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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.

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up vote 3 down vote accepted

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.

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This argument fails because the line of reasoning "$A$ is abelian implies $A\subseteq Z(G)$" is false. For instance, consider the subgroup of rotations in the group of rigid transformations of the square. It properly contains the center. – peoplepower Dec 4 '12 at 16:41
@peoplepower: spin is using the fact that $Z(G)=F(G)$ repeatedly here. – user641 Dec 5 '12 at 4:38
@SteveD Ah, thanks. – peoplepower Dec 5 '12 at 6:56

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.

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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.

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Well, this follows at once from the definition since $\,Z(G)\,$ is a nilpotent normal subgroup of $\,G\,$ , right?

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No. Because $Z(G)$ could be exactly equal to $F(G)$. I'm trying to see that is a proper subgroup. Thanks anyway. More ideas? – Mark_Hoffman Dec 2 '12 at 16:23
May I ask you to explain a bit more, Don? Thanks – Babak S. Dec 2 '12 at 16:28
But that's impossible both since (1) in a finite soluble group it is always true that $\,C_G(F(G))\leq F(G)\,$ , and (2) since $\,F(G)\,$ is normal nilpotent it can't then equal its own center! – DonAntonio Dec 2 '12 at 16:29
Sorry, but curiously in my task they ask me to prove $C_{G}(F(G))\leq F(G)$ as a consequence from the first question. There's another way to prove the first question then? – Mark_Hoffman Dec 2 '12 at 16:36
Hmmm...let me think a while about this and if I find a way I'll add an edit to my answer above. – DonAntonio Dec 2 '12 at 16:47

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