One parameter subgroup on a Lie group

Question

Let $G$ be a Lie group and $\{\psi_t\}$ a one-parameter subgroup of $G$ (i.e. $\psi:\mathbb R\times G\to G$ a smooth map, $\psi(t,\cdot):G\to G$ diffeomorphism, $\psi(0,\cdot)=\text{id}_G$, and $\psi(t,\cdot)\circ \psi(s,\cdot)=\psi(s+t,\cdot)$). Is true that $\psi(\mathbb R\times \{e\})$ is a subgroup of $G$?

Answer 1

First of all, $\psi(\mathbb R \times \{e\})$ is non-empty, because $e=\text{id}_G(e)=\psi(0,e)\in \psi(\mathbb R \times \{e\})$.

Furthermore, $\psi(\mathbb R \times \{e\})$ is closed under the operation $\cdot$ of $G$, because $$ \psi(s,e)\cdot \psi(t,e)=\psi(s+t,e)\in \psi(\mathbb R \times \{e\}) $$ for all $s,t\in\mathbb R$.

The last thing you need to check are inverses: For $g=\psi(s,e)\in\psi(\mathbb R \times \{e\})$ you have $$ \psi(s,e)\cdot\psi(-s,e)=\psi(0,e)=\text{id}_G(e)=e, $$ hence $g^{-1}=\psi(-s,e)\in\psi(\mathbb R \times \{e\})$.