Counterexample: G is matrix lie group such that there is $A\in G$ such that A can not be written as $A=e^{X_1}....e^{X_m} $ for some $X_1,..X_m\in \mathfrak g$ where $\mathfrak g $ is lie algebra of G.
I come across theorem which states following
G is connected matrix lie group such that $\forall A\in G$ such that A can be written as $A=e^{X_1}....e^{X_m} $ for some $X_1,..X_m\in \mathfrak g$.
I think this will not true in case of disconnected lie group.
So I wanted a construct example.
Please Help me.
Any Help will be appreciated.
In fact any Lie group $G$ that is not connected will do: For any $g \in G$, $X \in \mathfrak g$, $\gamma : [0, 1] \to G$, $$\gamma(t) = g \exp (t X)$$ is a path between $g$ and $g \exp X$, so for any $X_1, \ldots, X_n \in \mathfrak g$, induction gives that $\exp X_1 \cdots \exp X_m$ is in the identity component of $G$.
For a trivial example, take $G$ to be any nontrivial discrete group: Then $\mathfrak g = \{ 0 \}$, and so $\exp X_1 \cdots \exp X_m = e_G$.