Isomorphic Semi-Direct Products

[AKG]

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_φ1 K and H x_φ2 K are isomorphic if φ₁(K) and φ₂(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:

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(x^a))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 x^a for some a in Z

= (φ₂(xⁿ))^a​

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^a 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 x^a 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?

 

[AKG]

I think I might have an idea. If φ₁^-1(κ₁) contains a generator of K but φ₂^-1(κ₂) does not, but if κ₁ and κ₂ are still conjugate, then κ₂ is still a generator for φ₂(K) because κ₁ is. However, if κ₂ is a generator, then every element of φ₂(K) can be expressed as a power of κ₂, and so ¹the pre-image of any such element can be expressed as a power of φ₂^-1(κ₂). But since φ₂^-1(κ₂) contains no generators, ²neither does any of its powers. The collection of the pre-images for each element of φ₂(K) should fill out all of K', but if no coset in this collection contains a generator, then we are missing some cosets, and this is a contradiction. Therefore if φ₁^-1(κ₁) contains a generator of K then so does φ₂^-1(κ₂).

I'd like to prove points ¹ and ². Suppose k₂ is in φ₂^-1(κ₂). Since the identity is in K', k₂ is clearly in k₂K', and since cosets are either disjoint or identical, k₂K' = φ₂^-1(κ₂). It's obvious that the pre-image of a power of κ₂, κ₂ⁿ, is in (k₂K')ⁿ since (k₂K')ⁿ = k₂ⁿK' so k₂ⁿ is in (k₂K')ⁿ and φ₂(k₂ⁿ) = φ₂(k₂)ⁿ = κ₂ⁿ.

I have a feeling that ² may not even be true. Since K is cyclic, assuming it's finite, then K' is cyclic and has some element of lowest order m. φ₂^-1(κ₂) is some coset k₂K' for various possible choices of k₂, but let's choose the k₂ with the lowest possible order n. The order of the product of two elements of an abelian (and hence this is true for cyclic groups) is the least common multiple of the orders of the elements being multiplied. If lcm(m,n) is not equal to |K|, then knowing that the order of any power of k₂ will divide the order of k₂, we will have no element in any power of k₂K' whose order is |K|.

If K is infinite, then there actually can be cosets that contain no generators but whose powers do, for example if K = Z, K' = 5Z, and we take the coset 2+K' = {..., -10+2, -5+2, 0+2, 5+2, 10+2, ...} = {-8, -3, 2, 7, 12, ...}, we see it has no generators (it doesn't contain 1 or -1) but 2(2+K') = 4+K' = {..., -10+4, -5+4, 0+4, 5+4, 10+4, ...} = {-6, -1, 4, 9, 14, ...} does contains the generator -1. Note here that 2(2+K') is the second power of (2+K'), it is just written in additive notation instead of multiplicative notation since addition is the group operation. So this approach won't work here, but I have a feeling that if I go back to what I was originally trying to prove (with pre-images of conjugate automorphisms either both containing generators or neither doing so) is easier even without the above when K is simply Z.

If no generator of K maps to κ₂, then no generator maps to the other generator, κ₂^-1 since if x maps to κ₂^-1 then x^-1 maps to κ₂. But if no generator of K maps to κ₂ then nothing does, contradiction, so some generator must map to it. So again, we see that every generator is the image of some generator of K, which is what we wanted. Done.

 

[thepatientmental]

The proof provided is on the right track, but it can be made more clear by breaking it down into smaller steps. Here is a step-by-step explanation of the proof:

1. First, we need to show that the function ψ is well-defined, that is, it maps elements of H xφ1 K to elements of H xφ2 K. This is done by showing that σ(h) and ka are in H xφ2 K for any (h,k) in H xφ1 K.

2. Next, we need to show that ψ is a homomorphism. This is done by showing that for any (h1, k1) and (h2, k2) in H xφ1 K, ψ(h1, k1) * ψ(h2, k2) = ψ(h1h2, k1k2).

3. Now, we need to show that ψ is injective, that is, it maps distinct elements of H xφ1 K to distinct elements of H xφ2 K. This is done by showing that if ψ(h1, k1) = ψ(h2, k2), then (h1, k1) = (h2, k2).

4. Finally, we need to show that ψ is surjective, that is, every element of H xφ2 K is mapped to by ψ. This is done by showing that for any (h, k) in H xφ2 K, there exists an element (h', k') in H xφ1 K such that ψ(h', k') = (h, k).

5. To show that such an element (h', k') exists, we use the fact that K is cyclic, generated by some element x. Since φ1(K) and φ2(K) are conjugate, there exists an automorphism σ of H such that σφ1(x)σ-1 = φ2(y) for some y in K. This means that for any k in K, there exists an integer a such that σφ1(k)σ-1 = (φ2(k))a.

6. Now, if we choose k' = xa, we have σφ1(k')σ-1 = (φ2(xa))a = (φ2(k))a = σφ1(k)σ-1. This means that ψ(