Question: Here's what I've been able to find out so far to address this question:

1) $C^\*\_2$ = {1}, $C^\*\_3$ = {1, 2}, $C^\*\_4$ = {1, 3}, $C^\*\_6$ = {1, 5}, $C^\*\_8$ = {1, 3, 5, 7}, and $C^\*\_{12}$ = {1, 5, 7, 11}, $C^\*\_{18}$ ={1, 5, 7, 11, 13, 17}, and $C^\*\_{30}$ = {1, 7, 11, 13, 17, 19, 23, 29} are all prime.

2) All $C^\*\_p$ where p is a prime larger than 3 have {4} as an element and thus cannot be all prime.

What is the largest modulo multiplicative group with all primes or is that even possible and there is no "largest"?

Answer: 
**Hint:** The arithmetic mean of $\frac{x+y}2$, $\frac{x+z}2$, $\frac{y+z}2$ is $\frac{x+y+z}{3}$, whereas the geometric mean is $\sqrt[3]{\frac{(x+y)(y+z)(x+z)}{8}}$
Let $$P(x;y;z)=\frac{8}{27}(x+y+z)^3-(x+y)(y+z)(x+z)$$
$P(1;1;1)\ge0; P(1;1;0)\ge0$ and $P(1;0;0)\ge0$
Then $$P(x;y;z)\ge0$$
Hence,
$$(x+y)(y+z)(x+z) ≤ \frac{8}{27}(x+y+z)^3$$