Cyclic subgroups of semidirect products

Question

Let $H$ and $N$ be two finite groups and $\phi:H\to Aut(N)$ a homomorphism. Let $G=N\rtimes_\phi H$ and let $\pi:G\to G/N=H$ be the quotient map.

Let $p$ be a prime which does not divide $|N|$ but does divide $|H|$, and let $\gamma:C_p\to G$ be a homomorphism. ($C_p$ is the cyclic group of order $p$.) If $\gamma$ is nontrivial, $\gamma(C_p)$ is a cyclic subgroup of $G$, and I would like to understand such subgroups better.

Do we have $\gamma(C_p)\subset\{e_N\}\times H<G$, perhaps under some additional assumptions? That is, does the homomorphism $\gamma$ have to be trivial on $N$, so to say?

I know that the claim is true if $\phi|_{\pi(\gamma(C_p))}$ is trivial (which happens if, for example, $p$ does not divide $|\phi(H)|$), but I would prefer not to make assumptions on the homomorphism $\phi$ and $\gamma$ (especially $\gamma$). Assumptions on the groups $H$ and $N$ would be more convenient.

Edit: Since $|\phi(H)|$ divides $|H|$, it does indeed happen for some primes $p$ satisfying the division requirements that $p$ divides $|\phi(H)|$.

Refined question: Let $H,N,\phi$ be as above. As a set, we can write $G=N\times H$. Are there any assumptions on the groups $H$ and $N$ so that for every prime $p$ dividing $|H|$ but not $|N|$ the following is true: If $(n,h)\in N\times H=G$ has order $p$, then $n=e_N$?

Answer 1

Let $n$ be an odd number and $$G=D_{2n}=\langle x,y|x^n=y^2=1,yxy^{-1}=x^{-1}\rangle,$$ with $N=\langle x\rangle\cong \Bbb Z/n$ and $H=\langle y\rangle\cong \Bbb Z/2$.

Now $yxy^{-1}=x^{-1}=x^{n-1}$ implies $x^{-1}yx=x^{n-2}y$. That means $x^{n-2}y$ has order $2$ and of course $x^{n-2}\neq 1$.

Edit: I just realize that I didn't really answer the question. However, I think the example above represents somewhat the general idea.

One only needs to consider the case $H=\langle h\rangle\cong\Bbb Z/p$. Since $p$ does not divide $|N|$, $\langle h\rangle$ is a $p$-Sylow subgroup of $G$. The condition $$(nh)^p=1$$ then implies that $\langle nh\rangle$ is conjugate to $\langle h\rangle$ by $g\in N$. In particular $$nh=g^{-1}h^kg,$$ or $$gnhg^{-1}=h^{k}.$$ Equivalently one has $h^{k-1}\in N\cap \langle h\rangle=1$, or $$nh=g^{-1}hg.$$ Here, in general $n\ne1$ unless $H=\langle h\rangle$ is also normal in $NH$, which is equivalent to the trivial action of $h$ on $N$.

Answer 2

The element $n$ is not necessarily trivial, as the other answer points out.

Suppose that $nh$ has order $p$. Then $1=(nh)^p=nhnh\cdots nh=n^{\prime}h^p$ where $n^{\prime}\in N$. As $N\cap H=1$, $n^{\prime}=1=h^p$. Hence, $h$ has order dividing $p$, but we cannot conclude this about $n$.

The element $n^{\prime}$ has the form $\phi(n)\phi^2(n)\cdots\phi^{p}(n)$, so all you can really say is that this element is trivial, which is really a description of the automorphism $\phi$.