Let $M$ be a smooth manifold of dimension $m$ and $$\phi: \mathbb{R} \times M \rightarrow M : (t,x) \to \phi_t(x)$$ such that: $1)\ \phi_0(x) = x$ for all $x \in M$ and $2)\ \phi_t(\phi_s(x)) = \phi_{t+s}(x)$ for all $t, s \in \mathbb{R}$ and for all $x\in M$. Then $$G := \lbrace \phi_t \mid t\in \mathbb{R} \rbrace$$ is called the one parameter group of diffeomorphisms of $M$.
Show that $G$ is isomorphic to $(\mathbb{R}, +)$ or the circle group $T = \lbrace z\in \mathbb{C} \mid |z| = 1 \rbrace$
What I tried was first trying to construct candidates for isomorphism and then checking if everything holds. The candidates I came up with were
$$h_1 : \mathbb{R} \rightarrow G : h_1(t) = \phi_t$$ and $$h_2 : G \rightarrow T : h_2(\phi_t) = e^{it}$$ Now, $h_1$ and $h_2$ are both group homomorphisms, and both maps are surjective (correct me if I'm wrong). Now the issue here is the injectivity. So I tried to distinguish two cases depending on whether the map $\phi: \mathbb{R} \times M \rightarrow M$ is injective or not.
If $\phi$ is injective, then $(t,x) \neq (s,y)$ implies $\phi_t(x) \neq \phi_s(y)$. So if $t \neq s$, then $h_1(t) \neq h_1(s)$ by the infectivity of $\phi$ and hence $G$ would be isomorphic to $(\mathbb{R},+)$.
Now I'm stuck with the case where $f$ is not injective. I thought of considering the set $$H = \lbrace t\in \mathbb{R} \mid \phi_t(x_0) = x_0 \rbrace$$ for some fixed $x_0 \in M$. I know that if $H = \{0\}$, then the orbit of $G$ determined by $x_0$ is injective, and that if $H \neq \{0\}$, the orbit of $G$ determined by $x_0$ is diffeomorphic to a circumference. But since we are assuming that $\phi$ is not injective, it must be the case where $H \neq \{0\}$, so the orbit must be diffeomorphic to a circumference. I think this already proves the result, since we have a diffeomorphism, let's call it $g : \mathbb{S}^1 \rightarrow O(x_0)$, where $O(x_0)$ stands for the orbit of $x_0$, but I lack of a complete proof of this.
$\phi$ can be seen as a group homomorphism $\phi : \mathbb{R} \rightarrow \rm{Diff}(M)$. Assume that it's continuous and that it's not the trivial homomorphism $\phi_t=\mathrm{id}_M$ $\forall t\in \mathbb{R}$. We have two cases:
If there is a $s\in \mathbb{R}^+$ such that $\phi_s = \mathrm{id}_M$ then $\mathrm{Ker}(\phi)\cong\mathbb{Z}$ (see the proof below) and $\phi$ induces a group monomorphism $\tilde{\phi}:\mathbb{R}/\mathbb{Z}\rightarrow\rm{Diff}(M)$. As $\mathbb{R}/\mathbb{Z}\cong T$, we have that $G=\mathrm{Im} (\phi)=\mathrm{Im}(\tilde{\phi})\cong T$.
Otherwise, $\rm{Ker}(\phi)=\{0\}$ and $\phi$ is injective, so $G=\mathrm{Im}(\phi)\cong\mathbb{R}$.
If there is a $s\in \mathbb{R}^+$ such that $\phi_s = \mathrm{id}_M$ then $\mathrm{Ker}(\phi)\cong\mathbb{Z}$. Proof
Let $P=\{s\in\mathrm{Ker}(\phi):\;s>0\}$. Then $P$ has a minimum. If it didn't, then there would be arbitrarily small $s\in P$. By continuity, there is some neighborhood $[0,u)$ of $0$ inside which $\phi$ is constant and equal to $\mathrm{id}_M$. In turn, as $\mathbb{R}$ can be generated by sums of numbers in $[0,u)$, $\phi$ has to be $\mathrm{id}_M$ in the whole domain $\mathbb{R}$. This is a contradiction with our assumption that $\phi$ is not trivial. Let $r$ be the minimum of $P$.
For every $n\in \mathbb{Z}$ we have that $\phi_{n r} = \phi_r$. This generates the whole set $\mathrm{Ker}(\phi)$. To prove it, we have to show that there are no elements $s\in\mathrm{Ker}(\phi)$ between $nr$ and $(n+1)r$ for any $n\in\mathbb{Z}$. If there were such an $s$ then $s-nr\in P$ and $s-nr<r$, leading to contradiction with the assumption that $r$ is the minimum of $P$.