Prove that it is sufficient to map the generators and their relationship onto each other in order to show that we have an Isomorophism between two Groups.
Say we have $\langle g,h\mid g^3=1,h^2=1,hg=g^{-1}h\rangle $ and $\langle a,b\mid a^3=1,b^2=1,ba=a^{-1}b \rangle $. It is very obvious to me that there must be an Isomorphism because it simply means that it is esentially the same group but just with different symbols. If we let $\phi(g)=a$ and $\phi(h)=b$ this should get me an Isomorphism. But abstractly an Isomorphism means that $\phi(xy)=\phi(x)\phi(y)$ and this is by no means obvious.
Yes.
Use Tietze transformations (and their implicit isomorphisms) to introduce new generators equal to existing ones, rewrite relators/relations using those generators, and then to delete the old ones.
For example,
$$\begin{align} \langle g,h\mid g^3=1,h^2=1,hg=g^{-1}h\rangle &\cong \langle g,h, a\mid a=g, g^3=1,h^2=1,hg=g^{-1}h\rangle\\ &\cong\langle h, a, b\mid b=h, a^3=1,h^2=1,ha=a^{-1}h\rangle\\ &\cong\langle a,b\mid a^3=1,b^2=1,ba=a^{-1}b\rangle. \end{align}$$