# Quotient group of dihedral group

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 ?

• Note that in $G/N$, $\bar{r^3}=\bar e$, $\bar{r^4}= \bar r$, etc. – J. W. Tanner Apr 22 '19 at 16:42
• 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... – Mathbeginner Apr 22 '19 at 16:58
• It is also true that $\bar {r^3}=\bar {r^6}$ in $G/N$ – J. W. Tanner Apr 22 '19 at 16:59
• $r^3N=N$: don't you see that $r^3\in N$ ? – J. W. Tanner Apr 22 '19 at 17:03
• $N$ is the identity element in the quotient group – J. W. Tanner Apr 22 '19 at 17:05

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 $$G.$$

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?

• Yes, that makes sense. Nicely put, I wasnt familiar with "killing" but the terminology makes sense. – Mathbeginner Apr 23 '19 at 7:23
• @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 $r^9=e$ – J. W. Tanner Apr 28 '19 at 6:57
• You're right, @J.W.Tanner; thank you! I've corrected it now. – Shaun Apr 28 '19 at 8:22