<s>[INST] Order of group from its presentation Say we want to determine the order of a group generated by $x$ and $y$ who satisfy $x^2y = xy^3 = 1$.
Ok so it would be nice to know the order of $x$ and $y$ respectively. We can readily conclude from the equations above that $x^2 = y^{-1}$ and $x^{-1} = y^3$. 
Next we can derive the relationship $x = y^2$ and plugging this into the given equations we get $y^5 = 1$, hence $y$ is of order $5$. 
Now since $x = y^2 = y \cdot y$, we have $y^{-1} x = y$ and we knew $y^{-1} = x^2$ so we get $x^3 = y$. But since $y^5 = 1$ we get $x^{15} = 1$. Hence the $x$ is of order $15$.
$x$ and $y$ generate the group so every element can be written as $x^a y^b$ where $a$ and $b$ range from 0 to 14 and 0 to 5 respectively. But the group is not necessarily abelian, so are we not missing out elements of the form $y^kx^j$?
Not sure how to conclude the argument, help much appreciated!
 [/INST] $x^2 = y^{-1}, x = y^{-3} \Rightarrow x^6 = y^{-3} = x \Rightarrow x^5 = 1 \Rightarrow |x| = 5 \Rightarrow |x^2| = 5 \Rightarrow |y^{-1}| = 5 \Rightarrow |y| = 5.$ Now $y \in <x> \Rightarrow$ the group is the cyclic group of order $5$.
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