A group $G$ is called metacyclic if there is cyclic $N\unlhd G$ such that $G/N$ is cyclic as well. If $G$ is a metacyclic $p$-group, I know that there is a presentation
$$G\cong\langle x,y\mid\, x^{p^a}=1,\,y^{p^b}=x^{p^c},\,yxy^{-1}=x^k\rangle,$$
but is it always possible to realize $G$ as a semidirect product? Since it might make a difference: I do not care about the case $p=2$.
It turned out to be quite difficult to find a good link to answer this question. Therefore I will give here the full formulation of the theorem on metacyclic $p$-groups.
Theorem. For odd $p$, every metacyclic $p$-group $P$ has a presentation of the form $$ P =\{a,b\mid a^{p^\alpha}= b^{p^\beta}, b^{p^{\beta+\delta}}= 1,a^{-1}ba=b^{1+p^\gamma}\} $$ where $\alpha\geq\beta\geq\gamma\geq\delta\geq0$ and either $\gamma\geq1$ or $\beta=\gamma=\delta=0$. Conversely, each such presentation defines a metacyclic $p$-group of order $p^{\alpha+\beta+\delta}$, different values of the parameters $\alpha,\beta,\gamma,\delta$
(with the above condition) giving nonisomorphic groups.
Remarks.
The group $P$ is cyclic iff $\beta=\gamma=\delta=0$.
$P$ is split iff $\alpha=\beta$, or $\beta=\gamma$, or $\delta=0$.
This can all be found on page 27 of Hyo-Seob Sim's book Metacyclic groups of odd order There you can also find a rich bibliography on the subject.