1
$\begingroup$

I am asked to take a group $G$, define a new group $G^*$ that has the same elements of $G$ with operation $*$ defined as $a*b=ba$ for all $a$ and $b$ in $G^*$. Then, prove that the mapping from $G$ to $G^*$ defined by $\phi(x) = x^{-1}$ for all $x$ in $G$ is an isomorphism from $G$ onto $G^*$.

So, here is what I have so far. I know if I take the element $(ab)$ and perform the operation $\phi$, I get $$ \phi(ab)=(ab)^{-1}=b^{-1}a^{-1}=\phi(b)\phi(a)=\phi(a)*\phi(b)$$ which I believe shows that $\phi(x)$ is abelian. I'm kind of lost at this point of what I should do from here.

| cite | improve this question | | | | |
$\endgroup$
  • $\begingroup$ Good catch! Can't believe I missed it. $\endgroup$ – westhe32nd May 9 '14 at 21:08
  • $\begingroup$ Saying that $\phi(x)$ is abelian means nothing. A group can be abelian, not an element of a group. You are indeed finished, because $\phi$ is clearly bijective. $\endgroup$ – egreg May 9 '14 at 21:11
  • $\begingroup$ I have to ask (since I'm terrible at proofs), but is what I have written (minus the abelian part) enough to answer the question? $\endgroup$ – westhe32nd May 9 '14 at 21:12
  • $\begingroup$ No, you have to justify that $\phi$ is bijective, too. $\endgroup$ – Dietrich Burde May 9 '14 at 21:16
1
$\begingroup$

You are almost there. Just note that $\phi(b)\phi(a)=\phi(a)\ast \phi(b)$. Then $\phi(ab)=\phi(a)\ast \phi(b)$. Clearly the map $\phi$ is injective and surjective. So we obtain a group isomorphism $\phi\colon G\rightarrow G^*$.

| cite | improve this answer | | | | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.