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(adsbygoogle = window.adsbygoogle || []).push({}); _{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}K

defined 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}= σφensuring 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 inZsince 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 inZ

= (φ_{2}(x^{n}))^{a}^{a}is bijective. Well, what actually needs to be shown is that there issomea 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 that

σφ_{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 inother wordsbut 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}isnotthe image of a generator under φ_{2}then κ_{1}isnotthe 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?

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Isomorphic Semi-Direct Products

**Physics Forums | Science Articles, Homework Help, Discussion**