Let $G$ be finite group and $G$ has at least one sylow subgroup $K$ such that $K\ntriangleleft G$ . proof if every nonnormal and abelian subgroup of G be cyclic then $Z(G)$ is cyclic.
Theorem 1:Let $G$ be a finite $p$ -group and $H \ntriangleleft G$ be of order $p .$ If for some central element $z$ of order $p,\langle H, z\rangle \leq G$ then either $Z(G)$ is cyclic or for any central element $y$ such that $z \notin\langle y\rangle,\langle H, y\rangle \ntriangleleft G$
Theorem 2:Let $G$ be a finite non-nilpotent group. Thus there is some $P \in \mathcal{S} y \ell_{p}(G)$ such that $P \ntriangleleft G$
Theorem 3 : Let $G$ be a non-nilpotent group such that all the non-normal nilpotent subgroups of G are cyclic. Then $G$ is solvable with a cyclic center.
Also this is clear that $G$ is not abelian.i think we can show G is a non-nilpotent group .
Let $p$ be a prime such that a Sylow $p$-subgroup $P$ of $G$ is not normal in $G$.
Let $N = O_p(G)Z(G)$. Then $N$ is the product of two normal subgroups of $G$, and so $N$ is normal in $G$. Note that $N/O_p(G) \cong Z(G)/(Z(G) \cap O_p(G))$ is abelian.
We claim that $P$ is not contained in $N$. So suppose that $P \le N$. Since $O_p(G)$ is contained in all Sylow $p$-subgroups of $G$, we have $O_p(G) \le P$. Since $N/O_p(G)$ is abelian, $P/O_p(G)$ is normal in $N/O_p(G)$, so $P$ is normal in $N$. But a normal Sylow $p$-subgroup is characteristic, so $P$ is normal in $G$, contradiction.
Since $P$ is not contained in $N$, $PN/N \cong P/(P \cap N)$ is nontrivial, and hence has order divisible by $p$. So we can choose $g \in PN$ with $g \not\in N$ and $g^p \in N$. Let the order of $g$ be $p^aq$, where $q$ is not divisible by $p$. Then we can replace $g$ by $g^q$, and we still have $g \not\in N$ but $g^p \in N$, and now $g$ has order a power of $p$.
Let $H := \langle g, Z(G) \rangle$. Then $H$ is abelian. So $g \in O_p(H)$. If $H$ was normal in $G$ then we would have $O_p(H)$ normal in $G$ and hence $O_p(H) \le O_p(G)$ and $g \in O_p(G)$, contradiction. So $H$ is not normal in $G$.
So $H$ is a non-normal abelian subgroup of $G$. Hence $H$ is cyclic and hence so is $Z(G)$.