A question about groups generated by two elements. Suppose a group $G=\langle a,b \rangle$ and $|G|<\infty$ where $|a|=m_0$  and $|b| = m$. How is it that the operation table for $G$ can be completely determined just by knowing $ab=b^na$ for some $n$? 
I don't see how this is possible, but it says it is true in Fraleigh's introduction to abstract algebra book, but the does not have a proof.
I have played around a bit with looking at the coset $b\langle a \rangle$ but I am not guaranteed to get all the elements of G, so I look at another coset until you run out of non represented elements. Now I have partioned my group and I know every element must be of the form $a^kb^{k_0}$ or $b^ka^{k_0}$ but I don't see how knowing $ab=b^na$ helps defined the entire group.
Any help would be great.
 A: The wording of the statement is slightly unclear. The way you state it, it sound as though you are told what $m$ and $m_0$ are, but not what $n$ is - only that $n$ exists. Without knowing $n$, you cannot possibly determine the multiplication table, so I will assume we are told what $n$ is. In that case, you can use the group relation $ab=ba^n$ to write all elements in the form $a^kb^l$, and so $|G| \le mm_0$. But there is still some uncertainty, because there could be more than one possible intersection of the subgroups $\langle a \rangle$ and $\langle b \rangle$
To get more technical, we know that $G$ is a quotient group of the group $Z$ defined by the presentation $$Z=\langle x,y \mid x^{m_0}=y^m=1, xyx^{-1}=y^n\rangle.$$ 
Now $Y=\langle y \rangle$ is a normal subgroup of $Z$ of order $m$ (we know it must be exctly $m$, because we are told that $|b|=m$), and $Z/Y$ is cyclic of order $m_0$. So $|Z|=mm_0$, but $G$ could still be a proper quotient of $Z$.
If we know $|G|$, then its isomorphism type is completely determined, which is equivalent to saying that its multiplication table is determined by the conditions. But if we are not told $|G|$, then there could be more than one possible order of groups satisfying theses conditions. For example, if $|a|=|b|=4$ and $ab=b^3a$, then $|Z|=16$, and $|G$| could have order $8$ or $16$. (The quaternion group $Q_8$ satisfies these conditions, but it has the additional relation $a^2=b^2$.)
