Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$G$ 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

share|cite|improve this question
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

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 $$ ab=(a^{-1})^{-1}(b^{-1})^{-1}=(b^{-1}a^{-1})^{-1}=(f(b)f(a))^{-1}=(f(ba))^{-1}=((ba)^{-1})^{-1}=ba. $$

share|cite|improve this answer

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

share|cite|improve this answer

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, $$b^{-1}a^{-1}=(ab)^{-1}=\underset{\substack{\text{what it means for $f$}\\\text{to be a homomorphism}}}{\fbox{$f(ab)=f(a)f(b)$}}=a^{-1}b^{-1}$$

share|cite|improve this answer

Your Answer


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.