finding the order of an element in a group presentation 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?
 A: The relation $x^4=1$ guarantees that the order of $x$ in $G$ is at most $4$. The fact that $xG'$ generates $G/G'\cong C_4$ guarantees that the order of $x$ in $G$ is at least $4$.
A: 
What can we say about the order of $x$ in $G$?

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 :-)
A: 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 $G$.
