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In my group theory course, I'm trying to prove this statement. I don't know how to approach this kind of question, can someone help?

How do I prove that every group of order 391 is cyclic?

Math420
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2 Answers2

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You need the following (all groups finite here). Proofs (see also links by clicking on the light blue facts) can be found here on MathStackExchange or in any decent group theory textbook.

Fact 1 If $H$ is a subgroup of a group $G$ with $|G:H|$ being the smallest prime dividing the order of $G$, then $H$ is normal.

Fact 2 If $H$ is a subgroup of $G$, then $N_G(H)/C_G(H)$ embeds homomorphically into $Aut(H)$.

Now, by Cauchy you can find a subgroup $N$ of order $23$. By Fact 1, $N$ is normal. And by Fact 2, $N \subseteq Z(G)$ (note that $G=N_G(N)$, $Aut(N) \cong C_{22}$ and $gcd(22,391)=1$).

Again by Cauchy, there exists a subgroup $H$ of order 17. By Lagrange, $H \cap N=1$. Observe that $|HN|=\frac{|H||N|}{|H \cap N|}$. Hence $G=HN$ and since $N$ is central, $H$ is normal. This amounts to $G = H \times N \cong C_{391}$.

Note In general, it can be shown that if $gcd(n,\varphi(n))=1$, then a group of order $n$ is cyclic. Here $\varphi$ is Euler's totient function.

Nicky Hekster
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$391=17*23$. Apply Sylow theorems.

markvs
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  • This is the first route I thought of. The Sylow theorems tell you about subgroups. The information about subgroups needs to be leveraged into information about the whole group (a normal use of Sylow). In this case, this involves only basic results. – Mark Bennet Oct 07 '20 at 15:38
  • Sylow theorems are basic results. They imply that the group is the direct product of two cyclic groups of orders 17 and 23. Hence it is cyclic. – markvs Oct 07 '20 at 15:49