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Work in an inertial frame, take all masses to be constant, and assume the internal forces obey the pairwise relation from Newton's third law $F_{ij}=-F_{ji}$. For the angular momentum result, also assume these pair forces are central, so $F_{ij}$ is parallel to $r_i-r_j$ and its pair torque vanishes. The equations from Newton's second law areSumming them and pairing the internal forces gives

About a fixed point $a$, define the total angular momentum and external torque byDifferentiation produces no velocity term because ${\dot{r}}_i\times m_i{\dot{r}}_i=0$. Grouping internal terms into pairs givesEach final cross product is zero by centrality, proving $\boxed{d{L}_a/dt=G_a}$. Equal and opposite internal forces give momentum conservation when the total external force is zero, but do not in general make their internal torques cancel; the central-force assumption supplies that extra step.

Let $M={\sum }_i{m}_i$ and let the centre of mass be $R=M^{-1}{\sum }_i{m}_i{r}_i$, so $P=M\dot{R}$. For an arbitrary moving reference point $a(t)$, the same differentiation givesTaking $a=R$, the extra term is $-\dot{R}\times M\dot{R}=0$. Therefore the corresponding angular momentum about the centre of mass satisfiesIts definition using velocities relative to $\dot{R}$ agrees with the one above, since ${\sum }_i{m}_i(r_i-R)=0$ makes ${\sum }_i(r_i-R)\times m_i\dot{R}=0$. Thus this result remains true even when the centre of mass accelerates.

Finally take the common mass to be $m$ and $F_i=-k{\dot{r}}_i$, with constant $k$. About either a fixed point or the centre of mass,The angular-momentum equation becomes $\dot{L}=-(k/m)L$. Multiplying by the integrating factor $e^{kt/m}$ gives $\frac{d}{dt}(e^{kt/m}L)=0$, hence the angular momentum decay under uniform linear drag isThe common mass and common drag coefficient ensure that the same decay factor applies to every term.