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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!

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    $\begingroup$ I think you meant $r^{11}$ (r^{11}), not $r^11$ (r^11) $\endgroup$ – J. W. Tanner Apr 22 at 23:48
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    $\begingroup$ I've taken the liberty of apply the correction J.W. Tanner mentioned, as well as a few other minor fixes. $\endgroup$ – Travis Apr 23 at 1:23
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    $\begingroup$ I don’t understand the words, “because $\tilde5\in\Bbb Z_{10}$ has order $4$” $\endgroup$ – Lubin Apr 23 at 1:31
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    $\begingroup$ Doesn't $\tilde{5} \in \mathbb{Z}_{10}$ have order $2$? $\endgroup$ – Peter Shor Apr 23 at 1:49
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    $\begingroup$ And if you have a non-abelian group of order 22, isn't it easy to find one of order 44? $\endgroup$ – Peter Shor Apr 23 at 1:52
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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 $4$.

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  • $\begingroup$ is the automorphism between $U\mathbb{Z_{11}} \cong (\mathbb{Z_{10}},+)$ obvious? It would be nice to know about the elements of $U(\mathbb{Z_{11}})$ from knowing what they are in $(\mathbb{Z_{10}},+)$ $\endgroup$ – Mathematical Mushroom Apr 23 at 12:28
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    $\begingroup$ So since no matter what order of group four we have to pick, they will both have at least one element of order two, so we need to map them to elements of $Aut(Z_{11}}$ of order two, and there is only one. $\endgroup$ – Mathematical Mushroom Apr 23 at 12:31
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    $\begingroup$ You can think of an element of $\operatorname{Aut}(\mathbb Z_n)$ as determined by the image of $1$. For example, consider the map $1 \mapsto 2$. To make this a homomorphism, every other element of $\mathbb Z_{11}$ must be multiplied by $2$. Let's find out the order of this automorphism by listing the images of $1$: $2, 4, 8, 16\equiv 5, 10\equiv -1, -2, -4, -8, -5, -10\equiv 1$. You can check that every unit (generator) in $\mathbb Z_{11}$ is listed, so we have shown that $\operatorname{Aut}(\mathbb Z_{11})$ is cyclic of order $10$, generated by the automorphism $1\mapsto 2$. $\endgroup$ – Rylee Lyman Apr 23 at 12:51
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    $\begingroup$ So, while this setup generalizes to arbitrary finite cyclic groups, I wouldn't say that the isomorphism is "obvious" except once you know how to compute it in this way. $\endgroup$ – Rylee Lyman Apr 23 at 12:52
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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 $D_{44} \cong D_{22} \times \Bbb Z_2$.

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