We were told to give a proof for the following statement:
Let G be a group, where $ r \in G $, and suppose $ r \in G \setminus \langle s \rangle $. If $ \langle r \rangle \cap \langle s \rangle = \{e\}$ and $ |r|=k $, then $ \langle s \rangle $, $ r\langle s \rangle $, $r^2\langle s \rangle$,...,$r^{k-1}\langle s \rangle$ are distinct left cosets.
I was wondering if my proof below is a proper proof, and if there is perhaps a more simple, intuitive proof of this fact. Thanks.
$ Proof $. Let $ i=1,2,...k $, and suppose towards a contradiction that $ r^{i-1}\langle s \rangle = r^i\langle s \rangle$ for some $ i $. This implies that $ r^{i-1}s^t = r^is^u $, for some integers $ u $ and $ t $. This in turn implies that $ r^{-1} = s^{u-t} $. This is a contradiction as $ \{r,r^2,...,r^{k-1} \} \notin \langle s \rangle$, from $ \langle r \rangle \cap \langle s \rangle = \{e\}$. Therefore $ r^{i-1}\langle s \rangle \neq r^i\langle s \rangle$, or rather $ \langle s \rangle $, $ r\langle s \rangle $, $r^2\langle s \rangle$,...,$r^{k-1}\langle s \rangle$ are distinct left cosets.