Reason for why a physical quantity is zero in the below description If we assume the centre of mass to be the origin and the frame is the centre of mass frame, then we know that the total linear momentum of system from centre of mass frame is always zero.
But what about the total angular momentum?
Suppose that we have a body rotating about a fixed axis. Consider (i)th particle. I have posted the picture. If if the position vector is $r$ are and the linear momentum vector is $P$ and the angular momentum is $L$.
Then
L = Σr×P
= Σ(r'+rc)×(P'+Pc)=Σr'×P'+Σrc×P'
RC means the position vector of centre of mass. There will be also to other terms which I'm not writing because I know the meaning of those two rest terms. The first term here I wrote is is the the angular momentum of the body with respect to centre of mass as told by our teacher. But we know that the linear momentum from centre of mass frame is zero so in this case it is zero X something which should be zero. The total angular momentum from centre of maths should be zero but our teacher told it as non zero. Why?!
Also if possible kindly tell me that why the second term is zero as told by the teacher?

 A: I will show you something that maybe helps. Look at the total angular momentum as measured from the axis of rotation: $\vec{L} = \sum_i \vec{r}_i \times \vec{p}_i$. Now you can rewrite $\vec{r}_i$ as $\vec{r}_i = \vec{r}_{CM} + \vec{r'}_i$. Doing so, you can see that:
$\begin{align} \sum_i \vec{r}_i \times \vec{p}_i &= \sum_i (\vec{r}_{CM} + \vec{r'}_i) \times \vec{p}_i = \vec{r}_{CM} \times \sum_i \vec{p}_i + \sum_i \vec{r'}_i \times \vec{p}_i = \sum_i \vec{r'}_i \times \vec{p}_i  \end{align}$.
This means that the total angular momentum stays the same, no matter which reference you choose (as you should expect). The reason this is true is because the sum of the linear momenta is zero, as you pointed out!
Now looking at what you wrote, you also split up the linear momenta:
$L = \begin{align} \sum_i \vec{r}_i \times \vec{p}_i &= \sum_i (\vec{r}_{CM} + \vec{r'}_i) \times (\vec{p}_{CM} \times \vec{p'}_i) = \sum_i \vec{r'}_i \times \vec{p'}_i + \sum_i \vec{r}_{CM} \times \vec{p'}_{i} + \ldots \end{align}$
These two last terms are the ones you wrote down. The second term is zero for the reason I already explained: $\vec{r}_{CM}$ is independent of the index $i$, so you can pull it out of the summation so that $\sum_i \vec{r}_{CM} \times \vec{p'}_{i} = \vec{r}_{CM} \times \sum_i \vec{p'}_{i}$ and your object only rotates, it does not move, so $\sum_i \vec{p'}_{i} = 0$. That is why the second term is zero! Now the total angular momentum of the center of mass is non zero, as your teacher pointed out correctly. This is because your rigid body is spinning around an axis which is not going trough the center of mass, so your center of mass is rotating. That is why it SHOULD have a non zero angular momentum! I hope this makes it all clear, if not please ask again :)
