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# 数学代写|抽象代数代写Abstract Algebra代考|MATH411 Right cosets

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## 数学代写|抽象代数代写Abstract Algebra代考|Right cosets

In Section 19.1, we considered left cosets of the form $a H$, where we multiplied each element of $H$ on the left by $a$. We can also consider right cosets $H a$, as shown in the example below.

Example 19.7. Consider again the group $U_{13}$ and its subgroup $H={1,3,9}$. The left coset $6 H$ is given by $6 H={6 \cdot 1,6 \cdot 3,6 \cdot 9}={6,5,2}$, and we have the right coset $H 6={1 \cdot 6,3 \cdot 6,9 \cdot 6}={6,5,2}$. Observe that $6 H=H 6$; i.e., the left and right cosets are equal, because $U_{13}$ is commutative.

As seen in Example 19.7, the distinction between left and right cosets is irrelevant in a commutative group. In particular, additive groups are always commutative, so there is no distinction between the left coset $a+H$ and the right coset $H+a$. Thus, we will only consider left cosets $a+H$ with additive groups.
Here is a non-commutative example, where things get a bit more interesting.
Example 19.8. Let $H={\varepsilon, d}$ be a subgroup of $D_4$. Let’s compute and compare the left coset $r_{90} \mathrm{H}$ and the right coset $H r_{90}$ :

• $r_{90} H=\left{r_{90} \cdot \varepsilon, r_{90} \cdot d\right}=\left{r_{90}, h\right}$.
• $H r_{90}=\left{\varepsilon \cdot r_{90}, d \cdot r_{90}\right}=\left{r_{90}, v\right}$.
Therefore, the left and right cosets are not the same; i.e., $r_{90} H \neq H r_{90}$.
Example 19.9. Let $K=C(h)=\left{\varepsilon, r_{180}, h, v\right}$ be a subgroup of $D_4$. (It’s the centralizer of $h$ in $D_4$. See Section 5.3.) Let’s compute and compare the left coset $d K$ and the right $\operatorname{coset} K d$ :
• $d K=\left{d \cdot \varepsilon, d \cdot r_{180}, d \cdot h, d \cdot v\right}=\left{d, d^{\prime}, r_{270}, r_{90}\right}$.
• $K d=\left{\varepsilon \cdot d, r_{180} \cdot d, h \cdot d, v \cdot d\right}=\left{d, d^{\prime}, r_{90}, r_{270}\right}$
Thus we have a coset equality $d K=K d$, because these sets contain the same four elements. But this does not imply that we have an element-by-element equality; i.e., $d k=k d$ for all $k \in K$. Indeed, we have $d h \neq h d$ and $d v \neq v d$, where $h, v \in K$.

## 数学代写|抽象代数代写Abstract Algebra代考|Structure of Finite Fields

The next theorem tells us the additive and multiplicative group structure of a field of order $p^n$.
Theorem 21.9 Structure of Finite Fields
As a group under addition, $G F\left(p^n\right)$ is isomorphic to
$$\underset{n \text { factors }}{Z_p \oplus Z_p \oplus \ldots \oplus Z_p} .$$
As a group under multiplication, the set of nonzero elements of $G F\left(p^n\right)$ is isomorphic to $Z_{p^n-1}$ ( and is, therefore, cyclic).

PROOF Since GF $\left(p^n\right)$ has characteristic $p$ (Theorem 13.3), every nonzero element of GF $\left(p^n\right)$ has additive order $p$. Then by the Fundamental Theorem of Finite Abelian Groups, $\operatorname{GF}\left(p^{\prime}\right)$ under addition is isomorphic to a direct product of $n$ copies of $Z_p$.
To see that the multiplicative group $\operatorname{GF}\left(p^n\right)^$ of nonzero elements of $\mathrm{GF}\left(p^n\right)$ is cyclic, we first note that by the Fundamental Theorem of Finite Abelian Groups (Theorem 11.1), $\mathrm{GF}\left(p^n\right)^$ is isomorphic to a direct product of the form $Z_{n_1} \oplus Z_{n_2} \oplus \cdots \oplus Z_{n_m}$. If the orders of these components are pairwise relatively prime, then it follows from Corollary 1 of Theorem $8.2$ that $\operatorname{GF}\left(p^m\right)^$ is cyclic. Hence we may assume that there is an integer $d>1$ that divides the orders of two of the components. From the Fundamental Theorem of Cyclic Groups (Theorem 4.3) we know that each of these components has a subgroup of order $d$. This means that $\operatorname{GF}\left(p^n\right)^$ has two distinct subgroups of order $d$, call them $H$ and $K$. But then every element of $H$ and $K$ is a zero of $x^d-1$, which contradicts the fact that a polynomial of degree $d$ over a field can have at most $d$ zeros (Theorem 16.3).

## 数学代写|抽象代数代写Abstract Algebra代考|Right cosets

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$\mathrm{IE}, d k=k d$ 对所有人 $k \in K$. 确实，我们有 $d h \neq h d$ 和 $d v \neq v d$ ，在哪里 $h, v \in K$.

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