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?
Best Answer
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 $.