Showing $H=\langle a,b|a^2=b^3=1,(ab)^n=(ab^{-1}ab)^k\rangle$. 
Let $G=\langle a,b|a^2=b^3=1,(ab)^n=(ab^{-1}ab)^k \rangle$. Prove that $G$ can be generated with $ab$ and $ab^{-1}ab$. And from there, $\langle(ab)^n\rangle\subset Z(G)$.

Problem wants $H=\langle ab,ab^{-1}ab \rangle$ to be $G$. Clearly, $H\leqslant G$ and after doing some handy calculation which takes time I've got:


*

*$ab^{-1}=(ab^{-1}ab)(ab)^{-1}\in H$


*$b=b^{-2}=(ab)^{-1}ab^{-1}\in H$


*$a=(ab)b^{-1}\in H$

So $G\leqslant H$ and therefore $G=H=\langle ab,ab^{-1}ab\rangle$.
For the second part, I should prove that $N=\langle(ab)^n\rangle\leqslant Z(G)$.
Please help me.
Thanks.

The question is answered in the comments. However so that this question does not remain listed as unanswered forever, I will provide a solution. I will also give the details for the first part of the question. 
Part 1 Show $G= \langle ab, ab^{-1}ab  \rangle$
Let $H=\langle ab, ab^{-1}ab \rangle $. It is clear that $ H \leq G$. We will show that $G \leq H$ by showing that $a$ and $b$ are in $H$. 
We make the following observations which we will use in the argument:
1) $ (ab)^{-1}=b^{-1}a^{-1} \in H $
2) $ a^{-1}=a $
3) $ b^{-2}=b $
Then $(ab^{-1}ab)(ab)^{-1}=ab^{-1} \in H$. Thus $ab^{-1} \cdot b^{-1}a^{-1} = ab^{-2}a = aba \in H$. Hence $(ab)^{-1} \cdot (aba) = a \in H$. Thus we have that $a \in H$ and then multiplying $ab$ with $a^{-1}$ on the left gives that also $b$ is in $H$. Hence $ G \leq H$ and we now can conclude $G=H$. 
Part 2 Show $ \langle (ab)^{n} \rangle \leq Z(G)$
Since $G = \langle ab, ab^{-1}ab \rangle$ we will try to show $ab$ and $ab^{-1}ab$ commute with $(ab)^{n}$. Of course it is clear that $(ab)$ does. 
Now consider $$(ab^{-1}ab)\cdot (ab)^{n} = ab^{-1}ab(ab^{-1}ab)^{k}$$ (using a relation given in the presentation for $G$ in the question).
$$=(ab^{-1}ab)^{k+1}=(ab^{-1}ab)^{k}ab^{-1}ab = (ab)^{n} \cdot ab^{-1}ab$$
Hence we see that $(ab)^{n}$ commutes with both of our generators of $G$ and thus $ \langle (ab)^{n} \rangle $ is central. 
