Automorphism that is an Involution of a finite group

hbghlyj

icesheep 发表于 2011-11-27 00:35
$f$ 是有限群 $G$ 的一个自同构，$f \circ f = I$，若 $f$ 除了 1 之外没有不动点，证明：$f( x) = x^{ - 1}$

Abstract Algebra 3rd Edition by Dummit and Foote 第41页第1.6节第23题
Topics in Algebra by Herstein 第70页第11题

hbghlyj

这个MSE问题中的$\sigma$就是1#的$f$, 最终要证明$G$为奇数阶交换群, 而1#的$\sigma(x)=x^{-1}$包含在证明中.

The map $x^{-1}\sigma(x)$ is injective, proof: assume $x^{-1}\sigma(x)=y^{-1}\sigma(y)$ then $yx^{-1}=\sigma(y)\sigma(x)^{-1}=\sigma(yx^{-1})\implies yx^{-1}=e\implies y=x$ since there are no fixed points besides $e$. since $G$ is finite and $x^{-1}\sigma(x)$ is injective it is also surjective.

hence we can write every element of $G$ in the form $x^{-1}\sigma(x)$. Take $g$ in $G$, it can be written as $x^{-1}\sigma(x)$. thus $\sigma(g)=\sigma(x)^{-1}x$. Notice multiplying $g$ with $\sigma(g)$ gives us $e$. we conclude $\sigma(g)=g^{-1}$

Two things are now clear:

• $|G|$ cannot be even for if it were, it would have an element of order $2$, which would map to itself under $\sigma$ for it is its own inverse.
• $G$ is abelian, for $\sigma(gh)=g^{-1}h^{-1}$ since $\sigma$ is homomorphism, but $\sigma(gh)=(gh)^{-1}=h^{-1}g^{-1}$. thus $g^{-1}h^{-1}=h^{-1}g^{-1}\implies hg=gh$ (just take inverses on both sides).

hbghlyj

tchappy ha的回答:

Let $\phi$ be a mapping $\phi: G\ni x\mapsto x^{-1}(xT)\in G$.  
If $\phi(x)=\phi(y)$, then $x^{-1}(xT)=y^{-1}(yT)$.  
Then, $(xT)(yT)^{-1}=xy^{-1}$.  
Then, $(xT)(y^{-1}T)=xy^{-1}$.  
Then, $(xy^{-1})T=xy^{-1}$.  
By the assumption of Problem 10, $xy^{-1}=e$.  
So, $x=y$.  
So, $\phi$ is injective.  
Since $G$ is a finite set, $\phi$ is also surjective.  
So, for every $g\in G$, there exists $x\in G$ such that $g=x^{-1}(xT)$.  

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I will write Karthik Kannan's answer here:

Let $g$ be an arbitrary element of $G$.  
Then, by Problem 10, there exists $x\in G$ such that $g=x^{-1}(xT)$.  
Then, $gT=(x^{-1}(xT))T=(x^{-1}T)(xT^2)=(xT)^{-1}(xI)=(xT)^{-1}x=(x^{-1}(xT))^{-1}=g^{-1}$.  
So, $gT=g^{-1}$ holds for every $g\in G$.  
Let $a,b$ be arbitrary two elements of $G$.  
Then, $(ab)^{-1}=(ab)T=(aT)(bT)=a^{-1}b^{-1}=(ba)^{-1}$.  
So, $ab=ba$.

Czhang271828

有限群 $G$ 上的自同构 $f$ 满足 $f\circ f=\mathrm{id}_G$. 若 $f$ 唯一的不动点是单位元 $e$, 求证 $f(g)=g^{-1}$.

　

来源

　

注意到集合间映射 $\varphi:G\to G,g\mapsto g^{-1}\cdot f(g)$, 则 \[ [\varphi(g)=\varphi(h)]\Leftrightarrow [g^{-1}f(g)=h^{-1}f(h)]\Leftrightarrow [hg^{-1}=f(hg^{-1})]\Leftrightarrow [hg^{-1}=e]\Leftrightarrow [h=g]. \] 从而 $\varphi$ 是单射. 由于 $G$ 有限, 故 $\varphi$ 是双射, 因此任意 $g\in G$ 可写作 $x^{-1}\cdot f(x)$ 的形式. 此时 \[ f(g)=f(x^{-1}f(x))=f(x^{-1})x=f(x)^{-1}(x^{-1})^{-1}=(x^{-1}f(x))^{-1}. \]

　

此时 $G$ 交换. 因为 $gh=f((gh)^{-1})=f(h^{-1}g^{-1})=f(h^{-1})f(g^{-1})=hg$.