# Obtaining a presentation of the dihedral group from a semidirect product

I am working on classifying groups of order 44. I have shown that $G\cong P_{11} \rtimes_{\varphi} P_{2}$, where $P_p$ are Sylow p-subgroups and $\varphi:P_{2} \to$Aut$(P_{11})$ is a group homomorphism.

I am considering the case when $P_{2}\cong \mathbb{Z}/2 \mathbb{Z} \oplus \mathbb{Z}/2 \mathbb{Z}=\langle r,s \rangle$ and $|\varphi(r)|=|\varphi(s)|=2$. If we let $\mathbb{Z}/11\mathbb{Z}=\langle a \rangle$, then $r \cdot a =a^{-1}$ and $s\cdot a=a^{-1}$. Am I correct in saying that this gives a presentation $$G=\langle a,r,s \mid r\cdot a=a^{-1}. s\cdot a=a^{-1}, a^{11}=r^{2}=s^{2}=e \rangle?$$

Either way, I know that this semidirect product is isomorphic to the dihedral group $D_{44}$ and I want to see why. Can someone show me how to get the correct presentation?

• See number $3$ of this test solution. Jun 5 '15 at 19:30
• @DietrichBurde I have seen this. Thank you for linking me to how to classify groups of order 44, but this does not directly answer my question. Jun 5 '15 at 19:32
• Your displayed presentation of $G$ is garbled. The relations you want are $rar^{-1}=a^{-1}$ and $sas^{-1}=a^{-1}$. Jun 5 '15 at 20:43
• @DerekHolt By $ra$ and $sa$, I meant $r \cdot a$ and $s \cdot a$, the action of $r$ and $s$ on $a$. I just fixed it. Do they act by conjugation as you've written? If so, this answers my question. Jun 5 '15 at 20:46
• As Brillo points out, you are missing a relation $(rs)^2=1$. But you shouldn't invent your own notation for group presentations. As I said before you should write $rar^{-1}=a^{-1}$ not $r \cdot a = a^{-1}$ which doesn't mean anything. Jun 5 '15 at 21:46

Sticking with your notation, I claim $$|G|=|\langle a,r,s \mid ra=a^{-1}. sa=a^{-1}, a^{11}=r^{2}=s^{2}=e \rangle| =\infty$$ Why?
The element $$g=r\cdot s$$ has infinite order. From the relations given its easy to see $$(r\cdot s)^2 \neq (r\cdot s)^3 \neq (r\cdot s)^4 \cdots$$
I think the presentation you want is $$H=\langle r,a,s\ |\ rar^{-1}=a^{-1}, \ sas^{-1}=a^{-1}, \ rsr^{-1}=s,\ a^{11}=s^2=r^2=e\rangle.$$ With this presentation, it is not hard to show any element in $H$ can be written uniquely as $$h=r^is^ja^k \ \ \ \text{where }i,j\in{0,1} \text{ and } 0 \leq k \leq 10.$$