Center and Fitting subgroups I have to prove that, if G is a finite soluble and nonabelian group, its center is a proper subgroup of the Fitting subgroup of G.
In other words, that $Z(G)<F(G)$
Any ideas?
Thanks a lot.
 A: As DonAntonio mentioned, it is clear that $Z(G) \leq F(G)$. Assume $Z(G) = F(G)$. 
Since $G$ is solvable, there exists an integer $k$ such that the $k$th derived subgroup $G^{(k)} = 1$. Then $G^{(k-1)}$ is abelian, so $G^{(k-1)} \leq Z(G)$. Thus $G^{(k-2)}$ is nilpotent, because $$[G^{(k-2)}, G^{(k-1)}] \leq [G^{(k-2)}, Z(G)] = 1.$$ Hence $G^{(k-2)} \leq Z(G)$. Continuing this way we get $G^{(k-3)} \leq Z(G), \ldots, G^{(1)} \leq Z(G)$ and finally $G \leq Z(G)$, which is a contradiction because we assumed $G$ was nonabelian.
A: For any finite group, $F(G/Z(G))=F(G)/Z(G)$.  This follows from the fact that a group is nilpotent if and only if a central quotient of that group is nilpotent. Also, clearly $Z(G)\le F(G)$.
Now since $G$ is solvable, so is $G/Z(G)$, and since $G$ is nonabelian, $G/Z(G)$ is nontrivial.  But then $F(G/Z(G))\neq\lbrace 1\rbrace$, because, for example, it contains any minimal normal subgroup.
A: In fact, we construct a metabelian, normal subgroup which contains $Z(G)$ properly. Let $1=H_0\subset H_1\subset\dots\subset H_n=G/Z(G)$ be a chief series so that $H_1$ corresponds to a normal subgroup $H$ of $G$ which contains properly $Z(G)$. The factor $H/Z(G)$ is abelian since chief factors of finite, solvable groups are elementary abelian. Thus, $H'\subseteq Z(G)\subseteq Z(H)$, and $H$ is metabelian as desired.
A: Well, this follows at once from the definition since $\,Z(G)\,$ is a nilpotent normal subgroup of $\,G\,$ , right?
