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.
 A: $35 =7\cdot5$ , 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)
A: The fact that $x,y$ generate $G$ and $xyx^{-1}y^{-1}=1$--equivalently, $xy=yx$--means that $G$ is an abelian group. (Do you see why?) Thus, since $x^7=1$ and $y^5=1,$ then every element of $G$ can be written uniquely as $x^ky^m$ for some $k\in\{0,1,2,3,4,5,6\}$ and some $m\in\{0,1,2,3,4\}.$ (Can you prove this?) From this, we see that $G$ has $35$ elements. (Can you see why?) All that is left is to find some element of $G$ that has order $35.$ Fortunately, most of them do, so you shouldn't have to look for too long to find one.

Another approach we can take is to try to find some elements $a,b$ of $\Bbb Z_{35}$ such that the order of $a$ is $7,$ the order of $b$ is $5,$ and such that $a$ and $b$ commute with each other (simple, since $\Bbb Z_{35}$ is cyclic, and so abelian). In other words, $a$ and $b$ formally satisfy all the relations that $x$ and $y$ do. The idea is to show that $\Bbb Z_{35}$ is generated by $a$ and $b,$ and then let $\phi:G\to\Bbb Z_{35}$ be the unique homomorphism that sends $x\mapsto a$ and $y\mapsto b.$ (Do you see why such a homomorphism exists and is unique?) Note that $\phi$ is surjective. (Do you see why?) Finally, we prove that $\phi$ is injective, and so is an isomorphism, and we're done.

Yet another approach! Let $H$ be the subgroup of $G$ generated by $x$--that is, let $H=\langle x\rangle$--and let $K=\langle y\rangle.$ Show that the identity of $G$ is the only element of $H\cap K,$ that $H$ and $K$ are normal subgroups of $G,$ and that the product set $HK:=\{hk:h\in H,k\in K\}$ is all of $G.$ Next, we use this information to show that the map $G\to H\times K$ given by $hk\mapsto (h,k)$ is well-defined, and in fact is an isomorphism, so that $G\cong H\times K.$ Next, we can clearly see that $H\cong\Bbb Z_7$ and $K\cong\Bbb Z_5,$ so $G\cong\Bbb Z_7\times\Bbb Z_5.$ Finally, we use the theorem that says $\Bbb Z_m\times\Bbb Z_n\cong\Bbb Z_{mn}$ if and only if $\gcd(m,n)=1,$ so that $G\cong\Bbb Z_{35}.$
A: $x^7=1$ 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.
