Group Theory: Show that the following group is finite?? I am struggling with this one, I have no idea what the hint has to do with the problem or how to use it.
What we know: $a$, $b$ are from the General Linear Group over $\mathbb{C}$, so $a,b \in GL(2, \mathbb{C})$. And $a$ has order 3, $b$ has order 4. And 
$bab^{-1} = a^2$.

G is a subgroup of $GL(2, \mathbb{C})$ generated by $a$ and $b$. Show that G is a finite group and find it's order(Hint: Use $ba =a^2b$).

I think I can use Lagrange's theorem to find the order bit but I cannot show that it is finite.
I mean, I tried expanding for some $(ba)^n$ but that doesn't help, I can't find it's order and the same goes for $ab$ or basically most other combinations of $a$ and $b$.
Actually, I've never encountered a group with $two$ generators. How do I do this...???
Help really appreciated, thank you!
 A: The fact that $G$ is generated by $a$ and $b$ means that the elements of $G$ are exactly the elements of $\mathrm{SL}(2,\mathbb C)$ which can be made by multiplying $a$ and $b$.
If $\mathrm{SL}(2,\mathbb C)$ were abelian, a general element would be easy to write down succinctly: it would just be $a^nb^m$. But $\mathrm{SL}(2,\mathbb C)$ is not abelian; this is where the hint comes in.
An element of $G$ is of the form
$$a^{n_1}b^{n_2}\cdots a^{n_{k-1}}b^{n_k}$$ for some $n_i\in\mathbb Z$, $k\in\mathbb N$. Can you use the hint to simplify this expression to something of the form $a^nb^m$? What does this tell you about the order of $G$?
A: $G$ is the set of all words in $a,a^{-1},b,b^{-1}$.
Since $a^{-1}=a^2=aa$ and $b^{-1}=b^3=bbb$, $G$ is the set of all words in $a$ and $b$.
Since $ba =a^2b$, we can move all $a$'s to the left and so $G$ is the set of all words of the form $a^m b^n$. Therefore, because we it is enough to take $m<3$ and $n<4$, there is only a finite number of possible words and the group is finite (of order at most $12$).
To find the order, consider when two words the form $a^m b^n$ can be equal.
A: While the general element of $G$ s of the form $a^{n_1}b^{n_2}a^{n_3}b^{n_4}\cdots a^{n_{k-1}}b^{n_k}$, we can consider the subset $$S:=\{\,a^nb^m\mid n\in\Bbb N_0, m\in \Bbb N_0\,\}\subseteq G $$
and use $ba=a^2b$ to show that $S=G$:
Lemma 1. Let $A=\{\,k\in\Bbb N_0\mid b^ka\in S\,\}$. Then $A=\Bbb N_0$.
Proof.
Trivially, $0\in A$.
Assume $k\in A$, say $b^k=a^nb^m$. 
Then $$b^{k+1}a=b^kba=b^kab^2 =a^nb^mb^2=a^nb^{m+2}\in S$$
and hence also $k+1\in A$. The claim follows by induction. $_\square$
Corollay. $Sa\subseteq S$ and in fact $Sa^k\subseteq S$ for all $k\in\Bbb N_0$.
Proof. 
The first follows because in $a^nb^m\cdot a$ we have $b^ma\in S$ and clearly $a^nS\subseteq S$. Then the second part follows by induction. $_\square$
As also $Sb^k\subseteq S$ for all $k\in\Bbb N_0$, we conclude
$$ SS\subseteq S.$$
So if we now use that $S$ is actually finite (but nonempty) we conclude that $S$ is a subgroup (subgroup criterion for finite subgroups).
As it contains $a$ and $b$, we obtain
$$G=\langle a,b\rangle\le S\le G $$
and ultimately $G=S$.
A: Consider the groups $C_3=\langle a\rangle,C_4=\langle b\rangle$. There is an automorphism of $C_3$ that has order $2$; namely inversion $\eta:C_3\to C_3$ which sends $a\mapsto a^{-1}$, and there is a group morphism $\Phi:C_4\to {\rm Aut}(C_3)$ that sends $b\mapsto \eta$. We can thus form the semidirect product $G=C_3\rtimes C_4$, this has order $12$ (since its underlying set is just $C_3\times C_4$) and is generated by $a,b$, which have order $3$ and $4$ respectively. Moreover $bab^{-1}=\eta(a)=a^{-1}=a^2$, since $a$ has order $3$. This characterizes $G$ completely: $G$ is presented as the group $$G=\langle a,b\mid a^3,b^4,bab^{-1}=a^{-1}\rangle$$
You can now try to show that the subgroup of ${\rm SL}(2,\Bbb C)$ you've been given is isomorphic to $G$. 
