I have a semidirect product group $G=N\rtimes H$, where $N$ is a normal $p$-Sylow subgroup of $G$ of order $p^d$ and $H$ has order $m$. Additionally, $m$ and $p$ are coprime and the centralizer of $N$ in $G$ is $N$, meaning $C_G(N) = N$.
I would like to show the following:
If $C$ is an abelian subgroup of $G$ and the order of $C$ is coprime to $p$, then there exists $n\in N$ such that $n^{-1} C n \subseteq H$.
My idea is the this: Set $X := G/H$. It is clear that $X = N/H$ and $|X| = |N| = p^d$. $C$ operates on $X$ via multiplication from the left. Thus, we find $x_1,...,x_k\in X$ so that $X$ is the disjoint union of the orbits of the $x_i$. Then
$p^d = |X| = \sum\limits_{i=1}^k |C.x_i| = \sum\limits_{i=1}^k |C/C_{x_i}|,\tag{*}$
where $C.x_i$ denotes the orbit of $x_i$ (under the action of $C$ on $X$) and
$C_{x_i} = \{c\in C \mid c.x_i = x_i\}$
is the subgroup of all elements of $C$ that fix $x_i$.
To prove the claim, we need to find a fixpoint of the action, i.e. some $j\in\{1,...,k\}$ with $C_{x_j} = C$ or, equivalently, $C.x_j = x_j$. Since $x_j\in X$, we find $n_j\in N$ with $x_j = n_j H$, leading to $C n_j H = n_j H$. Therefore, $n_j^{-1} C n_j \subseteq H$.
But how can we find such a fixpoint $x_j$? My idea was to show this through a contradiction:
Assume $|C.x_i| > 1$ for all $i\in\{1,...,k\}$. I would like to show that
$\text{ggT}(|C.x_1|,...,|C.x_k|) =: q > 1.$
Because then, (*) implies that $q$ divides $p^d$ and is therefore a power of $p$. But by the orbit stabilizer theorem, it also divides $|C|$, which contradicts our assumptions.
The problem is that I have no idea how to show that $q > 1$. For all I know, $C$ could be isomorphic to $\mathbb{Z}/a\mathbb{Z} \times \mathbb{Z}/b\mathbb{Z}$ ($a,b$ coprime), $C_{x_1} = \mathbb{Z}/a\mathbb{Z}$ and $C_{x_2} = \mathbb{Z}/b\mathbb{Z}$. Then, $|C.x_1| = b$ and $|C.x_2| = a$ would be coprime, forcing $q=1$.
Is there any reason why this scenario cannot happen under my conditions (for example using $C_G(N) = N$ etc.)? Or is there another way to prove the claim? If not, are there any additional conditions that could force the claim to be true (in the case that the given conditions are not sufficient)?
Thank you for your help!
I found an answer myself (albeit using a completely different ansatz) which I would like to post in case someone is interested in this in the future. The conditions $C_G(N)=N$ and $C$ abelian are not needed for this to work!
Since $G = N\rtimes H$, we have $G = NH$. Then set
$\tilde{G} := NC = NC \cap NH = N (NC \cap H).$
We know that $N\cap C = \{e\}$ since every element of $N\cap C$ has order dividing both $|N|=p^d$ and $|C|$ and hence must be trivial. Additionally, $N\cap (NC \cap H)$ is trivial as well because each element in $N\cap (NC \cap H)$ must have order dividing both $|N|$ and $|H|$.
As a consequence, both $C$ and $NC\cap H$ are complements to $N$ in $\tilde{G}$. The orders of $C$ and $N$ are relatively prime, so by the Schur-Zassenhaus, all complements to $N$ in $\tilde{G}$ are conjugate. Thus, there exists $\tilde{g}\in\tilde{G}$ with
$\tilde{g} C \tilde{g}^{-1} = NC \cap H \subseteq H.$
But $\tilde{g}\in\tilde{G}$, so we can find $n\in N$, $c\in C$ with $\tilde{g} = n c$, leading to
$\tilde{g} C \tilde{g}^{-1} = n c\, C\, c^{-1} n^{-1} = n C n^{-1} \subseteq H.$