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Isomorphic Semi-Direct Products

  1. Aug 22, 2005 #1


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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 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-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))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 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?
    Last edited: Aug 22, 2005
  2. jcsd
  3. Aug 22, 2005 #2


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    I think I might have an idea. If φ1-11) contains a generator of K but φ2-12) 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 φ2-12). But since φ2-12) 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 φ1-11) contains a generator of K then so does φ2-12).

    I'd like to prove points 1 and 2. Suppose k2 is in φ2-12). Since the identity is in K', k2 is clearly in k2K', and since cosets are either disjoint or identical, k2K' = φ2-12). 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. φ2-12) 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.
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