I'm working on a problem that asks me to prove that, given a cyclic group K and an arbitrary group H, the semi-direct products H xφ1 K and H xφ2 K are isomorphic if φ1(K) and φ2(K) are conjugate by producing an explicit isomorphism. They suggest the function: ψ : H xφ1 K → H xφ2 K defined by: ψ(h, k) = (σ(h), ka) where σ is in Aut(H) and satisfies: σφ1(K)σ-1 = φ2(K) and a is an integer satisfying: σφ1(k)σ-1 = (φ2(k))a for any k in K. If x generates K, then every k in K can be expressed as xn for some integer n. Then we get: σφ1(k)σ-1 = σφ1(xn)σ-1= σφ1(x)nσ-1 = (σφ1(x)σ-1)n = (φ2(xa))n for some a in Z since σφ1(x)σ-1 is in φ2(K), every element of φ2(K) is φ2(k) for some k in K, and every k in K is xa for some a in Z = (φ2(xn))aensuring such an a exists. Now I've already proven that this mapping is a homomorphism, so it remains to show that it is a bijection. The "first component" of the homomorphism (which maps h to σ(h)) is clearly a bijection, so it remains to show that the mapping k to ka is bijective. Well, what actually needs to be shown is that there is some a satisfying: σφ1(k)σ-1 = (φ2(k))a for which the mapping is bijective, however it need not (and in some cases, can not) be the case that the mapping is a bijection for all a satisfying the above. Such a mapping is bijective if and only if xa is a generator of K. So it remains to prove that there is a generator y of K such that σφ1(x)σ-1 = φ2(y) In other words (and this seems to be all I've been able to do for this "second component" - continually phrase the problem in other words but not find a phrasing that makes the proof more evident), if an element κ1 of φ1(K) is conjugate to an element κ2 of φ2(K), then if κ1 is the image of a generator under φ1 then κ2 is the image of a generator under φ2. So it would suffice to prove that if an element κ1 of φ1(K) is conjugate to an element κ2 of φ2(K), then if κ2 is not the image of a generator under φ2 then κ1 is not the image of a generator under φ1. Since φ1(K) and φ2(K) are conjugate, they are isomorphic, hence φ1 and φ2 have identical kernels and K/Ker(φ1) = K/Ker(φ2) = K'. Well I have all these facts and rephrasings, I can't seem to figure out what to do though. Any help?