,5}=6 H=5 H$.

$4 H={4,12,10}=12 H=10 H$.

$7 H={7,8,11}=8 H=11 H$.
We can use the property $a \in a H$ (Theorem 19.12) and that distinct cosets do not overlap (Theorem 20.4) when finding these coset representatives. For instance, once we compute $2 H={2 \cdot 1,2 \cdot 3,2 \cdot 9}={2,6,5}$, then we know that this coset also equals $6 H$ and $5 H$, because 6 and 5 are contained in $2 H$.

Notation. We define $U_{13} / H$ (read ” $U_{13} \bmod H$ “) to be the set of distinct cosets of $H$. Thus,
$$U_{13} / H={1 H, 2 H, 4 H, 7 H} .$$
Since different coset representatives can generate the same coset (e.g., $2 H=6 H)$, we could have written $U_{13} / H$ slightly differently, perhaps like this:
$$U_{13} / H={1 H, 6 H, 10 H, 11 H} .$$

## 数学代写|抽象代数代写Abstract Algebra代考|Coset multiplication shortcut

As we saw in Examples 21.2, 21.3, and 21.4, coset multiplication can be a tedious process. Fortunately, there is a shortcut. To motivate this shortcut, let’s examine the earlier examples in more depth.

Example 21.7. Consider the subgroup $H={1,3,9}$ of $U_{13}$. We found the coset product $2 H \cdot 4 H=7 H$. But $7 H$ can be written as $8 H$ (i.e., they’re the same coset). Thus $2 H \cdot 4 H=8 H$, where we observe that $2 \cdot 4=8$ in $U_{13}$. We also found that $1 H \cdot 4 H=4 H$, where $1 \cdot 4=4$ in $U_{13}$. For the product $4 H \cdot 4 H=1 H$, we observe that $1 H$ can also be written as $3 H$. Hence we have $4 H \cdot 4 H=3 H$, where $4 \cdot 4=3$ in $U_{13}$.

Example $21.7$ above suggests that to find the coset product $a H \cdot b H$ in $G / H$, we can simply multiply their coset representatives in $G$; i.e., $a H \cdot b H=(a b) H$. The theorem below is for the case when $G$ is commutative, such as $G=U_{13}$. Its proof is left for you as an exercise at the end of the chapter.

Theorem 21.8 (Coset multiplication shortcut). Let $G$ be a commutative group, $H$ a subgroup of $G$, and $a, b \in G$. Define the coset product by $a H \cdot b H={\alpha \cdot \beta \mid \alpha \in a H$, $\beta \in b H}$. Then $a H \cdot b H=(a b) H$.

## 数学代写|抽象代数代写|将一个宇宙集变成一个群

$$1 H=1,3,9=3 H=9 H$$
$$2 H=2,6,5=6 H=5 H\text {. }$$
$$4 H=4,12,10=12 H=10 H\text {. }$$
$$7 H=7,8,11=8 H=11 H\text {. }$$

$$U_{13} / h=1 h, 2 h, 4 h, 7 h 。$$

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