Let $G$ be a group s.t. $|G|=44=2^211$. Using Sylow's Theorems, I have deduced that there is a unique Sylow $11$-subgroup of $G$; we shall call it $R$. Let $P$ be a Sylow $2$-subgroup of $G$. Then we have $G=P\rtimes R$ and a homomorphism
$$\gamma: P \rightarrow Aut(R)=Aut(\mathbb{Z_{11}})\cong(\mathbb{Z_{10}},+) .$$
Is this all correct so far?
So what about $\gamma(p)=\phi_p$ where $\phi_p(r)=r^5$. I thought this because $\tilde{5}\in\mathbb{Z_{10}}$ has order $4$ so the order of any element of $P$ could divide it... or something...
So I was thinking the group would be something like
$$G= \langle p,r | p^4=r^{11} prp^{-1}=r^5 \rangle .$$
Any insight is greatly appreciated! Thanks! I would like to know both where I went wrong and how to do it correctly.
Did I do the above right? Identifying $\mathbb{Z_{11}}$ with the additive group of $\mathbb{Z_{10}}$? Or should I look at it multiplicatively, because I don't understand how that isomorphism works so it doesn't make sense to define the conjugation that makes the semi-direct product well defined based on elements of the additive group $\mathbb{Z_{10}}$, but instead realize that $10 \in U(\mathbb{Z_{11}})$ has order $2$ so we can have a group presentation something like:
$G = \langle p, r | p^2=r^{11}=1 , prp^{-1}=r^{10} \rangle$
Insight appreciated!
I understand the dihedral group of the $22$-gon works now, thank you. Can somebody help me with my approach in constructing a non-abelian group of order $44$ via the methods I've been using? Thanks!