# For every group of odd order, is $F(\omega(G))$ abelian?

Here, $$F(G)$$ stands for the Fitting subgroup of a group, and $$\omega(G) = \cap N_G(H)$$, where $$H$$ are all the subnormal subgroups of $$G$$, is the Wielandt subgroup.

Since we're talking about groups of odd order, by the Feit-Thompson theorem $$G$$ is solvable, this must mean $$\omega(G)$$ is solvable and has an abelian series $$1=G_0 \lhd G_1 \lhd \cdots \lhd G_n=\omega(G),$$ where each $$G_{i+1}/G_i$$ is abelian.

We also have $$F(\omega(G))= \cap C_{\omega(G)}(G_{i+1}/G_{i})$$. I get stuck here and can't go further. How can I finish the proof?

And is there an alternative method to show $$F(\omega(G))$$ is abelian because I don't think it ought to be necessary to use such a strong result (Fiet-Thompson)?

• I am really not sure what you are asking. I have no idea what you should do next. Is there an alternate (alternative?) approach to what? – Derek Holt Jan 8 '17 at 7:13
• @DerekHolt Sorry, I will edit the question. – Maria-0 Jan 8 '17 at 7:27
• But why do you think that $F(\omega(G)$ is abelian when $|G|$ is odd. Have you been asked to prove it as an exercise? If not, what is the evidence for it being true? – Derek Holt Jan 8 '17 at 11:19
• @DerekHolt It was mentioned in class. This isn't homework that will add to the grade, but our prof. did wanted us to check if it was true or not. – Maria-0 Jan 8 '17 at 11:28

## 1 Answer

All subgroups of $F(\omega(G))$ are subnormal in $G$, and hence all such subgroups are normal in $F(\omega(G))$. So $F(\omega(G))$ is a Dedekind group. The structure of these groups is fully understood, and in particular all Dedekind groups of odd order are abelian. Note that this argument does not make any use of the Feit-Thompson theorem.