As the title says i have to find wheter the subgroup $H$ of $SO(3)$, defined by: $\omega\in H, \omega= \begin{pmatrix} 1 & 0 \\ 0 & \lambda \in SO(2) \end{pmatrix} \space (1)$
The definition of normal subgruop is that $\forall h_1,h_2 \in H , \exists g\in G $ such that $gh_1g^{-1}=h_2 \space(2)$. This, however led me nowhere, so i have two questions.
Q1) Is it okay to find an explicit matrix $g\in SO(3)$ that doesn't verify (2), meaning that the resulting matrix is not in the form (1)?
Q2) Another idea i had is using the isomorphism $SO(3)/SO(2)\simeq S^2$ and, since $S^2$ is not a group then $H$ is not normal in $SO(3)$
No, $H$ is not a normal subgroup of $SO(3)$. Let $a=\left(\begin{smallmatrix}0&-1&0\\1&0&0\\0&0&1\end{smallmatrix}\right)$ and let $h=\left(\begin{smallmatrix}1&0&0\\0&0&-1\\0&1&0\end{smallmatrix}\right)\in H$. Then$$a.h.a^{-1}=\begin{pmatrix}0&0&1\\0&1&0\\-1&0&0\end{pmatrix}\notin H.$$