quantum field theory - Question about the Noether charge algebra

I'm reading these notes - page 8 and 9 - and I'm a bit confused.

If we consider a field $\phi$ (which can be either bosonic or fermionic) transforming as: \begin{equation} \phi(x) \rightarrow \phi(x) + \delta \phi (x) \end{equation} with: \begin{equation} \delta \phi^a = t^a \phi(x) \end{equation} where $t^a$ is the generator of the transformation. The generators satisfy the Lie algebra: \begin{equation} [t^a,t^b] = if^{abc} t^c \tag{$*$} \end{equation} Let us suppose that the above transformation is a symmetry transformation such that the Noether charge corresponding to this symmetry is given by: \begin{equation} Q^a = \int \mathrm{d}^3 \mathbf{x} \; \pi \delta\phi^a = \int \mathrm{d}^3 \mathbf{x} \; \pi t^a \phi \end{equation} where $\pi$ is the canonical momentum density. It is then possible (but tedious) to show that the charges satisfy the so-called charge algebra: \begin{equation} [Q^a,Q^b] = i f^{abc} Q^c \tag{1} \end{equation} Until this point I understand it. But then the notes say on page 8:

[...] the charges generally have to satisfy the same algebras as the generators – in fact it is only because of this that the symmetry has any useful physical meaning. In particular it is the charges which are the physical observables that participate in interactions rather than gauge fields for example.

I don't really understand what is meant with the above statement. What does the quote have to do with the fact that Noether charges obey equation $(1)$?

Edit: I understand that the charges satisfy the same Lie algebra as the generators. But according to the quote above, if understand it correctly, we should also expect this based on logical/physical reasons. Apparently, according to the notes, "it is only because of this that the symmetry has any useful physical meaning." I don't understand why this is the case.