0

What steps do I take in approaching this question? First 39=3 x 13, both of which are prime. 3 divides 13-1. I'm not too sure where to go from here. I know the Sylow Theorems and the Theorem of Finite Abelian Groups are relevant.

I realize there's another similar question here, but it asks only about the presentation of the nonabelian group.

How do I know that there are only 1 abelian and 1 nonabelian group?

jacksonf
  • 526

1 Answers1

3

It follows from the Sylow theorems that every such group $G$ has a normal subgroup $N$ of order $13$. Then $H:=G/N$ is a group of order $3$ and $G$ is a semidirect product of $N$ and $H$. How many such semidirect products are there?

Servaes
  • 63,261
  • 7
  • 75
  • 163
  • I can see from the other post that there is 1 abelian and 1 nonabelian group.... how is this determined? – jacksonf Dec 01 '17 at 17:34
  • A semidirect product $N\rtimes H$ is determined by a group homomorphism $H\ \longrightarrow\ \operatorname{Aut}N$. In this case there are three homomorphisms. The trivial map gives the abelian group, and the other two maps give isomorphic groups. – Servaes Dec 01 '17 at 20:44