# Define group $G^*$ with same elements as $G$ and prove mapping from $G$ to $G^*$ is an isomorphism

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.

• Good catch! Can't believe I missed it. – westhe32nd May 9 '14 at 21:08
• 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. – egreg May 9 '14 at 21:11
• 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? – westhe32nd May 9 '14 at 21:12
• No, you have to justify that $\phi$ is bijective, too. – Dietrich Burde May 9 '14 at 21:16

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 $\phi\colon G\rightarrow G^*$.