Understanding circular convolution Let $a$ and $b$ $\in \mathbb C^N$ and $a[k] = a[k \mod N]$ (same with b). Then the circular convolution of $a$ and $b$ is defined by 

$$ (a * b)[n] = \sum_{p=0}^{N-1} a[p] b[n-p].$$

I have a problem understanding the flip operation on the convolution. Let $(Ja)[n] = a[-n]$, then

$$ J(a*b)[n] = \sum_{p=0}^{N-1} a[p]b[-n-p].$$

so we first apply the convolution and then we flip the n - is this correct? Or do we flip the whole index of $b$, meaning that

$$ J(a*b)[n] = \sum_{p=0}^{N-1} a[p]b[-n+p].$$
 On the other hand we have

$$((Ja)*(Jb))[n] = \sum_{p=0}^{N-1}a[-p]b[n+p].$$
so we first flip $a$ and $b$ and then apply the convolution. However, my book says $J(a*b)[n] = ((Ja)*(Jb))[n]$, which is not the case with my calculation, so I'm sure did something wrong. Can you help me finding the logical mistake here?
Your first version of $J(a*b)[n]$ is correct:

$$J(a*b)[n] = \sum_{p=0}^{N-1} a[p]b[-n-p]$$

because you simply have to flip $n$, and not the summation index.

The problem is in your formulation of the circular convolution of $(Ja)[n]$ and $(Jb)[n]$. It should be

$$((Ja)*(Jb))[n]=\sum_{p=0}^{N-1} a[-p]b[-n+p]$$

because you have to negate the arguments of both sequences, so $p$ becomes $-p$, and $n-p$ becomes $-n+p$. Now you can simply change the sign of $p$ because it doesn't matter in which order you compute the sum. This gives

$$((Ja)*(Jb))[n]=\sum_{p=0}^{N-1} a[p]b[-n-p]$$

which is identical to $J(a*b)[n]$ as given above, as it should be.