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.