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Question is :

Let $G$ be a finite group and suppose the automorphism $T$ sends more than three quarters of elements of $G$ onto their inverses.

Prove that $T(x)=x^{-1}$ for all $x\in G$ and that $G$ is abelian.

What I could see is...

Once I prove $T(x)=x^{-1}$ for all $x\in G$ then I would have that $G$ is abelian.

Because, for any $a,b\in G$ we have

$b^{-1}a^{-1}=(ab)^{-1}=T(ab)=T(a)T(b)=a^{-1}b^{-1}$

Thus, $ab=ba$ for all $a,b\in G$ Thus $G$ is abelian.

So, Only problem is to prove that $T(x)=x^{-1}$ for all $x\in G$

I am unable to use the fact that

"$T$ sends more than three quarters of elements of $G$ onto their inverses"

May be I should take : $A=\{x\in G : T(x)=x^{-1}\}$ and see if this is a subgroup (It is Not) or something like that.

Please help me to clear this.

I am very excited about this question So please do not spoil my excitement by posting full answer at once (This is a request cum order).

Please help me to do this by myself.

Thank you so much :)

  • @MartinBrandenburg : My bad... Thanks for the link..... I have showed my effort and that showed nothing.... I have to accept that this is asked before irrespective of its effort.... Thank you... –  Dec 27 '13 at 14:20
  • But this question has been asked twice in 24 hours. very strange! – Derek Holt Dec 27 '13 at 14:59
  • @DerekHolt : Really? I was seeing all questions but could not come across that... –  Dec 27 '13 at 15:46
  • It's http://math.stackexchange.com/questions/619075 – perhaps you missed it because it's been closed. – Derek Holt Dec 27 '13 at 18:25
  • @DerekHolt : Yes yes.. Thank you :) –  Dec 28 '13 at 02:46

1 Answers1

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This is an old problem from Herstein, and IMHO, hard to solve without some hints :

Let $A = \{x \in G : T(x) = x^{-1}\}$, and you want to prove that $A = G$ (and thus conclude that $G$ is abelian).

Take $a\in A$, and consider $K = A\cap a^{-1}A$. Given what you know about $|A|$, check that $$ |K| \geq |G|/2 $$ For any $x\in K$, since $x \in A$ and $x\in a^{-1}A$, check that $xa = ax$. Conclude that $K \subset C(a)$.

This is true for each $a \in A$.

Now what can you say about $Z(G)$?

Prahlad Vaidyanathan
  • 31,554
  • Is there any specific reason to consider $A\cap a^{-1}A$... I could not see why would $|K|\geq |G|/2$.. Please help me.... Is it a bit natural.... –  Dec 27 '13 at 14:10
  • Considering $A\cap a^{-1}A$ is the "trick" here. To see $|K| \geq |G|/2$, use http://en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle – Prahlad Vaidyanathan Dec 27 '13 at 14:12
  • $|A\cap a^{-1}A|=|A|+|a^{-1}A|-|A\cup a^{-1}A|\geq \frac{3}{4}|G|+\frac{3}{4}|G|-|G|=\frac{|G|}{2}$ Thanks for the link... so... now ? :) –  Dec 27 '13 at 14:22
  • So now what does Lagrange's theorem say about $|C(a)|$? From there, what do you conclude about $|Z(G)|$? – Prahlad Vaidyanathan Dec 27 '13 at 14:27
  • Yes Yes i got it... Only thing i could not understand is $K\subset C(a)$... Assuming $K\subset C(a)$ i am done, any proper subgroup can not be of order greater than half of the order of $|G|$.. –  Dec 27 '13 at 14:33
  • @ Praphulla Koushik : How could you say that $|a^{-1} A| \geq \frac{3}{4} |G|$ – Struggler Mar 07 '14 at 05:19
  • @PrahladVaidyanathan : All problems of Herstein are of the same age :P – C.S. Sep 29 '14 at 18:17
  • @PrahladVaidyanathan: Let me just complete the proof. Since $x \in K = A \cap a^{-1}A$ we have $x \in a^{-1}A \implies ax \in A \implies T(ax) =(ax)^{-1} \implies xa = ax \implies x \in N(a)$. This says that $|N(a)| \geq |K| \geq |G|/2 \implies N(a) = G \implies a\in Z(G)$. Since $a \in A \implies a\in Z(G)$ we have $|Z(G)| > |A| > 3/4 |G| \implies Z(G)=G$. Am I correct?? – C.S. Sep 29 '14 at 18:41
  • Dear Prahlad! Thanks for solution. Let me ask you the first question: Why are you considering the set $A\cap a^{-1}A$? What so special in this intersection? – RFZ Jan 06 '18 at 09:40
  • @PrahladVaidyanathan Now how to conclude $T(x)=x^{-1}$ for all $x \in G$ – Marco Polo Aug 14 '22 at 04:41