Assume that K is a cyclic group, H is an arbitrary group and f1 and f2 are homomorphisms from K into Aut(H) such that f1(K) and f2(K) are conjugate subgroups of Aut(H). If K is infinite, assume f1 and f2 are injective. Prove by constructing an explicit isomorphism that (H x_f1 K) is isomorphic (H x_f2 K)
K is a cyclic group, we can assume K=<k>. f1, f2 are homomorphisms from
K to Aut(H). Then f1(k)=<f1(k)>, f2(K)=<f2(k)>. Let g1=f1(k), g2=f2(k).
Since f1(K) and f2(K) are conjugate to each other, then we can find
some gamma in Aut(H), such that gamma*f1(k)*gamma^(-1) = f2(k). That is
gamma*g1*gamma^(-1)=g2, or gamma*g1=g2*gamma.
Now we define a map phi: (H x_f1 K) -> (H x_f2 K) as
phi(h,x)=(gamma(h),x), for ...
Direct Products of Groups, Conjugate Subgroups, Homomorphisms and Isomorphisms are investigated. The solution is detailed and well presented. The response was given a rating of "5/5" by the student who originally posted the question.