Let $G$ be a finite group. Suppose $Z \le Z(G)$ and $G/Z \cong A\times B$, where $A$ is a $\pi$-group and $B$ is a $\pi'$-group for some set of primes $\pi$. Is it true that $G\cong C \times D$, where $C$ is a $\pi$-group and $D$ is a $\pi'$-group for the same set of primes $\pi$?
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$\begingroup$ Are you assuming $G$ is finite? If not, how do you classify elements of infinite order? $\endgroup$– Arturo MagidinSep 20, 2011 at 20:58
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$\begingroup$ Yes, I am assuming that $G$ is finite. $\endgroup$– the_foxSep 20, 2011 at 21:00
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$\begingroup$ Consider the product of three distinct cyclic groups, each of prime order. $\endgroup$– Yuri SulymaSep 20, 2011 at 21:12
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$\begingroup$ @Yuri: Sorry... how would that be a counterexample? The conclusion holds, trivially, for any finite nilpotent group, because a finite nilpotent group is the product of its $p$-parts, so you can always express $G$ as a product of a $\pi$-group and a $\pi'$-group, for any set $\pi$ of primes... $\endgroup$– Arturo MagidinSep 20, 2011 at 21:20
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$\begingroup$ @YuriDelanghe: I didn't get that either. $\endgroup$– the_foxSep 20, 2011 at 21:26
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2 Answers
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Hall's version of Sylow's theorem handles this.
Background: A Hall π-subgroup is a π-subgroup whose index is a π′-number. If every composition factor of G is either a π-subgroup or a π′-subgroup (that is, G is π-separable), then G posseses Hall π-subgroups. If furthermore every composition factor is either a p-subgroup for some p in π or a π′-subgroup (that is G is π-solvable), then in fact all Hall π-subgroups of G are conjugate and all Hall π′-subgroups of G are conjugate.
Reduction: The condition is simply that both a Hall π- and π′-subgroup are normal. Without loss of generality, Z = Z(G), since if a group has a normal Hall π-subgroup, then so does every quotient.
Proof: Write $G/Z = A/Z \times B/Z$ where $\gcd(|A/Z|,|B/Z|)=1$. A has a composition series refining A ≥ Z ≥ 1 with factors that are π and abelian. Hence A is π′-solvable, and all of its Hall π-subgroups are conjugate. Similarly B is π-solvable, and all of its Hall π′-subgroups are conjugate.
Let A0 be a Hall π-subgroup of A, and let B0 be a Hall π′-subgroup of B. Every G-conjugate of A0 lies in A since A is normal, but then by Hall's theorem, such a G-conjugate is already A = A0Z-conjugate. Hence $A_0^g = A_0^{az} = A_0$, and A0 is normal in G. Similarly B0 is normal in G, and so G = A0 × B0. $\square$
answered Sep 21, 2011 at 2:50
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$\begingroup$ If $A/Z$ is just a p-group, you can work the same argument using Sylow's theorem. If anything about this version is unclear, it should be easier to understand in the π=p case. $\endgroup$ Sep 21, 2011 at 2:53
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$\begingroup$ Great answer! Here is a different way to say pretty much the same thing (I am only including this because the OP mentions he is doing an exercise in Isaacs's FGT, and in that book the "Hall-C" theorem is not proved, although it is mentioned): Write $Z(G)=M\times L$, where $M$ is a $\pi$-group and $L$ is a $\pi'$-group. Then $L$ is a normal Hall $\pi'$-subgroup of $A$, so by Schur-Zassenhaus $A$ has a Hall $\pi$-subgroup $H$. Since $A=HL$, $H$ is in fact normal (characteristic) in $A$, hence normal in $G$. Similarly we find a Hall $\pi'$-subgroup $K\le B$, normal in $G$.... $\endgroup$– user641Sep 21, 2011 at 19:35
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$\begingroup$ ...Now $H\cap K=\{1\}$ ($H$ is a $\pi$-group and $K$ is a $\pi'$-group), and $G=HK$ ($[G:H]$ is a $\pi'$-number and $[G:K]$ a $\pi$-number), so in fact $G=H\times K$. $\endgroup$– user641Sep 21, 2011 at 19:36
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$\begingroup$ Both really good answers. In retrospect, it is very, very disturbing that I didn't think of Schur-Zassenhaus (or Hall for that matter)... $\endgroup$– the_foxSep 21, 2011 at 21:05
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To provide some context for this problem, I needed it as a lemma in order to do part b) of problem 5A.8 in Martin Isaac's book "Finite Group Theory". In page 151 of his book Isaacs gives the definition for the Schur multiplier and states (without proof of course) the existence and uniqueness of a largest second component in a defining pair and also that any other second component is a homomorphic image of that one. Assuming this, he then goes on and asks to prove
- $|M(A\times B)| \geq |M(A)||M(B)|$
- if $(|A|,|B|)=1$, then $M(A \times B) \cong M(A) \times M(B)$
The first part is easy:
Suppose that $C$, $D$ are Schur representation groups for $A$ and $B$ respectively. Then
$C \times D/M(A) \times M(B) \cong C/M(A) \times D/M(B) \cong A \times B$.
Also, $(C \times D)’ = C’ \times D’$, while $Z(C \times D)=Z(C) \times Z(D)$.
Hence $(C \times D)’ \cap Z(C \times D) \cong (C’ \cap Z(C)) \times (D' \cap Z(D))$, which contains $M(A) \times M(B)$ as a subgroup. Thus we see that $(C \times D,M(A) \times M(B))$ is another defining pair for $A \times B$.
This shows that $M(A) \times M(B)$ is a homomorphic image of $M(A \times B)$. In particular, it shows that $|M(A) \times M(B)|$ divides $|M(A \times B)|$ and the desired inequality follows.
Under the assumption that $(|A|,|B|)=1$ the exercise can be used as a lemma to go all the way. If $E$ is a Schur representation group for $A \times B$, then there exist groups $C_0$, $D_0$ such that $E \cong C_0 \times D_0$. According to your proof, $C_0 \leq C$ and $D_0 \leq D$ so that $|E|$ divides $|C||D|$, which implies that $|M(A \times B)|$ divides $|M(A) \times M(B)|$. By the first part then, $|M(A \times B)|=|M(A) \times M(B)|$ and the homomorphism is forced to be an isomorphism.
There is of course some cheating behind all this, because nothing is said of the existence of a unique largest second component, such that all other second components are homomorphic images of that one. Nevertheless, I thought this was an interesting problem for someone to try.