jmc

We draw the medians as shown below.



From right triangles $ACM$ and $BCN$, we have $$\begin{array}{rl}A{C}^2+C{M}^2& =36,\\ B{C}^2+C{N}^2& =(2\sqrt{11}{)}^2=44.\end{array}$$

However, we have $CM=BC/2$ and $CN=AC/2$, so the equations above become $$\begin{array}{rl}A{C}^2+\frac{1}{4}B{C}^2& =36,\\ B{C}^2+\frac{1}{4}A{C}^2& =44.\end{array}$$

Adding these equations gives $$\frac{5}{4}(A{C}^2+B{C}^2)=80,$$ so $A{C}^2+B{C}^2=64$. But the Pythagorean Theorem gives us $A{B}^2=A{C}^2+B{C}^2$, so $A{B}^2=64$, which means $AB=\boxed{8}$.