Let $G$ be a group given by $$G \ =\ \langle \ x, \ y \mid x^4 =\ y^4=1, \ yx =\ x^2\ y^2 \rangle$$ I found $\ G/G'$ to be isomorphic to $C_4$. What can we say about the order of $\ x$ in $G$? The final answer is $4$. But $\ x^4 = 1$ in $G$ means that $x^4$ belong to the normal closure of $R$ where $R$ is the relations in $G$.
I can't see the connection. Can any one help?
On
Well, the normal closure is even bigger than the subgroup generated by the relation. The only thing that has to be shown is that the order is, in fact, 4 and not less. Since the order divides 4, this boils down to showing that $x^2$ is not in the normal closure. Let $N$ be this normal closure, and let $R \le N$ be the subgroup generated by the relations. By considering the powers of $x$ and $y$ occuring in any element of $R$, we find that $x^2 \notin R$. But the normal closure is precisely the union of $fRf^{-1}$ ($f \in F\langle x, y \rangle$, the free group over $x$ and $y$), and we see that $x^2$ is not in any of these sets. Thus, $x^2$ is nonzero in your group $G$.
If $x^n$ is a relator of $G$ then the only think you can say for certain is that $x$ has order dividing $n$.
For example, consider the following two presentations: $$ \langle a, b, c\mid a^3, b^4, c^{6}, a=b, b^2=c^2\rangle $$ and $$ \langle x_0, x_1, x_2, x_3, x_4\mid x_ix_{i+1}=x_{i+2}\rangle $$ (where subscripts are computed modulo $5$).
In the first group, we see that the elements $a$ and $b$ are the same and so must have order diving both $3$ and $4$. Hence, they have order $1$ and are trivial: $a=b=1$. Then $c^6=1$ and $c^2=b^2=1$ so $c^2=1$. It turns out that $c$ actually has order $2$.
For the second group, it was a famous "fun" problem of John Conway to work out what this group is. It "clearly" does not have any elements of finite order. However, it is actually cyclic of order $11$...
Group presentations can be tricky beasts :-)