Why is conservation of angular momentum considered a law? We know that angular momentum is defined as the cross product of position and linear momentum. By taking the time derivative, we can deduce that the time rate of change of the angular momentum equals the net torque.
Then why is the conservation of angular momentum considered a law, when we can easily show it is conserved when no net torque is applied, by using Newton's laws?
It truly confuses me.
 A: You are right by saying that Newton's law and conservation of angular momentum belong to different levels of knowledge.
Newton's law ($F=ma$) is an axiom, it cannot be deduced by other relations, and you can think to perfectly allowed worlds that could exist where it did not happen to be true.
Conservation of angular momentum, on the other hand, is derived by means of Newton's law (this can be done at different levels of abstraction, ie via direct computation, Noether's theorem, and so on, but nonetheless my statement remains basically true), and given the axioms is what you get without adding nothing new.
However, the word law in physics is used to indicate both of them.
A: Conservation of angular momentum really is a new phenomenon, one that does not follow from the Newtonian mechanics you already know; therefore it deserves its own place as a law. Specifically, you have proven that

If a system experiences no torque, then its angular momentum is conserved.

However, this statement by itself is useless. Maybe all systems always experience torque; maybe a system can exert a torque on itself. What we really want to say, i.e. the actual law of conservation of angular momentum, is more like

An isolated system's angular momentum is conserved.

To see how these are not equivalent, suppose we have a system of two isolated particles, one above the other. Newton's third law does not forbid the particles from pushing left and right on each other. But then the system will begin spontaneously rotating! It changes its own angular momentum by exerting a torque on itself.
To force conservation of angular momentum, we need to use the strong form of Newton's third law,

Forces between particles come in action/reaction pairs, and these forces are directed along the line of separation between the two particles.

This is a fundamentally new assumption, so angular momentum really is its own thing. On a deeper level, conservation of linear and angular momentum follow from translation and rotational symmetry of space, and it's possible to have spaces which only are translationally symmetric, or only are rotationally symmetric. The two are independent.
A: Just to expand a little on @knzhou's answer: If you have a system of particles of masses $m_i$ at positions $\mathbf r_i$ then they have external forces $\mathbf F_i$ acting on them as well as internal forces $\mathbf G_{ji} = -\mathbf G_{ij}$ by the third law; and the second law merely states,$$m_i \ddot {\mathbf r}_i = \mathbf F_i + \sum_j \mathbf G_{ij}.$$The great first idea about the properties of the bulk is to sum up over all $i$ to find that the $\mathbf G_{ij}$ all cancel leaving, $$\sum_i m_i \ddot {\mathbf r}_i = M \ddot {\mathbf R} = \sum_i \mathbf F_i. $$Above we define $M = \sum_i m_i$ and therefore $\mathbf R = \sum_i \frac{m_i}M ~ \mathbf r_i$ is the center of mass, per usual. So the awesome thing is that this $\mathbf G_{ij} = -\mathbf G_{ji}$ makes all of the internal forces cancel out, and we can pretend that the system has mass $M$ and lives at its center-of-mass and feels all the external forces $\sum_i \mathbf F_i.$
It's well worth taking a second to prove that, if you haven't already. If we really sum over all of these indices we have $\sum_{ij} \mathbf G_{ij} = \sum_{ij} \mathbf G_{ji},$ the order of summation does not matter. Because we have $A = B$ we can also rewrite these things as $A = (A + B)/2,$ so we rewrite as $\frac12 \sum_{ij} (\mathbf G_{ij} + \mathbf G_{ji}).$ Now we use the Third Law, that $\mathbf G_{ji} = -\mathbf G_{ij},$ to find out that this is $$\sum_{ij} \mathbf G_{ij} = \frac12 \sum_{ij} (\mathbf G_{ij} - \mathbf G_{ij}) = \frac12 \sum_{ij} 0 = 0.$$That's how we prove that this "antisymmetry" of $\mathbf G$ yields the 0 when we sum up over the $i$ index above.
Now we do the procedure that you're suggesting, taking the original equation and crossing with $\mathbf r_i$ to find that the external torques $\mathbf \tau_i$ come into the picture as,$$ m_i \mathbf{r}_i \times \ddot{\mathbf r}_i = \mathbf \tau_i + \sum_j \mathbf r_i \times \mathbf G_{ij}.$$ The left hand side works out, because $\dot{\mathbf r}_i \times \dot{\mathbf r}_i = 0,$ to $\frac{d}{dt}(m_i \mathbf r_i \times \dot{\mathbf r}_i)= d\mathbf L_i/dt.$ That's no problem.
The big question that remains: do the $\mathbf r_i \times \mathbf G_{ij}$ necessarily cancel out? Well, let's just try to repeat the same performance as above; we have $$\sum_i \dot{\mathbf L}_i = \sum_i \mathbf \tau_i + \frac12\sum_{ij}\Big(\mathbf r_i\times \mathbf G_{ij} + \mathbf r_j\times \mathbf G_{ji}\Big),$$and after applying the antisymmetry of $\mathbf G$ we only get to$$\sum_i \dot{\mathbf L}_i = \sum_i \mathbf \tau_i + \frac12\sum_{ij}\big(\mathbf r_i - \mathbf r_j) \times \mathbf G_{ij}.$$So in fact this is not zero in general and instead requires $\mathbf G_{ij}$ to be parallel to $\mathbf r_i - \mathbf r_j$ in general in order to make it zero. And this is precisely what was described by @knzhou; the internal forces need to point along the line that connects the particles, otherwise angular momentum is not conserved.
