# About finite group such that $Z(G)$ must be cyclic

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 $$Z(G)$$.

• why $G/O_p(G)$, and hence also $G/N$ with $N = Z(G)O_p(G)$ have order divisible by $p$? May 24, 2020 at 16:22
• Because otherwise $O_p(G)$ would be a normal Sylow $p$-subgroup of $G$. May 24, 2020 at 16:25
• why we can Choose $g \in G \setminus N$ with $g^p \in N$ ? why $g \in G$ exist with $g^p \in N$ ? May 24, 2020 at 16:30
• also why $H$ is abelian ? May 24, 2020 at 16:36
• If you know about Sylow's theorem, then you must know that every group of order divisible by a prime $p$ has an element of order $p$. So $G/N$ has an element of order $p$. You can prove for yourself that $H$ is abelian. May 24, 2020 at 16:41