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How does one apply Thompson's A×B lemma (Lemma 24.2 on page 112 of Aschbacher's Finite Group Theory) in order to prove this nice lemma (Lemma 31.16 on page 160)?

In the book, I basically don't understand what “G,” “A,” and “B” (from Thompson's A×B lemma) should be in the second sentence of the proof of lemma 31.14.1 on page 159. I would not mind help verifying that they satisfy the hypotheses of the A×B lemma, and (if its not obvious) that the conclusion really does imply what the proof says it implies. My direct attempts at this have failed, possibly because I'm using the wrong G,A,B, or possibly just a typo on my part, since it is hard to keep the letters disjoint.

Here are the lemmas in expressed in basic language if someone wants to tackle it more directly.

Thompson's A×B lemma: If $A,B,G$ are subgroups of a finite group with $[A,G], [B,G] \leq G$; $[A,B]=1=[C_G(B),A]$; $B$ and $G$ are $p$-groups; and $A$ has no normal subgroups whose index is a power of $p$; then $[A,G]=1$.

Definition: For a finite group $G$ and prime $p$, $O_p(G)$ is the intersection of the Sylow $p$-subgroups of $G$.

Nice lemma: If $C_G( O_p(G) ) \leq O_p(G)$, then for every $p$-subgroup $U \leq G$, one has $C_G( O_p(H) ) \leq O_p(H)$ where $H=N_G(U)$.

[ For a finite group $X$ and prime $p$, $C_X(O_p(X)) \leq O_p(X)$ iff $F^*(X) = O_p(X)$. This can be shown easily from the earlier results in that section of the book. Kurzweil–Stellmacher claim another form of the nice lemma: it works for all $H$ subnormal in $G$ and for all $O_p(G) \leq H\leq G$. ]

Jack Schmidt
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2 Answers2

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Let $C = C_G(O_p(H))$. Since $U \le O_p(H)$, $C \le H$, so $C = C_H(O_p(H)) \unlhd H$. So to prove that $C \le O_p(H)$, we need to prove that $C$ is a $p$-group. If not, then $O^p(C) \ne 1$.

Perhaps we can apply Thompson's lemma with the $A$, $B$, and $G$ in Thompson equal to $O^p(C)$, $O_p(H)$, and $O_p(G)$. If this works, then the conclusion will be that $O^p(C) \le C_G(O_p(G))$, contradicting the hypothesis that $C_G(O_p(G))$ is a $p$-group.

For the hypotheses of Thompson (back in the notation of Thompson), we cetainly get $[A,G],[B,G] \le G$, $[A,B]=1$, $B$ and $G$ are $p$-groups, and $A$ has no normal subgroup of index a power of $p$. That leaves $[C_G(B),A]=1$ to check, or in the notation of Nice Lemma, $[C_{O_p(G)}(O_p(H)),O^p(C)]=1$. But $C_{O_p(G)}(O_p(H)) = C \cap O_p(G) \le H \cap O_p(G) \le O_p(H)$, which centralizes $C$ by definition, so we have the final condition.

Derek Holt
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  • Perfect thanks! I completely misunderstood the goal (I think I had "B" and "G" switched). I'm reworking it now, but things are going much better. – Jack Schmidt Jan 02 '14 at 21:44
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Derek's answer is perfect. For my later memory, let me point out the key hypotheses:

Proposition: Suppose $X$ is a finite group, $H \leq X$ is a subgroup, such that $C_X(O_p(H)) \leq H$. Then $A=O^p( C_X(O_p(H)) )$, $B=O_p(H)$, and $G=O_p(X)$ satisfy the hypotheses of Thompson's lemma. Hence $[A,G]=1$ and $O^p( C_X(O_p(H)) ) \leq C_X( O_p(X) )$.

Proof: The only slightly tricky part is verifying $[A,C_G(B)]=1$: $$[A,C_G(B)] \leq [ C_X(O_p(H)), C_{O_p(X)}(O_p(H)) ] \leq \\ [C_X(O_p(H)), H \cap O_p(X) ] \leq [ C_X(O_p(H)), O_p(H) ] = 1.$$ Here $A \leq C_X(O_p(H))$ gives the first, next $C_Y(O_p(H)) \leq H \cap Y$ (the key hypothesis for $H$), then a general feature of subgroups $H \cap O_p(X) \leq O_p(H)$, and finally the definition of centralizer. $\square$

Now if $X$ is a finite group with $C_X(O_p(X))\leq O_p(X)$ and $H \leq X$ is a subgroup satisfying $C_X(O_p(H)) \leq H$, then the conclusion is stronger: $O^p( C_X(O_p(H)) ) \leq O_p(X)$ is a $p$-group (with no normal $p$-quotients) so $C_H(O_p(H)) \leq O_p(H)$ is a normal $p$-subgroup of $H$.

Corollary: Suppose $X$ is a finite group, $H \leq X$ is a subgroup, such that $C_X(O_p(H)) \leq H$ and $C_X(O_p(X)) \leq O_p(X)$. Then $C_X(O_p(H)) \leq O_p(H)$.

A few examples where the key hypothesis is satisfied:

  • $H=N_X(U)$ for some $p$-subgroup $U$ of $X$
    [ Since $U \leq O_p(H)$ and anything centralizing $U$ certainly normalizes it. ]
  • $H \geq O_p(X)$ in the corollary
    [ Since $O_p(H) \geq O_p(X)$ and $C_X(O_p(H)) \leq C_X(O_p(X)) \leq O_p(X) \leq O_p(H)$ ]

When $H\unlhd\unlhd X$, one gets that $F^*(H) = F^*(X) \cap H$, so the hypothesis is satisfied in the corollary, but rarely otherwise (take $H<X$ and $p$ not dividing the order of $\operatorname{Fit}(H)$ for instance).

Jack Schmidt
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