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We have that $|G|=66=2 \cdot 3 \cdot 11$, so we have 2,3,11-Sylow. The number of 11-Sylow $n_{11}$ is such that $n_{11} \equiv 1 \ \ (11)$ and $n_{11} \mid 6$, so we have that $n_{11}=1$, and this means that the only 11-Sylow is normal in $G$. So we can say that $\mathbb{Z}_{11}\mathbb{Z}_3<G$ (because the 11-Sylow, that is isomorphic to $\mathbb{Z}_{11}$, is normal). We have also that $\mathbb{Z}_{11} \cap \mathbb{Z}_3 =\{e\}$ for order reasons, and so we can say that $\mathbb{Z}_{11}\mathbb{Z}_3 \cong \mathbb{Z}_{11} \rtimes _{\varphi} \mathbb{Z}_3$ with $\varphi :\mathbb{Z}_3 \rightarrow \text{Aut} (\mathbb{Z}_{11}) \cong \mathbb{Z}_{10}$. There's only one possible homomorphism $\varphi$, that is the one such that $\varphi ([1]_3)=[0]_{10}$, and so we have that the only possible semidirect product is $\mathbb{Z}_{11} \times \mathbb{Z}_3 \cong \mathbb{Z}_{33}$. In other words, we have that $\mathbb{Z}_{33}<G$, and, because $[G:\mathbb{Z}_{33}]=2$, we can say that $\mathbb{Z}_{33}\triangleleft G$. Finally, we have that $G=\mathbb{Z}_{33}\mathbb{Z}_2$ because $\mathbb{Z}_{33} \cap \mathbb{Z}_2=\{e\}$, and so we have that $G \cong \mathbb{Z}_{33} \rtimes _\psi \mathbb{Z}_2$ with $\psi : \mathbb{Z}_2 \rightarrow \text{Aut} (\mathbb{Z}_{33}) \cong \mathbb{Z}_{20}$ homomorphism. So we end up with two homomorphisms that bring us to $\mathbb{Z}_{66}$ and $D_{33}$, but there are other two groups of order 66. What's wrong with my proof?

Luigi Traino
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    The automorphism group of $Z_{33}$ isn't cyclic. – Angina Seng Jul 01 '20 at 10:55
  • Why not? Don't we have that Aut$(\mathbb{Z}_m) \cong \mathbb{Z}_m^*$ for every natural $m$? – Luigi Traino Jul 01 '20 at 10:58
  • One thing is that you missed $S_3$ as a possible cofactor for $Z_{11}$. No idea, what led to the miss of the product of $Z_3$ and $D_{11}$ – Peter Jul 01 '20 at 11:07
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    $\mathbb{Z}m^*$ is cyclic if and only if $m= 1, 2, 4, p^k$ or $2p^k$, where $p$ is an odd prime and $k \gt 0$. $Aut(\mathbb{Z}{33}) \cong V_4 \times \mathbb{Z}_5$. – Nicky Hekster Jul 01 '20 at 11:10
  • Here it doesn't pose any conditions on $n$: https://math.stackexchange.com/questions/3013033/automorphisms-of-a-cyclic-group/3013085 – Luigi Traino Jul 01 '20 at 11:39
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    Correct, but it only discusses the fact that the automorphism group is isomorphic to the group of units of the ring $\mathbb{Z}_m$. That does not say anything about the internal structure of this group. – Nicky Hekster Jul 01 '20 at 12:22

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As you say, it's a semidirect product of $Z_2$ and $Z_{33}$. But there are other ways for $Z_2$ to act on $Z_{33}$. If $a$ and $b$ are generators of $Z_{33}$ and $Z_2$ then $bab^{-1}=a^r$ where $r^2\equiv1\pmod{33}$. But there are four possible $r$ modulo $33$ solving this, not just $\pm1$, there are $\pm10$ also. This gives two other groups, which are $Z_3\times D_{11}$ and $Z_{11}\times D_3$.

Angina Seng
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