How to calculate the order of commutator of two elements of a group $G$ in terms of their orders? if $G$ is a group , $x,y \in G$ and $[x,y]$ is the commutator of $x$ and $y$ 
so , $[x,y]=x^{-1}y^{-1}xy$
is there a formula to compute $|[x,y]|$ in terms of $|x| , |y|$ ? 
 A: No. If $|x|=|y|=2$ then $|[x,y]|$ can have any order $1,2,\ldots,\infty$.
Take $G$ dihedral of order $4n$ (with $4\infty = \infty$). Then the commutator of two generating reflections has order $n$.
Take $G = \langle a,b : a^2 = b^2 = (ab)^{2n} = 1 \rangle$ Then $[a,b] = a^{-1} b^{-1} a b = a b a b = (ab)^2$ should have order $n$ if the presentation is not “lying” about the order of $ab$.
However, the group of symmetries of a regular $2n$-gon satisfies this presentation with $a$ and $b$ “adjacent” reflections, and their product $ab$ a “primitive” rotation. Alternatively, you can start with the reflection $a$ and a primitive rotation $r$. Then define $b=ar$. By a fundamental geometric principal, $b$ is also a reflection, and $ab=r$ has order $2n$. At any rate, $[a,b] = [a,ar]$ is a “double” rotation, so only has order $n$.
If $n=\infty$ then $a=(x\mapsto -x)$ is a reflection (over the line $x=0$ if you imagine this in the Euclidean plane) and $b=(x\mapsto 1-x)$  is a reflection (over the line $x=\tfrac12$) and $[a,b]= x \mapsto -x \mapsto 1-(-x) \mapsto -(1-(-x)) \mapsto 1 - ( -(1-(-x))) = x+2$ is a translation by 2, so it has infinite order.
