Automorphisms. Abelian group.

[abhishekkgp]

Let $\displaystyle G$ be a finite group and suppose $\displaystyle T$ be an automorphism of $\displaystyle G$ which sends more than three quarters of the elements of $\displaystyle G$ onto their inverses. Prove that $\displaystyle T(x)=x^{-1} \, \forall \, x \in G$. (Hence $\displaystyle G$ is abelian).

ATTEMPT: the relevant stuff i found (after fiddling around a bit) is that if $\displaystyle T(x)=x^{-1}$ then $\displaystyle T(x^{-1})=x$. Now assuming the contradictory to what's to be proved we say that:
Let there exist $\displaystyle x \in G$ such that $\displaystyle T(x)=y \neq x^{-1}$. Then from the amazing discovery shown above we can conclude that $\displaystyle T(y) \neq y^{-1}$. But this doesn't seem to lead me anywhere.
Another thought i had was the use of induction since the group is given to be finite but i can't propose an appropriate inductive statement.
Someone please help.

 

[daon2]

Do you know that $\displaystyle Z(G)$ contains at most $\displaystyle 1/4|G|$ elements of $\displaystyle G$? (if $\displaystyle G\neq Z(G)$)

 

[abhishekkgp]

Do you know that $\displaystyle Z(G)$ contains at most $\displaystyle 1/4|G|$ elements of $\displaystyle G$? (if $\displaystyle G\neq Z(G)$)

I have never seen this result before. I'll give it a try.

 

[daon2]

I'm not sure that would actually help, I was just wondering because this seems like a counting problem. It might make use the class equation.

One way to use this (if at all possible) is to show that all such elements with $\displaystyle T(x)=x^{-1}$ belong to the center. They have to, because $\displaystyle G=Z(G)$ is what you need to prove. Then since you have the center having both: at most 1/4|G|, and at least 3/4|G|. A contradiction.

 

[abhishekkgp]

I'm not sure that would actually help, I was just wondering because this seems like a counting problem. It might make use the class equation.

One way to use this (if at all possible) is to show that all such elements with $\displaystyle T(x)=x^{-1}$ belong to the center. They have to, because $\displaystyle G=Z(G)$ is what you need to prove. Then since you have the center having both: at most 1/4|G|, and at least 3/4|G|. A contradiction.

This approach won't work, since my book says that the " three quarters " clause is tight. That is, there exist NON ABELIAN groups having an automorphism T which sends exactly three quarters of its elements to their inverses.

 

[daon2]

This approach won't work, since my book says that the " three quarters " clause is tight. That is, there exist NON ABELIAN groups having an automorphism T which sends exactly three quarters of its elements to their inverses.

That is the point. I assumed it was non-abelian, the goal is a contradiction.

 

[abhishekkgp]

That is the point. I assumed it was non-abelian, the goal is a contradiction.

What i understood was that you suggest we should try to prove that: "If $\displaystyle G$ is any finite group and $\displaystyle T$ an automorphism of $\displaystyle G$ then all $\displaystyle x \in G$ which satisfy $\displaystyle T(x)=x^{-1}$ are in $\displaystyle Z(G)$.

Assuming that $\displaystyle |Z(G)| \leq (1/4)|G|$ is true for all finite groups $\displaystyle G$, the above statement can't be true since there exist (non abelian) groups with an automorphism which sends three quarters (more that one fourth) of the elements of $\displaystyle G$ to their inverses.

Have i misunderstood something you have said??

 

[daon2]

What i understood was that you suggest we should try to prove that: "If $\displaystyle G$ is any finite group and $\displaystyle T$ an automorphism of $\displaystyle G$ then all $\displaystyle x \in G$ which satisfy $\displaystyle T(x)=x^{-1}$ are in $\displaystyle Z(G)$.

Assuming that $\displaystyle |Z(G)| \leq (1/4)|G|$ is true for all finite groups $\displaystyle G$, the above statement can't be true since there exist (non abelian) groups with an automorphism which sends three quarters (more that one fourth) of the elements of $\displaystyle G$ to their inverses.

Have i misunderstood something you have said??

Oh, I apologize, I misread this as as proving G is abelian.