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What steps do I take in approaching this question? First $39=3\cdot13$, 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?

Kan't
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jacksonf
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2 Answers2

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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
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  • 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
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To explicitly build up a (nontrivial) semidirect product $\mathbb Z_3\stackrel{\varphi}{\ltimes}\mathbb Z_{13}$, we must send $0$ to the identity map of $\mathbb Z_{13}$ ($Id_{\mathbb Z_{13}}$), and $1$ and $2$ to two distinct automorphisms of $\mathbb Z_{13}$ of order $3$$^\dagger$ such that $Id_{\mathbb Z_{13}}(=$ $\varphi_0=$ $\varphi_{1+2})=$ $\varphi_1\varphi_2$. Since $\operatorname{Aut}(\mathbb Z_{13})$ acts regularly (=transitively and freely) on the set $X$ of the generators of $\mathbb Z_{13}$, the only option is: \begin{alignat}{1} &{\varphi_1}_{|X} = (1,3,9)(2,6,5)(4,12,10)(7,8,11) \\ &{\varphi_2}_{|X} = (1,9,3)(2,5,6)(4,10,12)(7,11,8) \\ \end{alignat}

$^\dagger$Under a homomorphism, the order of the image of an element divides the order of the element.

Kan't
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