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AKG

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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 x

ψ : H x

defined by:

ψ(h, k) = (σ(h), k

where σ is in Aut(H) and satisfies:

σφ

and a is an integer satisfying:

σφ

for any k in K. If x generates K, then every k in K can be expressed as x

σφ

σφ

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 x

σφ

In other words (and this seems to be all I've been able to do for this "second component" - continually phrase the problem in

Since φ

Well I have all these facts and rephrasings, I can't seem to figure out what to do though. Any help?

_{φ1}K and H x_{φ2}K are isomorphic if φ_{1}(K) and φ_{2}(K) are conjugate by producing an explicit isomorphism. They suggest the function:ψ : H x

_{φ1}K → H x_{φ2}Kdefined by:

ψ(h, k) = (σ(h), k

^{a})where σ is in Aut(H) and satisfies:

σφ

_{1}(K)σ^{-1}= φ_{2}(K)and a is an integer satisfying:

σφ

_{1}(k)σ^{-1}= (φ_{2}(k))^{a}for any k in K. If x generates K, then every k in K can be expressed as x

^{n}for some integer n. Then we get:σφ

_{1}(k)σ^{-1}= σφ_{1}(x^{n})σ^{-1}= σφ

= (σφ

= (φ

= (φ

ensuring 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 k_{1}(x)^{n}σ^{-1}= (σφ

_{1}(x)σ^{-1})^{n}= (φ

_{2}(x^{a}))^{n}for some a in**Z**since σφ_{1}(x)σ^{-1}is in φ_{2}(K), every element of φ_{2}(K) is φ_{2}(k) for some k in K, and every k in K is x^{a}for some a in**Z**= (φ

_{2}(x^{n}))^{a}^{a}is bijective. Well, what actually needs to be shown is that there is**some**a satisfying:σφ

_{1}(k)σ^{-1}= (φ_{2}(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 x

^{a}is a generator of K. So it remains to prove that there is a generator y of K such thatσφ

_{1}(x)σ^{-1}= φ_{2}(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 κ_{1}of φ_{1}(K) is conjugate to an element κ_{2}of φ_{2}(K), then if κ_{1}is the image of a generator under φ_{1}then κ_{2}is the image of a generator under φ_{2}. So it would suffice to prove that if an element κ_{1}of φ_{1}(K) is conjugate to an element κ_{2}of φ_{2}(K), then if κ_{2}is**not**the image of a generator under φ_{2}then κ_{1}is**not**the image of a generator under φ_{1}.Since φ

_{1}(K) and φ_{2}(K) are conjugate, they are isomorphic, hence φ_{1}and φ_{2}have identical kernels and K/Ker(φ_{1}) = K/Ker(φ_{2}) = K'.Well I have all these facts and rephrasings, I can't seem to figure out what to do though. Any help?

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