Let $H \mathrel{\unlhd} SU(2)$ that contains an element $A$ such that $\text{tr}(A)=2\cos \varphi_0 \neq \pm2$. Use i),ii) to show that $H \mathrel{\unlhd} SU(2)$ contains matrices with all possible traces between $2$ and $2\cos \varphi_0 $.
i) Any element $A\in SU(2)$ is conjugate to a diagonal matrix $U_{\varphi}=\begin{bmatrix} e^{i\varphi }& 0 \\ 0& e^{-i\varphi }\\ \end{bmatrix}$. $A$ uniquely determines $\varphi \in [0,\pi]$, and $\varphi=\cos ^{-1}(\text{tr}(A/2))$.
ii) If $B = \begin{bmatrix} \alpha & -\beta \\ \overline{\beta} & \overline{\alpha}\\ \end{bmatrix}$, $\text{tr}(BU_{\varphi}B^{-1}U_{\varphi}^{-1})=2(\alpha \overline{\alpha} + \beta\overline{\beta} \cos 2\varphi)$.
I think I have to show that $U_{\varphi_0} \in H$ first. Then, it would be possible to say that $BU_{\varphi_0}B^{-1} \in H$ for any $B \in SU(2)$ so that $BU_{\varphi_0}B^{-1}U_{\varphi_0}^{-1} := C \in H$. And, as $B$ can be any matrix, we can let $||\alpha||^2 = t \in [0,1] $ vary, and I can get the desired result. But, how do I show that $U_{\varphi_0} \in H$?
Edit: I showed that $U_{\varphi_0} \in H$. And, got the desired result.
Now, I would like to show that $H = SU(2)$ with the result I got and i), ii). So, I'm trying to show that "If $B \in SU(2)$, then $B \in H$." But, I'm stuck. Can I get some help with this?
My Approach:
Let $B \in SU(2)$.
Then, $\exists U_{\theta}$ s.t $P^{-1}BP = U_{\theta}$ for some $P \in SU(2)$.
Let $G_{U_{\theta}} \leq SU(2)$ that is generated by $U_{\theta}$.
Then, for any $n \in \mathbb{N}$, $(U_{\theta})^n = P^{-1}B^n P \in G_{U_{\theta}}$.
But, note that $\exists n \in \mathbb{N}$ s.t $\text{tr} ((U_{\theta})^n)=\text{tr} (B^n) = 2 \cos n\theta \in [2\cos 2\varphi_0,2]$
Then, by the previous result, $B^n \in H$.
How do I show that $B \in H$ from here?