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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 $\mathbb{R}$?

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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.

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  • $\begingroup$ Oh right, I was adding a different H when t was the element in the group, thank you. $\endgroup$ May 22, 2020 at 9:22
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    $\begingroup$ @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. $\endgroup$
    – user750041
    May 22, 2020 at 12:02
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    $\begingroup$ That is a great tip, thank you $\endgroup$ May 22, 2020 at 12:52

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