Let $\hat k\in\mathbb R^3-\{0\}:||\hat k||=1$. We define the rotation group around the axis defined by $\hat k$ as $$\text{Rot}(\hat k,\mathbb R^3)=\Big\{\big(\begin{smallmatrix} A & 0 \\ 0 & 1 \end{smallmatrix}\big):A\in SO(2)\Big\}$$ I need to prove that:
Proposition: If $H$ is a closed connected and 1-dimensional lie subgroup of $SO(3)$, then there exists $\hat k\in\mathbb{R}^3$ such that $\text{Rot}(\hat k,\mathbb R^3)=H$.
My idea for prove it is by the following lemma:
Let $\{\psi_t\}$ be a one-parameter subgroup of $SO(3)$ and $e_{SO(3)}$ be the unitary element of $SO(3)$. Then there exists a unique $\hat k\in\mathbb R^3$ with $||\hat k||=1$ such that $\{\psi_t\}=\text{Rot}(\hat k,\mathbb R^3)$.
Proof idea of the proposition:
It is sufficient show that there exists a one-parameter subgroup $\{\psi_t\}$ of $SO(3)$ such that $\{\psi_t(e_{SO(3)})\}$ coincides with $H$ (by the lemma). The lie algebra $\mathfrak h$ of $H$ must be isomorphic to $\mathbb R$. Denote this isomorphisms by $\phi: \mathfrak h\to \mathbb R$. If we define $\psi:\mathbb R\times SO(3)\to SO(3)$, $(A,t)\mapsto \psi(A,t)=\varphi\cdot\big(\text{exp}\circ \phi^{-1}(t)\big)$, then $\psi$ induce a one-parameter subgroup $\{\psi_t=\psi(\cdot,t)\}$ of $SO(3)$. From here, is it possible show that $\{\psi_t\}=H$?