|
|
#4
07 กรกฎาคม 2009, 23:10
|
|||
|
|||
Let $n\in\mathbb{Z}_{10}$ and $f$ be an automorphism of $\mathbb{Z}_{10}$.
Then $f(n)=f(1+1+\cdots+1)=f(1)+f(1)+\cdots+f(1)=f(1)n$. This shows that $f(x)=f(1)x$ for all $x\in\mathbb{Z}_{10}$. Thus every automorphism (homomorphism) can be written in this form. Since $f$ is an automorphism (which is onto) there is some $y$ such that $f(y)\equiv 1\pmod{10}$ $f(1)y\equiv 1\pmod{10}$ This shows that $f(1)$ must have a multiplicative inverse which implies that $(f(1),10)=1$. Thus $f(1)\in U(10)$. Next, we show that if $f(1)\in U(10)$ then $f$ is an automorphism. It is enough to check that $f$ is one to one because on a finite set if $f$ is one to one then $f$ is also onto, and vice versa. It is not hard to check. Suppose $f(x)=f(y)$. Then $f(1)x \equiv f(1)y\pmod{10}$ $x\equiv y\pmod{10}$ since $(f(1),10)=1$. Thus $f$ is one to one (and hence onto) and ,of course, an automorphism. Therefore, all possible automorphisms of $\mathbb{Z}_{10}$ are the functions $f$ of the form $f(x)=ax$ where $(a,10)=1$. It is clear that any $f\in Aut(\mathbb{Z}_{10})$ corresponds uniquely to a number $f(1)\in U(10)$. Thus we can define a function $\phi:Aut(\mathbb{Z}_{10})\to U(10)$ by $\phi(f)=f(1)$. Then we can check that $\phi$ is a group isomorphism. In general, we can replace $10$ by any natural number $n$ and we can use the same idea to show that $Aut(\mathbb{Z}_{n})\cong U(n)$. nooonuii@school
__________________
site:mathcenter.net คำค้น 08 กรกฎาคม 2009 01:46 : ข้อความนี้ถูกแก้ไขแล้ว 2 ครั้ง, ครั้งล่าสุดโดยคุณ nooonuii 
|
  |
|
|
