How to find the order of the ring I have some examples solved about order of an element in a ring.
In $\mathbb Z_{5}$  find order of element $2$ in the ring.   Identity is $1$
$2^1 ≠ 1$
$2^2 ≠ 1$
$2^3 ≠ 1$
$2^4 = 16 = 1$ &lt;----------This is the part that I dont understeand
First of all, mention in the title that you would like to calculate order of an 'element' of a ring.
In $Z_5$, the multiplication is defined by multiplying two elements as integers and then reducing the result to modulo $5$ which means that you need to divide the result by $5$ and take remainder.
An example is $3\cdot 4=12$ which gives remainder $=2$ on dividing by $5$ and hence in $Z_5$, $3\cdot 4=2$.
Now in the last line of your question, the correct equation is $2^4=16\equiv 1 \pmod 5$ because $16$ yields remainder $=1$ when divided by $5$.
I think this is a matter of the rigor of definitions. Strictly speaking, $16$ is NOT an element of $\mathbb{Z}_5$, which as a set is $\{0,1,2,3,4\}$. Strictly speaking, we have a surjective homomorphism of rings $\pi:\mathbb{Z}\rightarrow\mathbb{Z}_5$ sending a number to its remainder modulo $5$, i.e., when dividing by $5$: $\pi(n):=r$ where $n=5q+r$.
So, strictly speaking we have $\pi(2^4)=\pi(16)=\pi(5\cdot 3+1)=1$ (in $\mathbb{Z}_5$), and also $2^4=\pi(2)^4=1$ in $\mathbb{Z}_5$ (since $\pi$ is a homomorphism such that $\pi(2)=2$), which we can write also as $2^4\equiv 16\equiv 1\pmod{5}$, where now $\equiv\pmod{5}$ is an equivalence relation for $\mathbb{Z}$. Nevertheless it is usual, but an abuse of notation, to write $16=1$ in $\mathbb{Z}_5$.