Abelian semidirect products

Algebra Level 2

Let $N$ and $H$ be groups , let $\phi : H \to \text{Aut}(N)$ be a homomorphism , and let $G = N \rtimes_\phi H$ be the (outer) semidirect product . Suppose that $G$ is abelian . Which of the following three statements must necessarily be true about $H,N,\phi$ ?

I. $H$ and $N$ are abelian

II. $\phi$ is the trivial homomorphism

III. $G$ is actually a direct product of groups

I and II I and III II and III I,II,III

1 solution

Patrick Corn
Oct 7, 2016

All three of the statements must be true.

Since any subgroup of an abelian group is abelian, and $H$ and $N$ are (isomorphic to) subgroups of $G,$ they must also be abelian.

Take two elements $(n,1)$ and $(1,h).$ Then $(n,1) \cdot (1,h) = (n,h)$ but $(1,h) \cdot (n,1) = (\phi_h(n),h).$ Since $G$ is abelian, $n = \phi_h(n)$ for all $n \in N,$ so $\phi_h$ is the trivial automorphism of $N.$ This is true for all $h \in H,$ so $\phi$ is the trivial homomorphism.

When $\phi$ is the trivial homomorphism, the semidirect product is actually a direct product of groups. (See the semidirect product wiki for an explanation.)

Of course, this is both necessary and sufficient. A semidirect product is abelian if and only if both groups are abelian and the action is trivial.

Mark Hennings - 4 years, 8 months ago

I do not think homomorphism is trivial. By definition, a homomorphism is trivial if and only if it takes every input to the identity element. Hence, we do not have trivial homomorphism. Actually, we have trivial action of group H on group N but this is not the same as a trivial homomorphism.

Dejan Stančević - 2 months, 3 weeks ago

This is incorrect; the problem is right. I made a more detailed comment about this in the report you filed.

Patrick Corn - 2 months, 3 weeks ago

Sorry for reporting it. I was unsure whether someone would look at the comment (to the problem that is more than 4 years old). Thank you for the explanation.

Dejan Stančević - 2 months, 3 weeks ago

Since $G$ is abelian, every element of $H$ maps to the identity automorphism of $N$ and hence to the identity element of $\mathrm{Aut}(N)$ . Thus the homomorphism $\phi \,:\, H \to \mathrm{Aut}(N)$ does map every element of $H$ to the identity of $\mathrm{Aut}(N)$ .By your own definition, this is what it means to be a trivial homomorphism.

Mark Hennings - 2 months, 3 weeks ago

Thank you. Somehow I overlooked that is a homomorphism to Aut(N).

Dejan Stančević - 2 months, 3 weeks ago

Abelian semidirect products

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