On the exponent of the product of two groups

REND. SEM. MAT. UNIV. PADOVA, Vol. 115 (2006)

On the Exponent of the Product of two Groups.

AVINOAMMANN(*)

To Guido Zappa on his 90th birthday

Let the groupGˆABbe the product of two subgroupsAandB, and assume that these subgroups have finite exponentseandf. IfAandBare abelian, thenGis metabelian, by a famous result of N.Ito [AFG, 2.1.1], and in that case R.W.Howlett has shown that the exponent of G satisfies exp(G)jef [AFG, 3.3.1]. Apart from that, it seems that very little is known about the exponent ofGin general. IfGis soluble, it has a finite exponent [AFG, 3.2.11]. From this we deduce

THEOREM1. Let the soluble group G of derived length d be the product of two subgroups A and B of finite exponents e and f . Then the exponent of G is bounded by some function of d;e,and f .

PROOF. If no such function exists, then we can find soluble groups GnˆAnBn, of derived lengthd, such thatAnandBnhave exponentseand f, and the exponents of the groups G_n are unbounded. Then the direct product of all theG_n's has derived lengthd, is the product of subgroups of exponentseandf, but has infinite exponent, a contradiction.

COROLLARY2. LetGˆAB,whereAandBhave exponentseandf, respectively. Assume thatAand[A; B]are soluble of lengthd. Then the exponent ofGis bounded by a function ofe; f,andd.

PROOF. We haveA^GˆA[A;B]. ThereforeA^Gis soluble of length 2d.

But also A^GˆAC, where CˆA^G\B has exponent f. By the previous theorem,exp(A^G) is bounded, whileexp(G=A^G)ˆexp(B=C)jf.

Using Howlett's result, we can give explicit bounds for two cases: when G is metabelian, or nilpotent-by-abelian, and when one of the factors is abelian.

(*) Indirizzo dell'A.: Einstein Institute of Mathematics, The Hebrew University of Jerusalem, Edmond J. Sofra Campus, Givat Ram Jerusalem, 91904, Israel.

THEOREM3. Let the metabelian group G be a product of two subgroups A and B,of exponents e and f . Then exp(G)j(ef)³. If either A or B is abelian,then exp(G)j(ef)².

This follows immediately by substitutingA₁ˆA⁰ andB₁ˆB⁰ in the following more elaborate result, in which no solubility assumption is made.

PROPOSITION4. LetGbe a productABof two subgroups,of exponents e and f. Assume that there exist two subgroups, A₁ andB₁,such that A⁰A₁A,B⁰B₁ B,and[A; B]normalizes bothA₁ andB₁. Then exp(G)j(ef)³. If eitherAorBis abelian,thenexp(G)j(ef)².

PROOF. As above, write A^GˆA[A;B]ˆAC. Then A₁/A^G, and we have a normal series A₁/A^G/G. Here A^GˆAC, for CˆA^G\B, G=A^GB=C, andA^G=A₁ˆA=A₁CA₁=A₁. Assume first thatBis abelian.

Thenexp(A^G=A1)jef, by Howlett, implyingexp(G)je²f². IfBis not abe- lian, then the factorizationA^G=A1ˆA=A1CA1=A1 satisfies the assump- tions of the proposition, withB1replaced byC1ˆB1\A^G, and withA=A1

abelian. By what was just proved,exp(A^G=A₁)je²f², andexp(G)je³f³. THEOREM5. Let GˆAB,where exp(A)ˆe and exp(B)ˆf . Assume that A is abelian,and that G is soluble of length d2. Then exp(G)j(ef)^{2d 2}.

PROOF. If dˆ2, this follows from Theorem 3. Otherwise, let NˆG^{(d 1)}. ThenNis abelian, andANis metabelian, and can be factored as ANˆAC, withCˆAN\B. Theorem 3 shows thatexp(AN)je²f², and in particularexp(N)je²f². Apply induction toG=N.

A similar argument establishes the following variation:

PROPOSITION6. Let GˆAB,whereexp(A)ˆeandexp(B)ˆf. As- sume that A is abelian,and that [A; B] is soluble of length d. Then exp(G)je^2df^2d‡1.

PROOF. LetNˆ[A;B]^{(d 1)}. ThenNis abelian, andANis metabelian, and can be factored asANˆAC, withCˆAN\B. Theorem 3 shows that exp(AN)je²f². If dˆ1, then Nˆ[A;B], and then ANˆA^G/G, with G=A^GB=C, and the result follows. In the general case we have exp(N)je²f², and we apply induction toG=N.

206 Avinoam Mann

PROPOSITION7. LetGˆAB be nilpotent-by-abelian,withcl(G⁰)ˆc, and letAandBhave exponentseandf. Thenexp(G)j(ef)^2c‡1.

PROOF. For cˆ1 this follows from Theorem 3. In the general case, writeZˆZ(G⁰), andAZˆAC, withCˆAZ\B. ThenA⁰/AZ, andAZ=A⁰ is metabelian, with A=A⁰ abelian, and so Theorem 3 implies that exp(AZ)je²f², thereforeexp(Z)je²f², and we can apply induction toG=Z.

NOTE. Theorem 5 yields a better bound than Proposition 7 whenever both are applicable.

REFERENCES

[AFG] B. AMBERG- S. FRANCIOSI- F.DEGIOVANNI,Products of Groups, Oxford Univ. Press, Oxford 1992.

Manoscritto pervenuto in redazione il 27 dicembre 2005.

On the exponent of the product of two groups 207