jmc

Let $M$ be the midpoint of $\overline{AB}$, so that the segment from $C$ to $M$ passes through $G$, by definition. One suspects that $\overline{CM}\perp \overline{AB}$, which can be confirmed by noting that $△AMG\cong △BMG$ since all corresponding sides are congruent. Since $AG=AB=2$ and $AM=\frac{1}{2}AB=1$, we can compute $MG=\sqrt{3}$ using the Pythagorean Theorem. We now recall an important property of the centroid: it lies on all three medians and divides each of them in a 2 to 1 ratio. In other words, $CG=2(MG)=2\sqrt{3}$. We deduce that $CM=3\sqrt{3}$, so we can finally compute length $AC$ by using the Pythagorean theorem in $△AMC$ to find $$AC=\sqrt{1^2+(3\sqrt{3}{)}^2}=\sqrt{28}=2\sqrt{7}.$$In the same manner $BC=2\sqrt{7}$ as well, giving a perimeter of $\boxed{2+4\sqrt{7}}$.