this question is related to the semidirect product of groups ,
so let $H,K$ are groups.
suppose , $f:K \rightarrow H$ is a homomorphism . so $H\rtimes_f K$ is a semidirect product .
the problem now is as follows , if i supposed that $f$ is the trivial homommorphism ,so $\forall k\in K, f(k)=1$
this guy , $f$ , can be extended to a nother function , say $g$ , whose codomain is $S_{|K|}$ and $g(k)=f(k)$ with no change with the domain .
so there is no change with the semedirect product , now , the definition of the product of the elements of the semidirect product uses the group action which is defined using this homomorphism $g$, but $g$ is trivial homomorphism as $f$ is the trivial homomorphism
so this group action from $K$ on $H $ is also trivial .
so for any\every $k\in K , h\in H $ we have $ k * h=f_k(h) = 1$ where * denotes operation of the the action and this is the trivial action.
but in Dummit and foote , it's defined that the trivial action is $\forall k \in K , h\in H , k*h=h $
and they are not the same clearly ! what is wrong ? how can this trivial homomorphism afford the group action like the text provide ? and where is my wrong ?