Angular momentum and external torque

In John Taylor - Classical mechanics it was mentioned that the equation

$$L' = \Gamma^{ext}$$

is true even with respect to the center of mass. Here we are working with a system of particles, $$L = \sum r_{\alpha} \times p_\alpha$$ is the net angular momentum, and $$\Gamma^{ext} = \sum r_\alpha \times F_\alpha^{ext}$$ is the net external torque.

I would appreciate if someone checked my attempt at showing this - especially whether I understand properly how we're switching the frames of reference.

Suppose that we are working with some inertial frame of reference where the position of particles is denoted by $$r_\alpha$$, i.e. $$F_\alpha =m_\alpha r_\alpha ''$$, and now consider a reparametrization $$\tilde{r}$$of the position of the particles with respect to the center of mass of the system $$c$$- i.e. $$\tilde{r}_\alpha = r_\alpha -c$$. We can assume that $$L' = \Gamma^{ext}$$ is true for the parametrization $$r$$ and we would like to show that $$\tilde{L}' = \tilde{\Gamma}^{ext}$$ in the frame of reference $$\tilde{r}$$.

$$\tilde{L} ' = \sum \tilde{r}_\alpha \times (m_\alpha \tilde {r}_\alpha'')$$

$$=\sum (r_\alpha -c) \times (m_\alpha (r_\alpha '' - c''))$$

$$=\sum (r_\alpha -c) \times m_\alpha r_\alpha '' - \sum (r_\alpha -c) \times m_\alpha c'' = \tilde{\Gamma}^{ext} - \sum (r_\alpha -c) \times m_\alpha c''$$,

so it only remains to show that $$\sum (r_\alpha -c) \times m c'' = 0$$. But that is clear from $$\sum (r_\alpha -c)m_\alpha = \sum r_\alpha m_\alpha - c \sum m_\alpha = 0$$, so $$\tilde{L}' = \tilde{\Gamma}^{ext}$$.

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What seems also true is a similar claim - suppose that we have a system of particles and an inertial frame of reference $$r$$, so $$L' = \Gamma^{ext}$$. Suppose that also that the center of mass $$c$$ with respect to $$r$$ satisfies $$c(t)=0$$ for all $$t$$, and consider any smooth curve $$g$$. Then $$\tilde{r} = r+g$$ also satisfies $$\tilde{L}' = \tilde{\Gamma}^{ext}$$, because similarly as above:

$$\tilde{L} ' = \sum (r_\alpha + g ) \times (m_\alpha r_\alpha'' ) + \sum (r_\alpha + g ) \times (m_\alpha g'' )$$

$$=\tilde{\Gamma}^{ext} + \sum (r_\alpha + g ) \times (m_\alpha g'' )$$,

but since $$\sum m_\alpha r_\alpha = 0$$ because $$c(t) = 0$$, we have $$\sum (r_\alpha + g ) \times (m_\alpha g'' ) = 0 + \sum g \times (m_\alpha g'' ) = 0$$.

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The first claim could be described as:

If a system of particles is inertial in some frame of reference $$r$$, then $$L' = \Gamma^{ext}$$ is true in this frame of reference, and it is also true in the frame of reference $$\tilde{r} = r-c$$.

And the second as (more precisely, the following is directly implied by it):

For a system of particles in some frame of reference $$r$$, if $$\tilde {r} =r-c$$ is inertial, then $$\tilde{L}' = \tilde{\Gamma}^{ext}$$ in the frame of reference $$\tilde{r}$$ and also $$L ' = \Gamma^{ext}$$ in the frame of reference $$r$$.

And together this can be described as: For a system of particles in a frame of reference $$r$$, if either $$r$$ or $$r-c$$ is inertial, then both satisfy $$L' = \Gamma^{ext}$$.

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In some sense, there seems to be a similarity of some kind - I mean the trick is pretty much the same in both the claims, so I would expect it would be possible to maybe show that the first claim is true, by somehow applying the result of the second, but I haven't been able to do so - I would be interested if this is true, or if perhaps both of these claims are implied by some generalization of both.

I'm also self studying and don't have much feedback, so just some feedback of whether my thoughts make sense/seem correct would be great too.