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!
Best Answer
No element of $\mathbb Z_{10}$ has order four (why not?) and there is one element of order 2 ($5\in\mathbb Z_{10}$ under addition, $10\cong -1\in \mathbb Z_{11}^x$ under multiplication.), so our possibilities are quite limited. There are two groups of order $4$, and either will work as our $R$. To give a nonabelian group, we need to pick a nontrivial homomorphism, as you pointed out.
So at least one generator of $R$ has to map to our order-two element. In the case of the Klein four group, there appear to be three possibilities, but I claim that up to an isomorphism of the Klein four group, there is only one possibility.
Actually, this exhausts the possibilities for groups of order 44: we have two abelian groups, $\mathbb Z_{11}\times\mathbb{Z}_4$, $\mathbb Z_{11}\times \mathbb{Z}_2\times \mathbb Z_2$, and two nonabelian groups: $\mathbb{Z}_{11} \rtimes \mathbb Z_4 = \langle a,b \mid a^{11}, b^4, b^{-1}ab = a^{-1}\rangle$ and $\mathbb Z_{11}\rtimes(\mathbb Z_2 \times \mathbb Z_2) = \langle a, b, c \mid a^{11}, b^2, c^2, [b,c], b^{-1}ab = c^{-1}ac = a^{-1} \rangle$. I think the latter is $D_{22}$ (for 22-gon, not order of group), while the former has an element of order $4$.