characterization of full families of groups

PropositionPlanetmathPlanetmath. Let 𝒢={Gk}kI be a family of groups. Then 𝒢 is full if and only if for any i,jI such that ij we have that any homomorphismMathworldPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath f:GiGj is trivial.

Proof. ,,” Assume that f:GiGj is a nontrivial group homomorphism. Then define


as follows: if tI is such that ti and gkIGk is such that gGt, then h(g)=g. If gkIGk is such that gGi, then h(g)(j)=f(g(i)) and h(g)(k)=0 for kj. This values uniquely define h and one can easily check that h is not decomposableMathworldPlanetmathPlanetmathPlanetmath.

,,” Assume that for any i,jI such that ij we have that any homomorphism f:GiGj is trivial. Let


be any homomorphism. Moreover, let iI and gkIGk be such that gGi. We wish to show that h(g)Gi.

So assume that h(g)Gi. Then there exists ji such that 0h(g)(j)Gj. Let


be the projection and let


be the natural inclusion homomorphism. Then πu:GiGj is a nontrivial group homomorphism. ContradictionMathworldPlanetmathPlanetmath.

Corollary. Assume that {Gk}kI is a family of nontrivial groups such that Gi is periodic for each iI. Moreover assume that for any i,jI such that ij and any gGi, hGj orders |g| and |h| are realitvely prime (which implies that I is countableMathworldPlanetmath). Then {Gk}kI is full.

Proof. Assume that ij and f:GiGj is a group homomorphism. Then |f(g)| divides |g| for any gGi. But f(g)Gj, so |g| and |f(g)| are relatively prime. Thus |f(g)|=1, so f(g)=0. Therefore f is trivial, which (due to proposition) completesPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath the proof.

Title characterization of full families of groups
Canonical name CharacterizationOfFullFamiliesOfGroups
Date of creation 2013-03-22 18:36:08
Last modified on 2013-03-22 18:36:08
Owner joking (16130)
Last modified by joking (16130)
Numerical id 9
Author joking (16130)
Entry type Derivation
Classification msc 20A99