Order of $D_4$ is $4$ or $8$? $D_4 = \{R_0,R_{90},R_{180},R_{270},H,V,D,D^{'}\}$
$|R_0|=1 ,\, |R_{90}|=4 ,\, |R_{180}|=2 ,\, |R_{270}|=4$ 
$|D|=|V|=|D|=|D^{'}|= 2$
So Does that mean order of $D_4$ is $4 = \text{lcm}\, (4,2,1).$
But answer  is $8$.
 A: I guess you are conflating two different meanings of the term order, plus the term exponent.
The order of a (sub)group is the number of its elements.
The order of an element $a$ is the least $k > 0$ such that $a^{k} = 1$. (If it exists, otherwise $\infty$.)
The link is that the order of the element $a$ is the order of the subgroup $\langle a \rangle$.
The least common multiple of the orders of the elements in a finite group $G$ is called the exponent of $G$. As hinted in other answers, the exponent of $G$ divides the order of $G$, but the two might be different.
A: Besides to other answers which contain complete aspects of the answer to your question; I suggest you to consider another presentation of $D_4$ (or $D_8$). I am sure via this equivalent presentation, you can  find the elements easier and find where you are confusing at. That is: $$D_4=\langle a,b\mid a^4=b^2=1, ba=a^{-1}b\rangle=\{1,a,a^2,a^3,b,ab,a^2b,a^3b\}$$ which has $8$ elements. Try this one too! 
A: The order of $D_4$ is 8.  This is because that list of elements that you gave,
$$ \{R_0,R_{90},R_{180},R_{270},H,V,D,D^{'} \}$$
contains 8 items.
The orders of the various elements, $|R_0|, |V|$, and so on, must divide the order of the whole group.  This follows from Lagrange's theorem. Here for example, each of 1, 2, and 4 divides the order of the group, 8.
But in general, there is no way to compute the order of a group just from the orders of its elements.  For example, there is a group whose elements (except the identity)  all have order 2  but which is infinite. Also for every $n$ there is a finite group of order $2^n$ whose elements (except the identity) all have order 2.
The Klein $V$-group has order 4, but its elements all have order 1 or 2:
$$\begin{array}{c|cccc}
\oplus & 0 & 1 & 2 & 3 \\ \hline
0 & 0 & 1 & 2 & 3 \\
1 & 1 & 0 & 3 & 2 \\
2 & 2 & 3 & 0 & 1 \\
3 & 3 & 2 & 1 & 0 \\
\end{array}$$
