How do I prove that the following is a presentation of $\mathbb{Z_{35}}$? Prove that $\mathbb{Z_{35}}$ is isomorphic to $G=\langle x, y \, | \, x^7 = y^5 = 1, xyx^{-1}y^{-1}=1\rangle$. I'm having a very difficult time with such problems and would be very thankful to anyone who would take the time to slowly explain how does one even begin to take a stab at answering such a question. Thank you for your help.
$35 =7\cdot5$ , the Chinese remainder theorem show that $\mathbb{Z}_{35}\simeq\mathbb{Z}_7\times \mathbb{Z}_5$. Take u = a generator of $\mathbb{Z}_7$ and v a generator of $\mathbb{Z}_5$ and set x=(u,0) and y =(0,v)
$x^7=1$ has 7 complex solutions that constitute a cyclic group. The Same reasoning can be used for $y^5=1$ (a cyclic group with 5 elements). Now, every finite cyclic group is isomorphic to $Zn$ (n is the number elements in the group). So, we have two groups (Z7 and Z5).A theorem says that the Cartesian product of groups is a group itself. If each group is cyclic and the greatest common divisor of the orders of every two of the groups is one, then the product group is cyclic. So, the product group is of the order 7*5=35 and as it is cyclic, it is isomorphic to Z35.