# Proving closure for a set with $e^{tH}$

To prove that a set forms a group under multiplication I know I must show that the set must be equipped with a multiplication law such that there is closure, associativity, identity and an inverse.

But I am unable to do this for a set $$G=\{ U(t), t \in \mathbb{R}\}\tag{1}$$ with $$U(t) = e^{tH}\tag{2}$$

where $$H$$ has been fixed to be a $$N \times N$$ anti-hermitian, traceless matrix.

I am especially struggling with closure:

I considered the product of $$e^{tH}$$ with $$e^{tJ}$$ where the J has equal properties to H. But this gave me :

$$e^{tH}e^{tJ}= e^{t(H+J+\frac{1}{2}[H,J])} \tag{3}$$

But I don't see how to proceed from here. Do I simply state that the resulting exponential in $$(3)$$ is also in $$\mathbb{R}$$?

Here, all your matrices are of the form $$e^{tH}$$ for a $$t$$. Thus, if $$M_1,M_2 \in G$$, there exist $$t_1,t_2 \in \mathbb{R}$$ such that $$M_1 = e^{t_1 H}$$ and $$M_2= e^{t_2H}$$. Consequently, as $$t_1H$$ ans $$t_2H$$ commute, you can say : \begin{align} M_1 \times M_2 = e^{t_1H}\times e^{t_2H} = e^{(t_1+t_2)H} \in G \end{align} The key is that $$e^{A+B} = e^A e^B$$ if $$A$$ and $$B$$ commute.
• @user7077252 It happens. That's why it may be useful to denote with subscripts: $G_H=\{ U_H(t), t \in \mathbb{R}\}$, where $U_H(t) = e^{tH}$; just to remind that $H$ is a "parameter", not a variable.