Let $G=\{e,r^{2},...,r^{8},s,sr,...,sr^{8}\}$ and let $N=\langle r^{3} \rangle.$ Now let $\pi(g)=\bar{g}=gN$ be surjective with kernel $N$.
I have to show that $G/N=\{\bar{e},\bar{r},\bar{r^{2}},\bar{s},\bar{sr},\bar{sr^{2}} \}$ and I also need to compute a group table.
Now as far as I know, $G/N$ is the quotient group, which means that $G/N$ is a homomorphism with, for all $u,v \in G$, $uNvN=uvN$. But I have no idea how to proceed. Any suggestions ?
On
This is known as "killing $r^3$," since by quotienting out by $N$ you are changing the presentation
$$\langle r, s\mid r^9, s^2, srs^{-1}=r^{-1}\rangle$$
to
$$\langle r, s\mid r^9=r^3=1, s^2, srs^{-1}=r^{-1}\rangle,$$
but then $r^9=(r^3)^3=1$, which gives the presentation
$$\langle r, s\mid r^3, s^2, srs^{-1}=r^{-1}\rangle,$$
whose elements of the group it defines are exactly the ones you need. Can you check that yourself?
Note that $\bar e =\bar {r^3}=\bar{r^6}, \bar r=\bar {r^4}=\bar{r^7}, \bar {r^2}=\bar{r^5}=\bar{r^8},$ and these equations can be left-multiplied by $\bar s.$
Thus $G/N=\{\bar{e},\bar{r},\bar{r^{2}},\bar{s},\bar{sr},\bar{sr^{2}} \}.$
$\bar e$ is the identity.
$\bar r \bar r=\bar {r^2}; \quad\bar {r^2} \bar {r^2} = \bar {r^4} = \bar r;\quad \bar r \bar {r^2}= \bar {r^2} \bar r = \bar e.$
You should be able to multiply $\bar r$ and $\bar {r^2}$ by $\bar s$, $\bar {sr}$, and $\bar {sr^2}$ and vice versa using these principles and knowledge of multiplication in $G.$