I’m currently studying for an exam and I was working on the problem below:
Let $G$ be a finite semidirect product group $N \rtimes L$, where $N$ is normal. Assume $L$ nontrivial, and for all $n \neq 1$ in $N$, $C_{G}(n) \subseteq N$. Then, prove $gLg^{-1} \cap L = \{1\}$ for all $g \notin L$ and furthermore, prove gcd$(|L|,|N|)=1$.
I was able to get started on the first part but wasn’t able to go much further. I will show my work up until I got stuck below.
Proof. First, assume $\phi$ is the homomorphism associated to $G = N \rtimes L$. Thus, $\phi: L \to Aut(N)$ and, furthermore, $\phi_l: N \to N$ is an automorphism for all $l \in L$.
Now, let $g = (a,b) \in G$ such that $(a,b) \notin L$. Then, $a \neq 1$. Now, let $(x,y) \in gLg^{-1} \cap L$. Hence, $(x,y) \in gLg^{-1}$ and $(x,y) \in L$. Thus, $x = 1$. So, $(1,y) \in gLg^{-1} = g^{-1}Lg$. Hence, \begin{align*} (1,y) =&(a,b)^{-1}(1,l)(a,b)\\ =&(\phi_{b^{-1}}(a^{-1}),b^{-1})(1,l)(a,b)\\ =&(\phi_{b^{-1}}(a^{-1}) \phi_{b^{-1}}(1),b^{-1}l)(a,b)\\ =&(\phi_{b^{-1}}(a^{-1}),b^{-1}l)(a,b)\\ =&(\phi_{b^{-1}}(a^{-1}) \phi_{b^{-1}l}(a),b^{-1}lb)\\ =&(\phi_{b^{-1}}(a^{-1}\phi_l(a)),b^{-1}lb). \end{align*}
Consequently, $\phi_{b^{-1}}(a^{-1}\phi_l(a)) = 1$ which implies $a^{-1}\phi_l(a) = 1$. Thus, $\phi_l(a)=a$. Since $a$ is arbitrary, $l \in$ ker($\phi$).
Now, this is where I am stuck. The only way for my proof to go forward is for ker($\phi$) $=\{1\}$. However, I believe that I am unable to really say anything about $\phi$ other than it is a homomorphism. I’m not sure how to prove $\phi$ is injective, which is how we would get ker($\phi$) $=\{1\}$, if that is what I should be going for. We also have that $C_G(n) \subseteq N$ for all nontrivial $n \in N$ but I don’t know where or how to use that in this problem.
Furthermore, I am just not sure how to even approach the second half of this problem in general either. I’ve been working on this for about a day and a half or so and would really like to know what I am missing so I can study more for my exam.
Thank you so much ahead of time. Any help would be greatly appreciated!