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$G$ is a group and $f:G \rightarrow G$ is a function defined as $f(a)=a^{-1}$ where $a^{-1}$ is the inverse of $a$ under the group operation. Prove that $f$ is an isomorphism if and only if $G$ is abelian.

I understand that I have to prove $f(ab)=(ab)^{-1}=b^{-1}a^{-1}$. How might I do that?

Reference: Fraleigh p. 49 Question 4.40 in A First Course in Abstract Algebra

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I bet this is the 583940531111102947583 time this question's been asked in the site in the last 6 months or so... –  DonAntonio Mar 10 '13 at 17:26
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3 Answers 3

First note that it is a bijection of $G$ onto $G$ no matter what. So this boils down to $G$ Abelian if and onbly if $f$ is a homorphism.

If $G$ is Abelian, I think you can show that $f$ is a homomorphism.

Now if $f$ is a homomorphism $$ ab=(a^{-1})^{-1}(b^{-1})^{-1}=(b^{-1}a^{-1})^{-1}=(f(b)f(a))^{-1}=(f(ba))^{-1}=((ba)^{-1})^{-1}=ba. $$

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HINT: You have to prove two things:

  1. If $G$ is Abelian, then $f(ab)=f(a)f(b)$ for all $a,b\in G$, which means that $(ab)^{-1}=a^{-1}b^{-1}$ for all $a,b\in G$.

  2. If $f(ab)=f(a)f(b)$ for all $a,b\in G$, i.e., if $(ab)^{-1}=a^{-1}b^{-1}$ for all $a,b\in G$, then $G$ is Abelian.

You need just one basic fact for both: that in any group $(ab)^{-1}=b^{-1}a^{-1}$.

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Hint: No, we always have that $f(ab)=(ab)^{-1}=b^{-1}a^{-1}$. (One of the directions of) what you have to prove is that, if $G$ is abelian, $$b^{-1}a^{-1}=(ab)^{-1}=\underset{\substack{\text{what it means for $f$}\\\text{to be a homomorphism}}}{\fbox{$f(ab)=f(a)f(b)$}}=a^{-1}b^{-1}$$

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