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Let $G$ be a group of order $7.11.17$. Show that $G$ is cyclic.

I tried to find a solution using Sylow theorems but I got stuck, here it goes:

We know that $$n_7 \equiv 1 (7) \space, n_7|11.17 \implies n_7=1,$$$$n_{11} \equiv 1 (11) \space, n_{11}|7.17 \implies n_{11}=1,$$$$n_{17} \equiv 1 (17) \space, n_{17}|7.11 \implies n_{17}=1$$

If we call $H,M,N$ to the unique 7-Sylow, 11-Sylow and 17-Sylow subgroups respectively, then $H,M,N \lhd G$. We have $H \cong \mathbb Z_7,M \cong \mathbb Z_{11}$ and $N \cong \mathbb Z_{17}$. Each of these groups is cyclic so $Z_7 \times Z_{11} \times Z_{17}$ is cyclic as well. I don't know how could I conclude from here that $G$ is cyclic, I've tried with semidirect products but I didn't arrive to anything concrete. Any suggestions would be appreciated.

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Since $H$, $M$ and $N$ are the unique Sylow subgroups (of their respective orders) of $G$, then they are normal, so in particular $MN$ is normal. Then $G$ is the semidirect product $H\rtimes_f MN$ under the action of conjugation. But there are only $6$ automorphisms of $H$ (i.e. $\operatorname{Aut}(H)$ has order $6$), and $MN$ has order $11\cdot 17$, which is coprime with $6$, so the only homomorphism $MN\to\operatorname{Aut}(H)$ is the trivial one, that is $G=H\times MN$. Similarly, $MN$ is a semidirect product $M\rtimes_g N$ under some action $g$ which has to be trivial since the order of $\operatorname{Aut}(M)$ is 16 and is coprime with the order of $N$, which is 11. Therefore $G\simeq H\times M\times N$.

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For example, you want to know if elements of $H$ and $M$ commute. $M$ acts on $H$ by conjugation. Especially, if $m$ generates $M$, then conjugatuon with $m$ is an automorphism of $H$ of order dividing $|M|$, hence is either the identity or of order $11$. But $H$ has no automorphism of order $11$ (for example, the automorphism group of $H$ is a subgroup of the symmetric group $S_{|H|}$). Hence $H$ and $M$ commute. By the same considerations, $M$ and $N$ as well as $H$ and $N$ commute. We conclude that their product is direct.

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There is only one group of order $n$ (and hence $\cong C_n$) if and only if gcd$(n,\varphi(n))=1$.

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With your efforts, you proved every Sylow subgroup is normal, (well known basic theory of sylow subgroups says they are all conjugates and there is one of them, so it's normal). Thus you can conclude it's isomorphic to the direct product of it's Sylow subgroups, that is, $G \simeq P_7 \times P_{11} \times P_{17}$. You can end the proof either by proving that $(1,1,1)$ generates $G$, which happens since it's order it's the least common multiple of $7,11,17$ which is their product, or use this, using it first with 7 and 11 because they are coprime, and then with $7*11$ with $17$.

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Well, certainly

$$\begin{cases}G=HMN\\{}\\|G|=|H||M||N|\end{cases}\stackrel{\text{why? Care here!}}\implies G\cong H\times M\times N$$

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