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**ThePerfectHacker** The above paragraph has some nice implications. You probably heard of "Group Theory" before. A group $\displaystyle G$ is a set with an operation $\displaystyle *$ so that: (i)$\displaystyle a*b\in G$ for any $\displaystyle a,b\in G$ (ii) $\displaystyle a*(b*c) = (a*b)*c$ for all $\displaystyle a,b,c\in G$ (iii) there is an "identity element" $\displaystyle e$ so that $\displaystyle a*e = e*a = a$ (iv) for any element $\displaystyle a\in G$ there is an "inverse" $\displaystyle b$ so that $\displaystyle a*b=b*a = e$. For example, the integers $\displaystyle \mathbb{Z}$ are a group under the operation $\displaystyle +$. Now if we let $\displaystyle \mathcal{M}$ be the set of all Mobius transformation on the Riemann Sphere then $\displaystyle \mathcal{M}$ is a group under $\displaystyle \circ$ (function composition). Note, we could have not made such a statement about a group if the Mobius transformations were defined only on $\displaystyle \mathbb{C}$ because as stated above we cannot necessarily compose them. If you want you can try proving that $\displaystyle \mathcal{M}$, the "Mobius group", is actually a group.