What is $(Z/(15))^×$ What is $(Z/(15))^×$?

Don’t confuse the groups $Z/(m)$ and $(Z/(m))^×$. The second, as a set, is a subset of the first, but it is not a subgroup and it is usually not cyclic (see Theorem A.2). For instance, $(Z/(15))^×$ is not cyclic.

I am confused because I thought $\mathbb{Z}_{15}^\times$ is not even a group since $15$ is not prime?
 A: Good question. The group $\mathbb{Z}/(n)=\{0,1,\ldots,n-1\}$ is the cyclic group of order $n$ (where 1 serves as a generator). There exists one and only one (up to isomorphism) such group for every integer $n$. We must be careful here though because I think the confusion is arising from what this group's operation is. It is addition, not multiplication, where  $0$ is the identity and every element now has a unique inverse modulo $n$. We can still multiply number here but you'll notice not every number has a multiplicative inverse. Thus we end up with a ring, not a field. If $n$ is prime, then we do have a field as every nonzero number is now invertible modulo $n$. 
The notation $\mathbb{Z}/(n)^\times$ means all the elements in the ring $\mathbb{Z}/(n)$ which are multiplicatively invertible, and now think about this set as a group with multiplication as the group operation. In the case when $n$ is prime, that's all the nonzero numbers. When $n$ is not prime, we must exclude other numbers as well (all numbers which are not coprime to $n$).
It's a bit of an involved proof to see when $\mathbb{Z}/(n)^\times$ ends up being a cyclic group--under multiplication, but since that's not what you're asking, I'll just leave it at that for now. But you can check out more here: https://en.wikipedia.org/wiki/Multiplicative_group_of_integers_modulo_n
A: The eight elements of $(Z/(15))^×$ are $\{1,2,4,7,8,11,13,14\}$ - the residue classes coprime to $15$. They form a group under multiplication modulo $15$. 
The full fifteen elements of $(Z/(15))$ form a monoid under the same operation (although they are a group under addition mod $15$).
