0
$\begingroup$

This question already has an answer here:

When using semidirect product to construct new groups based on smaller groups, we have to define a group homomorphism from the non-normal subgroup to the group of automorphism of the normal one, i.e. if $G = H \rtimes_{\phi} K$, then $\phi\colon K \to Aut(H)$. Being $\phi$ non-trivial, can we ensure $G$ is non-abelian?

$\endgroup$

marked as duplicate by José Carlos Santos, Community Dec 21 '18 at 16:17

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 1
    $\begingroup$ If $\phi$ is non-trivial, it's easy to show that the group is non-abelian $\endgroup$ – Don Thousand Dec 21 '18 at 16:17