I want to find the order of a group given by the presentation $\langle \{a\} \mid \{a^r,a^s\}\rangle $ where $r$ and $s$ are coprime. My reasoning is as follows: since $a^r=1$ and $a^s=1$, the order will have to divide both $r$ and $s$. Then since $(r,s)=1$ it has to be $1$, right? Is this correct?
A follow-up question: if we drop the $(r,s)=1$ condition, would the order be the greatest common divisor of $r$ and $s$?
Yes and yes.
Let $d=(r,s)$. Then by Bézout's Identity there exist $x,y\in\Bbb Z$ such that
$$xr+ys=d.$$
Therefore,
$$\begin{align} a^d&=a^{xr+ys}\\ &=(a^r)^x(a^s)^y\\ &=1^x1^y\\ &=1. \end{align}$$
Also $d\mid r$ and $d\mid s$, so there exist $n,m\in\Bbb Z$ such that $r=dm, s=dn$.
Let $b=a^d$. Then by Tietze transformations,
$$\begin{align} \langle a\mid a^r, a^s\rangle &\cong \langle a,b\mid a^r, a^s, b=a^d=1\rangle\\ &\cong \langle a,b\mid b^m, b^n, b=a^d=1\rangle\\ &\cong \langle a\mid 1^m,1^n, a^d=1\rangle\\ &\cong\langle a\mid a^d\rangle, \end{align}$$
which is cyclic of order $d=(r,s)$.