# Conjugation action on a semi-direct product $N\rtimes H$

Let $$G$$ be a finite group, with $$H\le G$$ and $$N\trianglelefteq G$$ such that $$G=N\rtimes H$$.
Suppose that the conjugation action of $$H$$ on $$N$$ induces $$2$$ orbits in $$N$$. Prove the following requests:
a) $$\exists p$$ prime $$\forall 1\ne n\in N$$ such that $$o(n)=p.$$
b) SHow that $$N$$ is abelian.

I said that if $$G=N\rtimes H$$, hence we have the conditions $$(*)\begin{cases}N \cap H=\{1_G\}\\NH=G \end{cases}$$. For a generic element, $$n \in N$$ the stabilizer of $$n$$ in the action $$\varphi:H\to\operatorname{Aut}(N)$$ is such that $$h^{-1}nh=n\iff h\in C_H(n)$$, whose order divides $$H$$.
Furthermore, if the number of orbits is two, the quotient $$H/Ker(\varphi)\cong \varphi(H)\le\operatorname{Aut}(N)$$ has order two, with $$Ker(\varphi):=\underset{n\in N}\bigcap C_H(n)$$. I'm confusing things because the map induced by the semi-direct product has the same form of the map induced by the conjugation action and I'm not sure about how using the Hypothesis $$(*)$$.
If $$N$$ is abelian it means that its centralizer is the entire $$G$$ (with $$C_G(H)\le Z(G)$$), and in particular, considering tha action restricted to $$H$$, $$C_H(N)$$ should corespond to $$H$$.

Thank for the help and for your patience.

• I think that your quantifiers are not correct. It's not "for all $n\not=1$ there exists $p$" but rather "there exists $p$ such that for all $n\not=1$". Jan 15, 2021 at 11:46

Under conjugation one orbit is just the identity. So all other elements of $$N$$ are conjugate in $$G$$ and therefore have the same order.

Let this order be a multiple of the prime $$p$$, then the $$p$$th power of an element of $$N$$ is still in $$N$$ but has a smaller order i.e. it has to be the identity.

Since $$N$$ is a $$p$$-group, $$Z(N)$$ is non-trivial and is a characteristic subgroup of $$N$$. Therefore $$Z(N)$$ is fixed by $$H$$ and so all elements of $$N$$ are in $$Z(N)$$.

• The Kernel of the map is in $H$ and so has no intersection with $Z(N)$ which is the subgroup of interest.
– user502266
Jan 15, 2021 at 12:33
• Sorry, the last thing. Did you state that N is a p group for Sylow's theorem? Jan 15, 2021 at 17:43
• Yes, you could use Sylow. But it's simpler than that really. By Cauchy's Theorem if a prime divides the order of the group then it has an element of order that prime. Therefore a finite group with all elements of order a power of a prime p must be a p-group.
– user502266
Jan 15, 2021 at 17:48
• I think that $\nexists$ a non trivial subgroup $L \trianglelefteq G$ contained in $N$... This should be true (?) Jan 15, 2021 at 21:31
• That is correct. Such a subgroup $L$ would be fixed by the action of $H$ and we know this cannot be the case because of the number of orbits.
– user502266
Jan 15, 2021 at 21:36