Let $G=\mathbb Z\rtimes\mathbb Z$ and $H=\mathbb Z\rtimes7\mathbb Z$. I want to check if $H◁G$.
I know I need to calculate $N_G(H)$, and I think this is $$N_G(H)=\{(m,n) ∈ G \mid (m,n)(l,7k)(m,n)^{-1} ∈ H, \forall l,k\}$$
but I'm not sure when to go from there. I wasn't given great notes for this.
I think I can use $$(m,n)(l,7k)(m,n)^{-1}=(m+(-1)^ml, n+7k)(-(-1)^nm,-n) =(m+(-1)^nl+(-1)^n+7k)(-1)^(n+1)m,7k) =(m(1-(-1)^k)+(-1)^nl,7k)$$
But what does this mean? Thanks
I think you had better not to make those calculations (although they certainly give you the result eventually). Because $G$ is semi-direct product $\mathbb{Z}\rtimes \mathbb{Z}$ you can take $\pi$ the projection of $G$ onto the second factor this gives you :
$$\pi: \mathbb{Z}\rtimes \mathbb{Z}\rightarrow \mathbb{Z} $$
$$(a,b)\mapsto b$$
Now this is a surjective group morphism and from the very definition of $H$ you have that $\pi^{-1}(7\mathbb{Z})=:H$. It is clear that $7\mathbb{Z}\triangleleft \mathbb{Z}$, then you can use the following fact (easy exercise) the inverse image of a normal subgroup by a surjective morphism is a normal subgroup. Hence $H\triangleleft G$.