How many elements are there in $\mathbb{Z}_5$? It is easy to verify that $\mathbb{Z}_5$ forms a group with $+$ operation and $\mathbb{Z}_5 \setminus \{[0]\}$   with $\times $ operation. I know that there are infinite elements in $\mathbb{Z}$.
Question: How many elements are there in $\mathbb{Z}_5$? Is it $5$ or finite? I know it is a group of order $5$ so it should have $5$ elements, but I also know that there will be classes and in each there will be infinite elements. I am always confused in these things.
 A: There are $5$ elements in $\Bbb Z_5$ (which by the way is not a group with $\times$, although $\Bbb Z_5\setminus\{[0]\}$ is). Each of those five elements is an infinite set of integers.
A: Note that the multiplicative group of integers modulo $5$ is the following group : 
$$\mathbb Z_5 = \{\bar{0}, \bar{1},\bar{2},\bar{3},\bar{4}\}$$
Hence, there are $5$ elements in it, also denoted by the order of it, $|\mathbb Z_5| = 5$. Now, regarding your question about each element, it's true that each one of the elements elaborated in the expression of $\mathbb Z_5$ consists of a set of infinite integers.
Take notice, that $\mathbb Z_5$ is a group with the operation $+$. A group with the operation $\times$ regarding a group of integers modulo $5$, would be $\mathbb Z_5 \setminus \{\bar{0}\} = \mathbb Z_5^*.$ 
A: In $\mathbb{Z}_n$ with $n\in\mathbb{N}$ there are $n$ number if it is an additive group and it is:
$$\mathbb{Z}_n=\left\{\bar{0}, \bar{1},\dots,\overline{n-2},\overline{n-1}\right\}$$
Instead if it is a multiplicative group there are $\phi(n)$ elements, where $\phi(\cdot)$ is the Euler’s totient function and it is:
$$\mathbb{Z}^*_n=\left\{\overline{a}\in\mathbb{Z}_n|\gcd(a,n)=1\right\}$$
In your case $\mathbb{Z}_5$ as an additive group is made of 5 elements and it is 
$$\mathbb Z_5 = \{\bar{0}, \bar{1},\bar{2},\bar{3},\bar{4}\}$$
As a multiplicative group it has cardinality $\phi(5)=4$ and it is 
$$\mathbb Z_5^*= \{\bar{1},\bar{2},\bar{3},\bar{4}\}$$
One one hand there are only a limited number of elements. On the other hand there are unlimited number on element inside because you group them in base of the remainder with the division modulo $n$. 
In fact there are, for example, infinite numbers $k\in\mathbb N$ such as $k\equiv 1 \pmod n$.
