The group you are looking at is the multiplicative group modulo $17$ which the powers of $3$ generate.
As a set, for general $n$ this does not include $0$ and is usually written as
$$
(\mathbb{Z}_n^\ast,\cdot)
$$
where $\mathbb{Z}_n^\ast \subseteq \{1,2,\ldots,n-1\}$
for any positive integer $n\geq 2.$
If $n=p$ is a prime then this set is actually all of $\{1,2,\ldots,p-1\}$ Otherwise, it is just the set of elements in $\mathbb{Z}_n$ that is relatively prime to $n.$ If $n=p$ a prime, then the group is also cyclic meaning a single element $g$ can generate all its members as powers $g^i\pmod p.$
For your example $p=17,$ and $g=3.$
Edit: If $n$ is nonprime, say $n=pq$ where $p\neq q$ are primes then there are
$n/p$ elements in $\{0,1,\ldots, n-1\}$ that are divisible by $p.$ There are
$n/q$ elements in $\{0,1,\ldots, n-1\}$ that are divisible by $q$. Since zero is divisible by both we get
$$
n\left(1-\frac{1}{p}\right)\left(1-\frac{1}{q}\right)
$$
elements that are relatively prime to $n$ which is the size of the multiplicative group.
For general $n$ we have $\varphi(n)$ elements in the group, where $\varphi(\cdot)$ is Euler’s totient function.
