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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 ?

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  • $\begingroup$ Note that in $G/N$, $\bar{r^3}=\bar e $, $\bar{r^4}= \bar r$, etc. $\endgroup$ – J. W. Tanner Apr 22 at 16:42
  • $\begingroup$ How come? I would say $\bar{r^{3}} = r^{3}N=r^{3}r^{3}=r^{6}$ if one would take $r^{3}$ from $N$ for example... $\endgroup$ – Mathbeginner Apr 22 at 16:58
  • $\begingroup$ It is also true that $\bar {r^3}=\bar {r^6}$ in $G/N$ $\endgroup$ – J. W. Tanner Apr 22 at 16:59
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    $\begingroup$ $r^3N=N$: don't you see that $r^3\in N$ ? $\endgroup$ – J. W. Tanner Apr 22 at 17:03
  • $\begingroup$ $N$ is the identity element in the quotient group $\endgroup$ – J. W. Tanner Apr 22 at 17:05
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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 $G.$

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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?

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    $\begingroup$ Yes, that makes sense. Nicely put, I wasnt familiar with "killing" but the terminology makes sense. $\endgroup$ – Mathbeginner Apr 23 at 7:23
  • $\begingroup$ @Shaun: are you saying that $r^8=e$ in $G$? In OP's question, it appears to me that $r^8$ is distinct from $e$, so I thought $r^9=e$ $\endgroup$ – J. W. Tanner Apr 28 at 6:57
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    $\begingroup$ You're right, @J.W.Tanner; thank you! I've corrected it now. $\endgroup$ – Shaun Apr 28 at 8:22

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