Consider the Harmonic oscillator.

**Some considerations:**
$H_{xyz}=H$  is not a C.S.C.O. in $\mathcal{E}=\mathcal{E}_{xyz}$  by itself: even if we know the eigenvalue (the energy) of $H$ , we may not not the state of the system: *e.g* if $E=1$
 , all we know is that the state is a linear combination of $\vert100\rangle,\vert010\rangle$  and $\vert001\rangle$ .

However, $\{H_{x},H_{y},H_{z}\}$  is a C.S.C.O. This is easy to see.

Now, $[H,L_{i}]=0$ since $H$  is rotationally invariant. 
One can prove that $\{H,L^{2},L_{z}\}$ is a C.S.C.O.

Thus, we get two basis for $\mathcal{E}$ : $\{\vert n_{x}\,n_{y}\,n_{z}\rangle\}$
  and $\{\vert n,l,m\rangle\}$, one for each of the C.S.C.O.s found.
We can of course go from one basis to the other.

A common question asked is: **for a given n, what are the possible values of l?** 

I will try to answer this using a method which I haven't found anywhere, so I would like someone to confirm it.

For example, for $n=1$, I know that $\mathcal{E}(n=1)$ is three-dimensional, since it is spanned by $\{\vert100\rangle,\vert010\rangle,\vert001\rangle\}$. Thus, one of the following is true:

$$\mathcal{E}(n=1)=3\mathcal{E}(n=1,l=0)$$
or 
$$\mathcal{E}(n=1)=\mathcal{E}(n=1,l=1)$$
 
So, if I find one state of $\mathcal{E}(n=1)$
  with $m=1$, then that state cannot belong to $\mathcal{E}(l=0)$
  and so $\mathcal{E}(n=1)=\mathcal{E}(n=1,l=1)$
  holds.

Another example: $n=2$
 . I know that $\mathcal{E}(n=2)$
  is six-dimensional, since it is spanned by $\{\vert200\rangle,\vert020\rangle,\vert002\rangle,\vert110\rangle,\vert101\rangle,\vert011\rangle\}$
 . Thus, one of the following is true:

$$\mathcal{E}(n=2)=6\mathcal{E}(n=2,l=0)$$
or 
$$\mathcal{E}(n=2)=\mathcal{E}(n=2,l=0)\oplus\mathcal{E}(n=2,l=2)$$

So, if I find one state of $\mathcal{E}(n=2)$
  with $m=-1$, for instance, then that state cannot belong to $\mathcal{E}(l=0)$  and so $\mathcal{E}(n=2)=\mathcal{E}(n=2,l=0)\oplus\mathcal{E}(n=2,l=2)$  holds.

Is this argument correct? The results are, but I'm not sure the reasoning is: I assumed that, if I have a state with a certain l
  in $\mathcal{E}(n)$
 , then all the $2l+1$
  states $\vert n,l,m\rangle$
 are in $\mathcal{E}(n)$. I do not know how to justify this.

If the method or the question are still unclear, please tell me and I'll try to clarify them.

----------------------------------
After reading what Emilio and ZeroTheHero and some other things, I tried another method which seems more general. Here it goes.

Let
$$
\begin{cases}
V_{1}=-\frac{a_{x}+ia_{y}}{\sqrt{2}}\\
V_{0}=a_{z}\\
V_{-1}=\frac{a_{x}-ia_{y}}{\sqrt{2}}
\end{cases}\mbox{ and }\begin{cases}
V_{1}^{\dagger}=-\frac{a_{x}^{\dagger}-ia_{y}^{\dagger}}{\sqrt{2}}\\
V_{0}^{\dagger}=a_{z}^{\dagger}\\
V_{-1}^{\dagger}=\frac{a_{x}^{\dagger}+ia_{y}^{\dagger}}{\sqrt{2}}
\end{cases}
$$

We have 
$$
[J_{z},V_{q}^{\dagger}]=q\hbar V_{q}^{\dagger}
$$
$$
[J_{z},V_{q}]=q\hbar V_{q}
$$
$$
[N,V_{q}^{\dagger}]=V_{q}^{\dagger}
$$
$$
[N,V_{q}]=-V_{q}
$$

So the “daggered” operators increase the eigenvalue $n$
  of $N$
  by one (and thus the energy by $\hbar\omega$
 ), while the “undaggered” operators decrease it by one. Also, the $V_{q}^{(\dagger)}$
  change the eigenvalue of J_{z}
  by $q\hbar$
 .

With this in mind, we can face the problem.

*Assumption: All states of the system can be obtained by applying these operators to the ground state $\vert0\rangle$
 . I don't know why!*-Please explain it if you do.

For $n=1$
 , the biggest value for $m$
  one can get for a state of $\mathcal{E}(n=1)$
  is $1$, since $(V_{1}^{\dagger})^{1}\vert0\rangle$
 is the only way to get the maximum $m$
  while (increasing $n$
  to $1$) by application of the operators defined above, and indeed $L_{z}V_{1}^{\dagger}\vert0\rangle=\hbar V_{1}^{\dagger}\vert0\rangle$
 . Now, this state, having $m=1$
 , must have $l\geq1$
 . It cannot be $l>1$
  since this would imply that I had states with $m>1$
  (obtained by applying $L_{+}$
  to $V_{1}^{\dagger}\vert0\rangle$
 ) in $\mathcal{E}(n=1)$
  (the Hamiltonian commutes with $L_{+}$
  because of rotational invariance and so does N
 ), which is absurd. For $m=0$
 , we easily see that there is only one state: $V_{0}^{\dagger}\vert0\rangle$
 . Since there is only one, it must be the one of $\mathcal{E}(l=1)$
 . There is no need to analyze other values for m
  (negative values will bring nothing new). In conclusion, the only value for $l$
  for states with $n=1$
  is $1$
 .

Let us try something harder.

For $n=3$
 , the biggest value for $m$
  for a state of $\mathcal{E}(n=3)$
  is three, corresponding to the state $(V_{1}^{\dagger})^{3}\vert0\rangle$
 . Clearly, there are no other states with $m=3$
  in $\mathcal{E}(n=3)$
 . Just like before, this means that the space $\mathcal{E}(l=3)$
  is in the direct sum decomposition of $\mathcal{E}(n=3)$
 . Now, for $m=2$
  we only have one state $V_{0}^{\dagger}(V_{1}^{\dagger})^{2}\vert0\rangle$
 , so there is no $l=2$
  representation in the direct product decomposition. For $m=1$
 , we have $V_{-1}^{\dagger}(V_{1}^{\dagger})^{2}\vert0\rangle$
  but also $(V_{0}^{\dagger})^{2}V_{1}^{\dagger}\vert0\rangle$
 , so there is one more representation (besides $l=2$
 ) with states with $m=1$
 . It can't be $l=2$
  because there are no $m=2$
  states. It is therefore an $l=1$
  representation. Finally, for $m=0$
  I only have two states: $V_{-1}^{\dagger}V_{0}^{\dagger}(V_{1}^{\dagger})^{1}\vert0\rangle$
  and $(V_{0}^{\dagger})^{3}
 \vert0\rangle$
 , which must belong to the two representations already found. Therefore, $l\in\{1,3\}$
 .