Classify, up to isomorphism, all groups of order $5\cdot 7 \cdot 19$.

I know that the general process for these types of problems is to first apply the Fundamental Theorem of Finite Abelian Groups (FTFAG), then find some semi-direct products.

First, if we assume $$G$$ is abelian, then it is isomorphic to $$\mathbb{Z}_5 \times \mathbb{Z}_7 \times \mathbb{Z}_{19}$$ by FTFAG. There is no other abelian groups of the proper order.

I am stuck, now, on finding the appropriate semi-direct products to find the non-abelian groups of the proper order. By the Sylow counting theorems, I know that the Sylow-$$5$$, Sylow-$$7$$, and Sylow-$$19$$ subgroups are all normal in $$G$$, but I can't seem to come up with a proper homomorphism to construct a semidirect product.

In addition to this specific problem, I would appreciate any help when it comes to constructing groups of a certain order as a semidirect product in general.

• GAP might help. NB: I haven't checked, as my laptop is elsewhere. Jan 6, 2023 at 23:03
• If all Sylow subgroups are normal, then the group is a direct product of them. In this case, it would necessarily be abelian because all Sylow subgroups are abelian. Jan 6, 2023 at 23:15
• that's a useful fact to keep in mind. I forgot that all normal Sylow subgroups implies the group is abelian. Thanks! Jan 6, 2023 at 23:27
• The number is an abelian number (all groups with this order are abelian) because it satisfies the conditions for an abelian number and moreover squarefree. That implies that there is only one group with this order, the cyclic group. Jan 6, 2023 at 23:42
• "I forgot that all normal Sylow subgroups implies the group is abelian". No! The conclusion that the group is abelian here is also because the Sylow subgroups are abelian. All Sylow subgroups of the dihedral group of order 8 are normal, but the group is not abelian. Jan 7, 2023 at 1:14

If the Sylow subgroups are all normal, it follows that $$G\cong C_5×C_7×C_{19}$$.
The Sylow subgroups in this case are all cyclic, hence abelian. Thus $$G$$ is abelian. (More can actually be said: by the Chinese remainder theorem, $$G$$ is cyclic.)