I'm wondering if using the 4 statements below to help prove that Mobius Transformations form a group is enough. If so, how can I use M(z)=(az+b)/(cz+d) to do that?
Thank you in advance, and please feel free to fix any formatting as I'm not great with that. :)
For example, we can show closure as follows. Let $M_1,M_2$ be the Mobius transformations $$ M_1(z) = \frac{a_1 z + b_1}{c_1 z + d_1}, \quad M_2(z) = \frac{a_2 z + b_2}{c_2 z + d_2}. $$ To show that condition 1 holds, show that the function $M_1 \circ M_2$ is a Mobius transformation. $$ M_1(M_2(z)) = \frac{a_1 \frac{a_2 z + b_2}{c_2 z + d_2} + b_1}{c_1 \frac{a_2 z + b_2}{c_2 z + d_2} + d_1} = \cdots = \frac{[a_1a_2 + b_1 c_2]z + [a_1b_2 + b_1 d_2]} {[c_1 a_2 + d_1 c_2]z + [c_1 b_2 + d_1 d_2]}. $$ We can see that this could be a Mobius transformation because it can be expressed in the form $$ M(z) = \frac{az + b}{cz + d} $$ for some complex numbers $a,b,c,d$. In fact, to properly show that $M$ is a Mobius function, we should also argue that $ad \neq bc$, which is to say that there is no cancelation and $M$ is not a constant function.
One way to argue that $M$ is not a constant function is to come back once you've proven point 4 (the existence of inverse). If $M(z)$ where a constant function, then the function $M_1^{-1}(M(z)) = M_2(z)$ would also be a constant function. That can't be true because, by the definition of the Mobius group, $M_2(z)$ was not a constant function (i.e. $a_2 d_2 \neq b_2 c_2$).
For 2, note that for two Mobius functions $M_1,M_2$: by the definition of your group, if $M_1(z) = M_2(z)$ for all complex numbers $z$, then $M_1 = M_2$. With that in mind, we note that for any $z \in \Bbb C$, we have $$ [(M_1 \circ M_2)\circ M_3](z) = (M_1 \circ M_2)(M_3(z)) = M_1(M_2(M_3(z))),\\ [M_1 \circ (M_2 \circ M_3)](z) = M_1((M_2 \circ M_3)(z)) = M_1(M_2(M_3(z))). $$ That is, we have $$ [(M_1 \circ M_2)\circ M_3](z) = [M_1 \circ (M_2 \circ M_3)](z) $$ for all complex numbers $z$, so that $(M_1 \circ M_2)\circ M_3 = M_1 \circ (M_2 \circ M_3)$. In other words, the defined multiplication is associative.
For 3, let $I$ denote the Mobius function $$ I(z) = z = \frac{1\cdot z + 0}{0\cdot z + 1}. $$ Verify that for any Mobius function $M$, we have $M \circ I = I \circ M = M$.