Why is the group $Z_6$ under addition isomorphic to the group $Z_7^*$ under multiplication?
So I'm trying to answer this question:
Q. Which of the set are isomorphic to each other?
$S_3$, $Z_6$, $Z_3 \times Z_2$ and $Z_7^*$
Now I know that $S_3$ is not because its not abelian while $Z_6$, $Z_3 \times Z_2$ are since my professor said this:
The group $Z_n$ under addition is an abelian group which means $Z_6$ is abelian.
Now another idea is: The direct product of two cyclic groups $Z_m$ and $Z_n$ is isomorphic to ($Z_{mn}$,+) iff m and n are relatively prime.
For $Z_3 \times Z_2$, 3 and 2 are relatively prime which means $Z_6$ and $Z_3 \times Z_2$ are isomorphic.
Now, whats throwing me off is $Z_7^*$. I know $Z_7^*$ is composed of the elements {1,2,3,4,5,6}. So i've researched a bit online, and $Z_7^*$ has an order of 6 which I'm not understanding.
Also, it was mentioned that $Z_7^*$ is abelian. Therefore, are all groups $Z_n^*$ under multiplication an abelian group? and why does $Z_7^*$ have an order of 6? Also, I've seen people say $Z_7^*$ is cyclic and has a generator such as 3, I'm not understanding that concept or in other words, how is 3 a generator? because if i do $3^6$=729 and that is not divisible by 7.
Or if anyone has a different way to do this problem, that would help as well. I'm open to any ideas. thanks!