$$\exists n_1, n_2 \in N, h_1, h_2 \in H, \quad g_1=n_1 h_1, \quad g_2=n_2 h_2 .$$

\begin{aligned} g_1 g_2 & =n_1 h_1 n_2 h_2 \ & =n_1 h_1 n_2 h_1^{-1} h_1 h_2 \ & =n_1 \gamma_{h_1}\left(n_2\right) h_1 h_2 \in N H . \end{aligned}

UpriviateTA有若干位非常擅长Abstract algebra的老师可以提供抽象代数的代写，代考和辅导服务，可以确保您在学习Abstract algebra的过程中取得满意的成绩。

Problem 1.

we prove that is $p$ is an odd prime then $U\left(p^k\right)$ is a cyclic group of order $p^{k-1}(p-1)$.
(a) Prove that if $k \geq 2$ and $a \in \mathbb{Z}$ with $p \nmid a$, then $(1+a p)^{p^{k-2}} \equiv 1+a p^{k-1}$ $\left(\bmod p^k\right)$.
(b) Deduce that for any $a$ with $p \nmid a$, the element $1+a p$ has order $p^{k-1}$ in $U\left(p^k\right)$.
(c) By Proposition 5.6.2, $U\left(\mathbb{F}_p\right)=U(p)$ is a cyclic group. Show that there exists $g \in \mathbb{Z}$ that is a generator of $U(p)$ and such that $g^{p-1} \not \equiv 1\left(\bmod p^2\right)$.
(d) Prove that a $g$ found in the previous part generates $U\left(p^k\right)$ to deduce that $U\left(p^k\right)$ is cyclic.
[Hint: Recall that $p \mid\left(\begin{array}{c}p \ j\end{array}\right)$ for all $j$ with $1 \leq j \leq p-1$.]

Problem 2.

Guides a proof that $\operatorname{Aut}\left(S_n\right)=S_n$ for all $n \neq 6$.
(a) Prove that for all $\psi \in \operatorname{Aut}\left(S_n\right)$ and all conjugacy classes $\mathcal{K}$ of $S_n$, the subset $\psi(\mathcal{K})$ is another conjugacy class.
(b) Let $\mathcal{K}$ be the conjugacy class of transpositions and let $\mathcal{K}^{\prime}$ be another conjugacy class of elements of order 2 (e.g., cycle type like $(a b)(c d)$ ). Prove that $|\mathcal{K}| \neq\left|\mathcal{K}^{\prime}\right|$, unless possibly if $n=6$.
(c) Prove that for each $\psi \in \operatorname{Aut}\left(S_n\right)$ and for all $k$ with $2 \leq k \leq n$, we have $\psi((1 k))=\left(a b_k\right)$ for some distinct integers $a, b_2, b_3, \ldots, b_n$ in ${1,2, \ldots, n}$.
(d) Show that the transpositions (12), (13), ., (1 $n)$ generate $S_n$.
(e) Deduce that Aut $\left(S_n\right)=\operatorname{Inn}\left(S_n\right) \cong S_n$.

## Semidirect Product

Suppose a group $G$ contains a normal subgroup $H$. Suppose also that $K$ is another subgroup of $G$ with $H \cap K=1$. Then $K$ acts on $H$ by conjugation. This action defines a homomorphism $\varphi: K \rightarrow \operatorname{Aut}(H)$. Furthermore, $H K$ is a subgroup of $G$ such that
$$k h k^{-1}=\varphi(h) \Longleftrightarrow k h=\varphi(h) k .$$
This remark sets up the construction of the semidirect product.

Let $H$ and $K$ be groups and let $\varphi: K \rightarrow \operatorname{Aut}(H)$ be a homomorphism. The Cartesian product $H \times K$, equipped with the operation
$$\left(h_1, k_1\right) \cdot\left(h_2, k_2\right)=\left(h_1 \varphi\left(k_1\right)\left(h_2\right), k_1 k_2\right),$$
is a group. Furthermore, if $H$ and $K$ are finite, then this group has order $|H||K|$.

semidirect product definition

Suppose that $H$ and $K$ are groups such that there exists a homomorphism $\varphi: K \rightarrow \operatorname{Aut}(H)$. The Cartesian product $H \times K$, equipped with the operation defined in (6.6), is called the semidirect product of $H$ and $K$ with respect to $\varphi$ and is denoted by $H \rtimes_{\varphi} K$.

## Relationship of $H$ and $K$ inside $H \rtimes_{\varphi} K$

Before developing this example further and presenting more examples, it is useful to explore the relationship of $H$ and $K$ inside $H \rtimes_{\varphi} K$. However, implicit in Example 6.6.5 is that there always exist a homomorphism $K \rightarrow$ Aut $(H)$, namely the trivial homomorphism, which maps all elements in $K$ to the identity automorphism. The following proposition describes this situation.
Proposition
The following are equivalent.
(1) $\varphi$ is the trivial homomorphism into $\operatorname{Aut}(H)$;
(2) $H \rtimes_{\varphi} K \cong H \oplus K$ with the isomorphism being the set identity function;
(3) $K \unlhd H \rtimes_{\varphi} K$.
Proof. (1) $\Longrightarrow(2)$ If $\varphi$ is trivial, then $\varphi(k)$ is the identity function. Hence,
$$\left(h_1, k_1\right) \cdot\left(h_2, k_2\right)=\left(h_1 \varphi\left(k_1\right)\left(h_2\right), k_1 k_2\right)=\left(h_1 h_2, k_1 k_2\right) .$$
Thus, $H \rtimes_{\varphi} K \cong H \oplus K$.
(2) $\Longrightarrow$ (3) We know that $K \unlhd H \oplus K$.
(3) $\Longrightarrow$ (1) Suppose that $K \unlhd H \rtimes_{\varphi} K$. Then for all $h_2 \in H$ and all $k_1, k_2 \in K$, the following element is in $K$ :
\begin{aligned} \left(h_2, k_2\right) \cdot\left(1, k_1\right) \cdot\left(h_2, k_2\right)^{-1} & =\left(h_2 \varphi\left(k_2\right)(1), k_2 k_1\right) \cdot\left(\varphi\left(k_2^{-1}\right)\left(h_2^{-1}\right), k_2^{-1}\right) \ & =\left(h_2 \varphi\left(k_2 k_1\right)\left(\varphi\left(k_2^{-1}\right)\left(h_2^{-1}\right)\right), k_2 k_1 k_2^{-1}\right) . \end{aligned}
Thus, $h_2 \varphi\left(k_2 k_1 k_2^{-1}\right)\left(h_2^{-1}\right)=1$ for all $h_2$ so $\varphi\left(k_2 k_1 k_2^{-1}\right)$ is the identity function for all $k_1, k_2$. Setting $k_2=1$ shows that $\varphi\left(k_1\right)$ is the identity function for all $k_1 \in K$ so $\varphi$ is trivial.

## Wreath Product

Consider the standard action of $S_9$ on $X={1,2, \ldots, 9}$. We propose to find a subgroup $H \leq S_9$ that acts transitively on $X$ and has ${1,2,3}$ as a block.

As a subgroup of $S_9$ it has generators
$$G=\langle(123),(147)(258)(369)\rangle .$$
It consists of permutations that cycle within the blocks ${1,2,3},{4,5,6}$, and ${7,8,9}$ and permutations that cycle through the three blocks. The action of $H$ that stays within the blocks is the subgroup
$$H=\langle(123),(456),(789)\rangle$$
and is isomorphic to $Z_3 \times Z_3 \times Z_3$. The generating permutation $\sigma=$ $(147)(258)(369)$ satisfies
$$\sigma(123) \sigma^{-1}=(456), \quad \sigma(456) \sigma^{-1}=(789), \quad \text { and } \quad \sigma(789) \sigma^{-1}=(123)$$
Hence, $H \unlhd G$. Setting $K=\langle\sigma\rangle$, the subgroups also satisfy $G=H K$. By Proposition 6.6.7, $G=H \rtimes_{\varphi} K$ where $\varphi$ corresponds to $K$ acting on $H$ by conjugation. If $x$ is a generator of $Z_3$, we can describe $G$ as
$$\left(Z_3 \oplus Z_3 \oplus Z_3\right) \rtimes_{\varphi} Z_3,$$
where $\varphi: Z_3 \rightarrow \operatorname{Aut}\left(Z_3 \oplus Z_3 \oplus Z_3\right)$ is defined by $\varphi(x)\left(g_1, g_2, g_3\right)=\left(g_3, g_1, g_2\right)$.
This is an example of a more general construction.
Let $K$ and $L$ be groups and let $\rho: K \rightarrow S_n$ be a homomorphism. Consider the action of $S_n$ on

$\overbrace{L \oplus L \oplus \cdots \oplus L}^{n \text { times }}$

by
$$\sigma \cdot\left(x_1, x_2, \ldots, x_n\right)=\left(x_{\sigma^{-1}(1)}, x_{\sigma^{-1}(2)}, \ldots, x_{\sigma^{-1}(n)}\right),$$
which corresponds to moving the $i$ th entry to the $\sigma(i)$ th location.

#### wreath product

The wreath product of $K$ on $L$ by the homomorphism $\rho: K \rightarrow S_n$ is the semidirect product
$$L \imath_\rho K=(L \oplus L \oplus \cdots \oplus L) \rtimes_{\varphi} K,$$
where $\varphi: K \rightarrow \operatorname{Aut}\left(L^n\right)$ is the homomorphism $\varphi(k)\left(x_1, x_2, \ldots, x_n\right)=$ $\rho(k) \cdot\left(x_1, x_2, \ldots, x_n\right)$.

We point out that the order of the wreath product is $\left|L<\rho K\right|=|L|^n|K|$ and that elements of a wreath product are $(n+1)$-tuples in the set $L^n \times K$. It is possible to give an alternative approach to the wreath product. Set $\Gamma=$ ${1,2, \ldots, n}$. Consider the isomorphism between $\operatorname{Fun}(\Gamma, L)$ and $L^n$ defined by $f \mapsto(f(1), f(2), \ldots, f(n))$ and where the group operation on functions $f, g \in \operatorname{Fun}(\Gamma, L)$ is $$(f \cdot g)(i)=f(i) g(i),$$ where the latter operation is in the group $L$. Then elements of the wreath product $L \imath\rho K$ are pairs $(f, k) \in \operatorname{Fun}(\Gamma, L) \times K$. The operation between elements in the wreath product is
$$\left(f_1, k_1\right) \cdot\left(f_2, k_2\right)=\left(i \mapsto f(i) g\left(\rho\left(k_1\right)^{-1}(i)\right), k_1 k_2\right) .$$
This is called the functional form of the wreath product.
In the scenario when $n=|K|$ and $\rho: K \rightarrow S_K$ corresponds to the action of $K$ on itself by left multiplication, the wreath product is called the standard wreath product of $L$ by $K$ and is denoted $L \nmid K$.

## The Hölder Program

A particular type of classification theorem involves finding all groups with a given cardinality.
With the concept of quotient groups, the effort to find all finite groups leads naturally to the Jordan-Hölder Program, which involves a two-pronged effort.
(1) Classify all finite simple groups.

(2) Find all methods such that given two groups $H$ and $K$, we can construct $G$ such that $G$ contains a normal subgroup $N \cong H$ such that $G / N \cong K{ }^1$
The group $G$ arising in the second part of the Program satisfies $|G|=$ $|H||K|$. Suppose we had solved the Jordan-Hölder Program. Given a positive integer $n$, we could use the first part of the Program to list all the simple groups of order $n$. Then, to find all non-simple groups, $G$ would have a normal subgroup $N$ with $|N|=d$ and $d \mid n$. If we knew all groups of order $d$ and of order $n / d$, we could use part (2) of the Program to find all non-simple groups of order $n$ with a normal subgroup of size $d$. So the construction of all groups of order $n$ builds up recursively.

The Jordan-Hölder Program is very difficult but drove much of the 20th century research in finite group theory. The first part of the Program is completed and carries the name of “The Classification of Finite Simple Groups.”

Theorem 1.

Every finite simple group is isomorphic to one of the following:
(1) A cyclic group $Z_p$ of prime order;
(2) An alternating group $A_n$ with $n \geq 5$;
(3) A member of one of 16 infinite families of groups of Lie type over a finite field;
(4) One of 26 sporadic groups not isomorphic to any of the above groups; or
(5) The Tits group, sometimes called the 27 th sporadic group.

The classification of finite simple groups “is generally regarded as a milestone of twentieth-century mathematics” . Because of the extreme length and the number of disparate results necessary for a full classification, the realization that the classification of finite simple groups was within reach arose slowly. It was estimated at the time that the work spanned 15,000 pages of articles both published and unpublished.

The form itself of the classification theorem is surprising. In retrospect, it is almost more surprising that so much work can be summarized in so brief a statement. Of course, to understand all parts of the theorem requires considerable advanced study. As of the publication of this textbook, group theorists believe that a complete proof might take around 5,000 pages. Efforts to find shorter proofs of the Classification of Finite Simple Groups continues to drive research.

abstract algebra代写请认准UpriviateTA. UpriviateTA为您的留学生涯保驾护航。