# Concrete example to illustrate the theory about blocks of groups with cyclic defect groups

I'd like to to have a concrete example to illustrate the theory about blocks of groups with cyclic defect groups.

Thus, I am looking for a finite group $$G$$ and a prime $$p$$ dividing $$|G|$$ satisfying all of the following properites:

• The defect group $$D$$ of a $$p$$-block $$B$$ of $$G$$ is cyclic and of order $$p^r$$, where $$r\geq 2$$
• $$G$$ is neither $$p$$-solvable nor solvable (or at least not solvable)
• The shape of the graph of the Brauer tree associated to $$B$$ is not only a (part of a) star, but more complicated (and, if it doesn't make things too complicated, the Brauer tree associated to $$B$$ has an exceptional vertex)

• All inclusions are proper in the chain $$1 \leq D_1 \leq D\cong C_{p^r} \leq N_G(D)\leq N_G(D_1)\leq G$$, where $$D_1$$ is the unique subgroup of $$D$$ of order $$p$$

• All inclusions are proper in the chain $$C_G(D_1) \leq T(c) \leq N_G(D_1)$$

• It is still possible to express these groups relatively nicely, such that one does not have to say that some group is an extension of a semidirect product of the double cover of the first non-split extension of... with ...of...

Remark for the penultimate point: Let $$b$$ be the block of $$N_G(D_1)$$ which is Brauer-correspondent to $$B$$, let $$c$$ be a block of $$C_G(D_1)$$ covered by $$b$$ and let $$T(c)$$ be the inertial goup.

Unfortunately, I'm not aware of such an example. Neither was I able to construct one.

If it is not possible to find an example with all inclusions proper, then maybe with not all but many of them?

Thanks for the help.

• @Geoff Robinson: Thank you very much for the comment. I fixed the typo. Jun 5 '19 at 18:52
• I think the fourth bulletpoint, especially the requirement that that $N_{G}(D) < N_{G}(D_{1})$ may be the most difficult to realise (given that $G$ is not $p$-solvable). I may write a formal answer later. Jun 5 '19 at 19:49

You probably know this, but your conditions can't be met by any principal $$p$$-block of any finite group $$G$$ which is not $$p$$-solvable, although I think the Classification of Finite Simple Groups (CFSG) is necessary for that. For suppose that $$G$$ is a finite group with cyclic Sylow $$p$$-subgroup $$D$$ and with $$|D| >p,$$ but that $$G$$ is not $$p$$-solvable. Then the principal $$p$$-block $$B$$ of $$G$$ has defect group $$D$$, and this does not change on passage to $$G/O_{p^{\prime}}(G),$$ so we might as well suppose that $$O_{p^{\prime}}(G) = 1.$$ Now $$O_{p}(G) = 1,$$ for otherwise we have $$D_{1} \lhd G,$$ and then $$C_{G}(D_{1}) \lhd G.$$ But $$C_{G}(D_{1})$$ has a (characteristic) normal $$p$$-complement ( which must be trivial) as $$O_{p^{\prime}}(G) = 1.$$ Then $$C_{G}(D_{1}) = D \lhd G,$$ and $$G$$ is certainly $$p$$-solvable, a contradiction.

Let $$H$$ be a Hall $$p^{\prime}$$-subgroup of $$N_{G}(D).$$ Then $$D = [D,H] \times C_{D}(H)$$ since $$D$$ is Abelian of order coprime to $$|H|$$. Since $$D$$ is cyclic, we either have $$C_{D}(H) = D$$ or $$C_{D}(H) = 1.$$ In the former case, $$N_{G}(D)$$ has a normal $$p$$-complement, and then so does $$G,$$ contrary to the fact that $$G$$ is not $$p$$-solvable. Hence $$D = [D,H] \leq G^{\prime}.$$

Now $$G^{\prime}$$ is not $$p$$-solvable, as $$G$$ is not, so we then obtain $$D \leq G^{\prime \prime}$$ by the same argument, and ultimately $$D \leq G^{(\infty)},$$ the terminal member of the derived series for $$G$$, by repeating the argument.

Let $$M$$ be a minimal normal subgroup of $$G.$$ Then $$M$$ has order divisible by $$p$$, so that $$D_{1} \leq M.$$ Also $$M$$ is not a $$p$$-group, so $$M$$ is non-Abelian simple (using the fact that $$M$$ has cyclic Sylow $$p$$-subgroup). Then $$G = MN_{G}(D_{1})$$ by a Frattini (type) argument.Now $$MC_{G}(D_{1}) \lhd G$$ and $$G/MC_{G}(D_{1})$$ is a homomorphic image of the Abelian $$p^{\prime}$$-group $$N_{G}(D_{1})/C_{G}(D_{1}),$$ so is itself Abelian of order prime to $$p$$.

Now we have $$D \leq MC_{G}(D_{1})$$ and $$MC_{G}(D_{1})/M$$ (being a homomorphic image of $$C_{G}(D_{1})$$ ) has a normal $$p$$-complement. But $$MC_{G}(D_{1})$$ is certainly not $$p$$-solvable, so arguing as before, we have $$D \leq [MC_{G}(D_{1})]^{\prime}.$$ Hence $$MC_{G}(D_{1})$$ has no factor group of order $$p$$, so that $$MC_{G}(D_{1})/M$$ is a $$p^{\prime}$$-group. Now $$G/M$$ is a $$p^{\prime}$$-group, so that $$D \leq M.$$

Now $$G = MN_{G}(D)$$ by a Frattini argument. I claim that we now have $$N_{G}(D) = N_{G}(D_{1}).$$ This is the (first and only) time we require CFSG, though the previous argument could have been shortened considerably using CFSG.

Note that $$N_{G}(D_{1}) = N_{M}(D_{1})N_{G}(D),$$ so to prove the claim, it suffices to prove that $$N_{M}(D_{1}) = N_{M}(D).$$ But it is a Theorem of H. Blau that whenever a finite non-Abelian simple group $$X$$ has a cyclic Sylow $$p$$-subgroup $$P \neq 1$$, then $$P$$ is TI in $$X$$, that is, $$P \cap P^{x} = 1$$ for all $$x \in X \backslash N_{X}(P)$$.

In our situation, this immediately yields that $$N_{M}(D_{1}) = N_{M}(D),$$ as required. Hence your fourth bulletpoint can't be satisfied (for principal blocks of non-$$p$$-solvable groups) as remarked in comments.

• Thank you very much for the answer, the explanations and the very elegant proof. May I kindly ask, whether you happen to know an example, where not all, but many inclusions are proper? For instance, I considered $G=$PSL(3,8) and $p=3$, but there $[N_G(D_1):C_G(D_1)]=2$, so I don't have that the inclusions in the second chain are proper, what I'd really be fond of to have. Jun 6 '19 at 18:20
• I'll think a little, but you might find an example of a block with cyclic defect group of a simple group $S$ such that the block is not stable in ${\rm Aut}(S)$ (such a block would have to be non-principal). Jun 6 '19 at 19:09