We konw that in quantum harmonic oscillator $H=a^\dagger a$, $a^\dagger$, $a$, $1$ will span a Lie algebra, where $a, a^\dagger$ is annihilation and creation operator, $H$ is the Hamiltonian operator. The algebraic relation is following

$$[H,a^\dagger\ ]= a^\dagger$$ $$[H,a]=-a$$ $$[a,a^\dagger]=1$$ $$[H,1]=[a,1]=[a^\dagger,1]=[a,a]=[a^\dagger,a^\dagger]=[1,1]=[H,H]=0$$ So these four operators, $H=a^\dagger a$, $a^\dagger$, $a$, $1$, can span a lie algebra, because the commutator satisfies closure and Jacobi's identity.

We konw that for any lie algebra $\mathscr{G}$ there exist only one lie group $G$ up to the difference of the topology, whose lie algebra is $\mathscr{G}$.

So what is this Lie group whose lie algebra spaned by $\{H=a^\dagger a , a^\dagger ,a ,1\}$ ?