How to show $\exp(tX)\exp(tY)=\exp(t(X+Y)+tR(t))$ with $\displaystyle \lim_{t\to 0} R(t)=0$?

Question

Let $X\in GL(n, \mathbb R)$. The exponential of $X$ is the matrix given by $$\exp(X)=\sum_{n=0}^\infty \frac{X^n}{n!}.$$

I need some help for showing the following result: $$\exp(tX)\exp(tY)=\exp(t(X+Y)+tR(t)),\quad t\in\mathbb R,$$ with $\displaystyle\lim_{t\to 0} R(t)=0$.

Thanks

Remark: I need this result for showing that if $H$ is a closed subgroup of a Lie group $G\leq GL(n, \mathbb R)$ then $H$ is itself a lie group.

Answer 1

Because the exponential map is invertible near zero, you can consider the smooth matrix-valued function $\phi(t) = \exp^{-1}(\exp(tX)\exp(tY))$. A computation shows that $\phi(0)=0$ and $\phi'(0) =X+Y$, and therefore the first-order Taylor expansion of $\phi$ is $$ \phi(t) = t(X+Y) + tR(t), $$ where $R(t)\to 0$ as $t\to 0$. For a more complete proof, see Proposition 20.10 in my Introduction to Smooth Manifolds (2nd ed.).

Answer 2

Or simpler $e^{tX}e^{tY}=(I+tX+O(t^2))(I+tY+O(t^2))=I+t(X+Y)+O(t^2)=e^{t(X+Y)}+O(t^2)$.