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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.

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    $\begingroup$ @Geoff Robinson: Thank you very much for the comment. I fixed the typo. $\endgroup$ – Bernhard Boehmler Jun 5 at 18:52
  • $\begingroup$ 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. $\endgroup$ – Geoff Robinson Jun 5 at 19:49
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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.

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  • $\begingroup$ 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. $\endgroup$ – Bernhard Boehmler Jun 6 at 18:20
  • $\begingroup$ 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). $\endgroup$ – Geoff Robinson Jun 6 at 19:09

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