Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $G$ be a finite group, $T$ an automorphism of $G$ with the property that $xT=x$ for $x$ in $G$ if and only if $x=e$. Prove that every $g$ in $G$ can be represented as $g=x^{-1}(xT)$ for some $x$ in $G$.

I don't understand how I should start to pursue a solution. Can you please give me a hint instead of a whole solution? Which properties of automorphisms should I use?

Thanks.

share|improve this question
2  
It is quite unnecessary to say that the problem comes from Herstein, as the notation almost makes that clear :D –  Mariano Suárez-Alvarez Feb 16 '13 at 4:10
1  
I can't make sense of this: the OP wrote $$xT=T\,\,,\,\,\forall\,\,x\in G\,\Longleftrightarrow\,x=e$$Did he actually mean $\,xT=x\ldots$ ? –  DonAntonio Feb 16 '13 at 9:51
    
@DonAntonioSorry I made a mistake it has to be xT=x. And I guess I did this one but there is other question parallel with this question. Same introduction plus it says suppose further that T^2=Id. Prove G must be abelian. I assume we will use g= x^{-1}(xT) and somehow find G is abelian. But how? Do you have an idea on this? –  khankrt Feb 16 '13 at 15:12
add comment

3 Answers

Try investigating the map $U:G\to G$ given by $xU=x^{-1}(xT)$.

share|improve this answer
    
@JOnathanSo should I demonstrate for every g in G there exist a x in G? and I will do it by finding U is both 1-1 and onto? is it enough? –  khankrt Feb 16 '13 at 4:27
    
@khankrt yes, exactly. You can show that $U$ is surjective, so every element of $G$ can be realized in the desired form. –  Jonathan Feb 16 '13 at 5:22
    
@JonathanThanks for help. –  khankrt Feb 16 '13 at 5:48
    
@khankrt: Since you're fairly new here, I thought I'd just point out that one of the usual ways to express that an answer is helpful to you is to vote it up, especially when it's an answer to your own question. –  Tara B Feb 17 '13 at 18:03
add comment

Show that if a representation exists, it must be unique. Since $G$ is finite, you can conclude that every element has such a representation.

share|improve this answer
add comment

Just want to add that there is another way to word your question.

Suppose that $T$ is a fixed point free automorphism of $G$. Then the map $x\mapsto x^{-1+T}$ is a bijection from $G$ to $G$.

This is employed often in finite group theory. You can find it in Robinson or Huppert's books followed by many applications.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.