## 数学代写|抽象代数作业代写abstract algebra代考|MATH2701

2023年2月7日

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## 数学代写|抽象代数作业代写abstract algebra代考|Definition and Examples

Since a subgroup $H$ of a group $G$ must be nonempty, there exists some $x \in H$. Since $H$ is closed under taking inverses, then $x^{-1} \in H$. Since $H$ is closed under the group operation, then $e=x x^{-1} \in H$. Hence, a subgroup contains the identity element. The property of associativity is inherited from associativity in $G$ and so since $e \in H$ and $H$ is closed under taking inverses, $H$ equipped with the binary operation on $G$ is a group in its own right. (As a point of terminology, it is important to understand that we do not say that “a group is closed under an operation.” Such a statement is circular since a binary operation on $G$ by definition maps any pair of elements in $G$ back into $G$. The terminology of “closed under an operation” is a matter of concern only for strict subsets of $G$.)

Example 1.6.2. With the usual addition operation, $\mathbb{Z} \leq \mathbb{Q} \leq \mathbb{R} \leq \mathbb{C}$. With the multiplication operation we have $\mathbb{Q}^* \leq \mathbb{R}^* \leq \mathbb{C}^$. However, $\mathbb{Z}^$ is not a subgroup of $\mathbb{Q}^$, written $\mathbb{Z}^ \leq \mathbb{Q}^$, because even though $\mathbb{Z}^$ is closed under multiplication, it is not closed under taking multiplicative inverses. For example $2^{-1}=\frac{1}{2} \notin \mathbb{Z}^*$

Example 1.6.3. Any group $G$ always has at least two subgroups, the trivial subgroup ${e}$ and all of $G$.

Example 1.6.4. If $G=D_n$, then $R=\left{\iota, r, r^2, \ldots, r^{n-1}\right}$ is a subgroup. This is the subgroup of rotations. Also for all integers $i$ between 0 and $n-1$, the subsets $H_i=\left{\iota, s r^i\right}$ are subgroups. These subgroups of two elements correspond to reflection about various lines of symmetry.

Example 1.6.5. Let $G=S_n$ and consider the subset of permutations that leave the elements ${m+1, m+2, \ldots n}$ fixed. This is a subgroup of $S_n$ that consists of all elements in $S_m$.

## 数学代写|抽象代数作业代写abstract algebra代考|Subgroups Generated by a Subset

By virtue of Proposition 1.7.2, $\langle S\rangle$ is called the subgroup generated by $S$. (In the analogy with vector spaces, a subgroup generated by a subset is like the span of a set of elements in a vector space, which is a subspace.)

Example 1.7.3. Let $G=D_6$ be the dihedral group on the hexagon. The subgroup $\langle r\rangle$ consists of all powers of $r$, so is $\langle r\rangle=\left{\iota, r, r^2, r^3, r^4, r^5\right}$. Notice that $\langle r\rangle$ is the subgroup of rotations.

The subgroup $\langle s\rangle={1, s}$ consists of only two elements, reflection across the reference axis of symmetry and the identity transformation.

The subgroup $\left\langle s, r^2\right\rangle$ contains the elements $\iota, s, r^2$, and $r^4$, by virtue of taking powers of elements in $\left{s, r^2\right}$. However, $\left\langle s, r^2\right\rangle$ also contains $s r^2$ and $s r^4$. The defining relation on $s$ and $r$ give $r^a s=s r^{6-a}$. Hence, as we apply this relation, the parity on the power of $r$ does not change. Hence,
$$\left\langle s, r^2\right\rangle=\left{\iota, r^2, r^4, s, s r^2, s r^4\right} .$$
Finally, consider the subgroup $\langle s, s r\rangle$. Obviously this subgroup contains $s$ but it also contains $r=s(s r)$. Hence, $\langle s, s r\rangle=D_6$ because it contains all rotations and all reflections.

For any element $a$ in a group $C$, the subgroup $\langle a\rangle$ is a cyclic subgroup of $G$ whose order is precisely the order $|a|$. It is important to note that distinct sets of generators may give that same subgroup. In the previous example, we noted that $\langle r, s r\rangle=\langle r, s\rangle$. This occurs even with cyclic subgroups. For example, in $D_6$, the rotation subgroup is $\langle r\rangle=\left\langle r^5\right\rangle$. In $D_6$, we also have $\left\langle r^2\right\rangle=\left\langle r^4\right\rangle$.

## 数学代写|抽象代数作业代写abstract algebra代考|Definition and Examples

Imathbb ${Q}^{\wedge} \backslash \backslash e q \backslash m a t h b b{R}^{\wedge} \backslash l e q \backslash m a t h b b{C}^{\wedge}$. 然 而， $\backslash \operatorname{lmathbb}{Z}^{\wedge}$ 不是的子群 $\backslash m a t h b b{Q}^{\wedge}$ ，写
Imathbb ${Z}^{\wedge} \backslash$ leq $\left.\backslash m a t h b b{Q}^{\wedge}\right}$, 因为即使 $\backslash m a t h b b{Z}^{\wedge}$

$\mathrm{R}=$ Vleft:{iota, $r, r \wedge 2$, lddots, $r \wedge{n-1} \backslash r i g h t}$ 是一个子群。这 是旋转子群。也适用于所有整数 $i$ 介于 0 和 $n-1$, 子集 H_i $=\backslash$ left{iota, s r^ilright $}$ 是子群。这两个元素的子群对 应于关于各种对称线的反射。

## 数学代写|抽象代数作业代写abstract algebra代考|Subgroups Generated by a Subset

Vangle r rrangle= $=\backslash$ left{{iota, $\left.r, r^{\wedge} 2, r^{\wedge} 3, r \wedge 4, r \wedge 5 \backslash r i g h t\right}$. 请 注意 $\langle r\rangle$ 是旋转子群。

Veft $\backslash a n g l e ~ s, r^{\wedge} 2 \backslash r i g h t \backslash r a n g l e=\backslash l e f t\left{\backslash\right.$ iota, $r^{\wedge} 2, r^{\wedge} 4, s, s r$.

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## MATLAB代写

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