I need help in proving that every group of order $5865$ is cyclic.

I thought at the beginning that if I show that every such group is simple and abelian, It will be isomorphic to some $\mathbb{Z}/p\mathbb{Z}$ where $p$ is some prime.

The problem is, $5865 = 3*5*17*23$, and $n_{23}=1$, where $n_{23}$ is the number of $23$-sylow subgroups in $G$.

Therefore, there is only one $23$-sylow subgroup, and it is normal, and $G$ is not simple.

Could you give me some hints on other ways to prove that $G$ is cyclic?

  • 2
    $\begingroup$ $C_{5865}$ for one example, is definitely not simple. $\endgroup$ – Henning Makholm Jun 23 '18 at 18:09
  • 1
    $\begingroup$ You can't show that G is isomorphic to some Z/pZ, because the order of Z/pZ is p. And 5865 is not prime $\endgroup$ – Tom Jun 23 '18 at 18:14

Here's an idea, if you can show that the 3, 5, 17, and 23 Sylow subgroups are all normal, then $ G $ would be the direct product of these Sylow subgroups. As each of these Sylow subgroups have prime order, they are cyclic. From here, use the Chinese Remainder Theorem to conclude that $ G $ is cyclic.

Edit: I'll explain why $ G $ is the direct product of its Sylow subgroups if they are all normal.

Recall the criterion that $ G \cong H \times K $ if $ H, K \trianglelefteq G $, $ H \cap K = \{e\} $, and $ G = HK $. Now suppose the distinct Sylow subgroups of $ G $ are $ P_1, \ldots, P_n $ and that they are all normal subgroups. $ P_i \cap P_1 \ldots P_{i-1} P_{i+1} \ldots P_n = \{e\} $ by looking at the order of the elements in these subgroups. Then $ |P_1 \ldots P_n| = |G| $, so $ P_1 \ldots P_n = G $. We also have, by the criterion, $ P_1 P_2 \cong P_1 \times P_2 $. By a similar cardinality/order argument as above, we see $ P_1 P_2 P_3 \cong (P_1 P_2) \times P_3 \cong P_1 \times P_2 \times P_3 $. Then by induction, $ G = P_1 \ldots P_n \cong P_1 \times \ldots P_n $.

  • $\begingroup$ Could you explain why if " the $3, 5, 17,$ and $23$ Sylow subgroups are all normal, then G would be the direct product of these Sylow subgroups. "? $\endgroup$ – ChikChak Jun 23 '18 at 18:21
  • $\begingroup$ I will edit that into the post. $\endgroup$ – paul blart math cop Jun 23 '18 at 18:33

There are numbers $n$ such that every group of order $n$ is cyclic. These are called cyclic numbers and are characterized by $\gcd(n,\phi(n))=1$, where $\phi$ is Euler's function. See also this question.

For $n=5865$, we have $5865 = 3 \cdot 5 \cdot 17 \cdot 23$ and $\phi(5865)=2816= 2 \cdot 4 \cdot 16 \cdot 22 = 2^8 \cdot 11$.

Therefore $\gcd(5865,\phi(5865))=1$ and every group of order $5865$ is cyclic.

  • $\begingroup$ Good job finding that general result. $\endgroup$ – Jyrki Lahtonen Jun 23 '18 at 20:03
  • $\begingroup$ @JyrkiLahtonen, I learned about this result here in MSE and have used it in several answers. $\endgroup$ – lhf Jun 23 '18 at 20:50

You started out well by observing that the group $G$ has a normal cyclic subgroup $N$ of order $23$. One way of making progress is to use the known structure of the group of automorphisms of a cyclic group of prime order: $Aut(\Bbb{Z}_p)=\Bbb{Z}_p^*\simeq C_{p-1}$.

So the automorphism group of $N$ is cyclic of order $22$. As $22$ is coprime with the order of $G$ we can conclude that the conjugation action of $G$ on $N$ must be trivial. In other words $N\le Z(G)$ is contained in the center.

Consider the quotient group $G/N$ of order $3\cdot5\cdot17$. Sylow theory tells us that the subgroup $M$ of order $17$ is unique, hence normal. Again, $Aut(M)$ has order $16$. Because $\gcd(16,3\cdot5\cdot17)=1$ we, again see that $M\le Z(G/N)$.

It is a standard exercise to show that a group of order $15$ is necessarily cyclic. Hence $(G/N)/M$ is cyclic.

Then you need to recall the fact that if $H$ is any group with such a subgroup $K$ that i) $K\le Z(H)$, and ii) $H/K$ is cyclic, then we can conclude that $H$ must be abelian.

Apply this fact. Twice. You know that $G$ is abelian. By structure theory of finite abelian groups an abelian group of square free order is cyclic.

  • $\begingroup$ After you have shown that $G$ is abelian you can also follow the route in user 571438's answer (+1) to conclude that $G$ is cyclic. $\endgroup$ – Jyrki Lahtonen Jun 23 '18 at 19:07

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