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abstract algebra代写|Group Theory代写|抽象代数代写|陪集代写

Abstract algebra不算是一门简单的学科,这门学科在国内叫做抽象代数,经常有很多学生在学linear algebra或者analysis(advance calculus)的时候觉得并不困难,但是却觉得Abstract algebra很难,这是因为没有找到正确的方法学习Abstract algebra,UpriviateTA有一系列非常擅长Abstract algebra的老师,可以确保您在Abstract algebra取得满意的成绩。

以下是UCLA的一次assignment.更多的经典案例请参阅以往案例,关于abstract algebra的更多的以往案例可以参阅相关文章。abstract algebra代写请认准UpriviateTA.

问题 1. Suppose that $G$ is an infinite simple group. Show that for every proper subgroup $H$ of $G,$ the index $[G: H]$ is infinite.
证明 . If $[G: H]=n<\infty,$ then $G$ can be embedded in $S_{n},$ so $G$ is finite, a contradiction.
问题 2. Let $G$ act on left cosets of $H$ by multiplication. Show that the kernel of the action is
N=\bigcap_{x \in G} x H x^{-1}
证明 . $g(x H)=x H$ iff $x^{-1} g x \in H$ iff $g \in x H x^{-1}$
问题 3. Let $H$ be a subgroup of $G$ of finite index $n,$ and let $G$ act on left cosets $x H$ by multiplication. Let $N$ be the kernel of the action, so that $N \leq H$. Show that $[G: N]$ divides $n !$
证明 . By the first isomorphism theorem, $G / N$ is isomorphic to a group of permutations of $L$, the set of left cosets of $H .$ But $|L|=[G: H]=n,$ so by Lagrange’s theorem, $\mid G / N$ divides $\left|S_{n}\right|=n !$
问题 4. Let $H$ be a subgroup of $G$ of finite index $n>1 .$ If $|G|$ does not divide $n !$, show that $G$ is not simple.
证明 . Since $n>1, H$ is a proper subgroup of $G,$ and since $N$ is a subgroup of $H, N$ is a proper subgroup of $G$ as well. If $N=\{1\},$ then $|G|=[G: N],$ so by Problem 8 , $G$ divides $n !$, contradicting the hypothesis. Thus $\{1\}<N<G,$ and $G$ is not simple.
问题 5.

Here is some extra practice with left cosets of various subgroups. Let $H$ and $K$ be subgroups of $G,$ and consider the map $f$ which assigns to the coset $g(H \cap K)$ the pair of cosets $(g H, g K)$. Show that $f$ is well-defined and injective, and therefore
[G: H \cap K] \leq[G: H][G: K]
Thus (Poincaré) the intersection of finitely many subgroups of finite index also has finite index.

证明 .

$g_{1}(H \cap K)=g_{2}(H \cap K)$ iff $g_{2}^{-1} g_{1} \in H \cap K$ iff $g_{1} H=g_{2} H$ and $g_{1} K=g_{2} K,$ proving
both assertions.

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