Prove that every finite subgroup of $SL_2(\mathbb{C})$ is conjugate to a finite subgroup of $SU_2$

Question

I want to prove this proposition. For any finite subgroup $G$ of $SL_2(\mathbb{C})$, I showed that $G$ is unitary with respect to an inner product that we define properly. But I couldn't find the conjuction element.

Answer 1

There is such a Moore's theorem from 1898 (Eliakim Hastings Moore: 1862-1932). Every finite group $G\leq GL_n(\mathbb{C})$ is conjugate to a subgroup of the unitary group $U_n$. In your case $n=2$.

One of the proofs of this theorem explicitly states the conjugate element. Let $$ a=\sum_{g\in G}gg^*. $$ The matrix $a$ is positive-definite, so $a=bb^*$ and $b^{-1}Gb<U_n$.

For large groups this is not so easy to calculate, but you have $n=2$ and a small group. Try it.

Addition.

Let us prove that $b^{-1}xb\in U_n$ for every $x\in G$.

Let $y=b^{-1}xb$. Since $(uv)^*=v^*u^*$ and $a=bb^*$, we have $$ yy^*=(b^{-1}xb)(b^{-1}xb)^*=b^{-1}xbb^*x^*(b^*)^{-1}=b^{-1}xax^*(b^*)^{-1}. $$ Now notice that $xax^*=a$. Indeed, $$ xax^*=x\left(\sum_{g\in G}gg^*\right)x^*=\sum_{g\in G} xgg^*x^*=\sum_{g\in G}(xg)(xg)^*=\sum_{xg\in G}(xg)(xg)^*=a. $$ Now we can complete the proof $$ yy^*=b^{-1}xax^*(b^*)^{-1}=b^{-1}a(b^*)^{-1}=b^{-1}bb^*(b^*)^{-1}=e. $$ Here $e$ is the identity matrix. We see that $yy^*=e$, hence $y$ is a unitary matrix.