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# 数学代写|抽象代数代写Abstract Algebra代考|Cayley’s theorem

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## 数学代写|抽象代数代写Abstract Algebra代考|Cayley’s theorem

What began as a simple example of permutations from geometry gave rise to an interesting concept: recognizing a given group as a subgroup of a symmetric group. Is that possible for all groups? Not only is it true, but the theorem that answers our question is a famous result in abstract algebra.

Lemma 10.14. Let $G$ be a group. For each $g \in G$, define $\lambda_g: G \rightarrow G$ by $\lambda_g(x)=g x$, and let $\Lambda=\left{\lambda_g \mid g \in G\right}$. Likewise, for each $g \in G$, define $\rho_g: G \rightarrow G$ by $\rho_g(x)=x g$, and let $P=\left{\rho_g \mid g \in G\right}$. Then both $\Lambda$ and $P$ are subgroups of $S_G$.

Theorem 10.15 (Cayley’s Theorem). Let $G$ be a group. Then $G$ is isomorphic to a subgroup of $S_G$. In particular, every finite group of order $n$ is isomorphic to a subgroup of $S_n$

Notice that what Cayley’s theorem says is that every group $G$, no matter how large, is a subgroup of the (rather large) symmetric group $S_G$. You might be able to find a subgroup isomorphic to $G$ in a “smaller” symmetric group, and there might be multiple subgroups of $S_G$ isomorphic to $G$. In fact, if you used the lemma in your proof of Cayley’s theorem, then you showed that there are, in fact, at least two ways to view $G$ as a subgroup of $S_G$ : as the subgroups $\Lambda$ and $P$ of $S_G$.

Exercise 10.16. Let’s apply Cayley’s theorem to some known groups. For each small group $G$ below, find a collection of permutations of the elements of $G$ that correspond to $\Lambda$ and to $P$ as given by Cayley’s theorem.
(1) $G=\mathbb{Z}_3$
(3) $G=\mathbb{Z}_2 \times \mathbb{Z}_2$
(2) $G=\mathbb{Z}_4$
(4) $G=S_3$

## 数学代写|抽象代数代写Abstract Algebra代考|Orbits and cycles

Our ability to recognize the dihedral groups as subgroups of the symmetric groups suggests that we might be able to think of any permutation of a set $A$ as “moving” the elements of $A$ around. Let’s see if we can develop this idea by considering what repeated application of an individual permutation in $S_A$ does to an element of the set $A$.

Definition 11.1. Let $A$ be a set, let $a \in A$, and let $\sigma \in S_A$. Then the orbit of $a$ under $\sigma$ is the $\operatorname{set}\left{\sigma^n(a) \mid n \in \mathbb{Z}\right}$
Theorem 11.2. Let $A$ be a set, and let $\sigma \in S_A$. Then the relation $\sim$ on $A$ defined by $a \sim b$ if and only if $b$ is in the orbit of $a$ under $\sigma$
is an equivalence relation on $A$ whose equivalence classes are the orbits of the elements of A under $\sigma$.

Definition 11.3. Let $A$ be a set. A permutation of $A$ is a cycle if and only if the permutation has at most one orbit with more than one element, and the length of a cycle is the number of elements in its largest orbit. A cycle of length one is the trivial cycle, and a cycle of length two is a transposition. We say two nontrivial cycles are disjoint if their largest orbits are disjoint.

Exercise 11.4. List all the orbits of the given element of $S_6$. Which are cycles? Which are transpositions?
(1) $f(1)=5, f(2)=2, f(3)=1, f(4)=6, f(5)=3, f(6)=4$.
(2) $f(1)=2, f(2)=4, f(3)=1, f(4)=5, f(5)=6, f(6)=3$.
(3) $f(1)=1, f(2)=5, f(3)=3, f(4)=6, f(5)=2, f(6)=4$.
(4) $f(1)=4, f(2)=2, f(3)=3, f(4)=1, f(5)=5, f(6)=6$.

(5) $f(1)=3, f(2)=1, f(3)=6, f(4)=4, f(5)=5, f(6)=2$.
(6) $f(1)=1, f(2)=2, f(3)=3, f(4)=4, f(5)=5, f(6)=6$.

## 数学代写|抽象代数代写Abstract Algebra代考|Cayley’s theorem

(1) $G=\mathbb{Z}_3$
(3) $G=\mathbb{Z}_2 \times \mathbb{Z}_2$
(2) $G=\mathbb{Z}_4$
（4） $G=S_3$

## 数学代写|抽象代数代写Abstract Algebra代考|Orbits and cycles

11.1.定义设$A$为一组，设$a \in A$，设$\sigma \in S_A$。那么$\sigma$下面的$a$轨道就是$\operatorname{set}\left{\sigma^n(a) \mid n \in \mathbb{Z}\right}$

11.3.定义设$A$为集合。$A$的排列是一个循环当且仅当该排列最多有一个包含多个元素的轨道，周期的长度是其最大轨道上的元素数。长度为1的循环是平凡循环，长度为2的循环是转置。我们说两个非平凡循环是不相交的如果它们最大的轨道是不相交的。

(1) $f(1)=5, f(2)=2, f(3)=1, f(4)=6, f(5)=3, f(6)=4$。
(2) $f(1)=2, f(2)=4, f(3)=1, f(4)=5, f(5)=6, f(6)=3$。
(3) $f(1)=1, f(2)=5, f(3)=3, f(4)=6, f(5)=2, f(6)=4$。
(4) $f(1)=4, f(2)=2, f(3)=3, f(4)=1, f(5)=5, f(6)=6$。
(5) $f(1)=3, f(2)=1, f(3)=6, f(4)=4, f(5)=5, f(6)=2$。
(6) $f(1)=1, f(2)=2, f(3)=3, f(4)=4, f(5)=5, f(6)=6$。

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