Mrthfx

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 $G=Prtimes R$ 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 $tilde{5}inmathbb{Z_{10}}$ 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!

No element of $mathbb Z_{10}$ has order four (why not?) and there is one element of order 2 ($5inmathbb Z_{10}$ under addition, $10cong -1in mathbb Z_{11}^x$ 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, $mathbb Z_{11}timesmathbb{Z}_4$, $mathbb Z_{11}times mathbb{Z}_2times mathbb Z_2$, 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 $4$.

You're on the right track (but NB your semidirect product is written in the wrong order). To analyze the possible maps $gamma : P to operatorname{Aut}(Bbb Z_{11}) cong (Bbb Z_{10}, +)$, we consider separately the cases $P cong Bbb Z_2 times Bbb Z_2$ and $P cong Bbb Z_4$.

In the case $P cong Bbb Z_4$, $P$ is generated by a single element, $[1]$, of order $4$, and so $$operatorname{id}_{Bbb Z_{11}} = gamma([0]) = gamma([1] + [1] + [1] + [1]) = gamma([1])^4 .$$ So, $gamma([1])$ has order dividing $4$, and the only such elements of $operatorname{Aut}(Bbb Z_{11}) cong (Bbb Z_{10}, +)$ are $operatorname{id}_{Bbb Z_{11}} leftrightarrow [0]$ and $(x mapsto -x) leftrightarrow [5]$.

• If $gamma([1]) = operatorname{id}_{Bbb Z_{11}}$, then $gamma$ is the trivial homomorphism $Bbb Z_4 to operatorname{Aut}(Bbb Z_{11})$. This gives the direct product $G cong Bbb Z_{11} times Bbb Z_4 cong Bbb Z_{44} .$




• If $gamma([1]) = (x mapsto -x)$, then $gamma([b])([c]) = (-1)^b [c]$, and the semidirect product $G = Bbb Z_{11} rtimes_{gamma} Bbb Z_4$ is defined by
  $$([a], [b]) cdot ([c], [d]) = ([a] + (-1)^b [c], [b] + [d]) .$$ It's apparent from the multiplication rule that this group is nonabelian. The fact that the group is generated by $u := ([1], [0])$ and $v := ([0], [1])$ can be used to construct an explicit presentation of this group and to show that $G$ is the dicyclic group of order $44$.

One can analyze the case $P cong Bbb Z_2 times Bbb Z_2$ similarly, and this case gives rise to two more groups up to isomorphism, namely the abelian group $Bbb Z_{11} times Bbb Z_2 times Bbb Z_2$ and the (nonabelian) dihedral group $D_{44} cong D_{22} times Bbb Z_2$.