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Thread: Classifying Groups

  1. #1
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    Classifying Groups

    In my course, we didn't have time to cover classifying groups but we are expect to know it. I need some help in how this is done.

    |G|=70=2\times 5\times 7

    Let n_i be the number of \text{Syl}_p(G) for i=2,5,7

    n_i\equiv 1 \ (\text{mod} \ i) \ \text{and} \ n_i|\frac{70}{i}

    n_2=1,5,7,35
    n_5=1
    n_7=1

    So G is not simple and the Sylow 5 and 7 are normal in G. Now how do I classify the groups of order 70?
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    MHF Contributor Drexel28's Avatar
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    Re: Classifying Groups

    Quote Originally Posted by dwsmith View Post
    In my course, we didn't have time to cover classifying groups but we are expect to know it. I need some help in how this is done.

    |G|=70=2\times 5\times 7

    Let n_i be the number of \text{Syl}_p(G) for i=2,5,7

    n_i\equiv 1 \ (\text{mod} \ i) \ \text{and} \ n_i|\frac{70}{i}

    n_2=1,5,7,35
    n_5=1
    n_7=1

    So G is not simple and the Sylow 5 and 7 are normal in G. Now how do I classify the groups of order 70?
    This is not a nice problem. Let P,Q be the Sylow 5,7-subgroups of G respectively. Note that they are both normal, and so their product N is normal in G and of of order 35. Now, note that trivially if T denotes the Sylow 2 subgroup of G then N\cap T=\{1\}, NT=G where both follow from order arguments. Thus, G\cong T\rtimes_\varphi N. Thus, we are left to classify homomorphisms \varphi:T\to\text{Aut}(N) which induce different semidirect products. But, note that \text{Aut}(N)\cong \text{Aut}(\mathbb{Z}_5)\times\text{Aut}(\mathbb{Z  }_7)\cong \mathbb{Z}_4\times\mathbb{Z}_6, since any group of order 35 is cyclic, and so isomorphic to \mathbb{Z}_5\times\mathbb{Z}_7, \text{Aut} splits accross the product of cyclic groups, and \text{Aut}(\mathbb{Z}_p)\cong\mathbb{Z}_{p-1} for every prime p . And, since T\cong\mathbb{Z}_2 we can count the number of possible homomorphisms T\to\text{Aut}(N) by counting the number of possible homomorphisms \mathbb{Z}_2\to\mathbb{Z}_4\times\mathbb{Z}_6. But, this clearly corresponds to the number of elements of order two in \mathbb{Z}_4\times\mathbb{Z}_6 which is four. Thus, there are at most four homomorphisms \varphi:T\to\text{Aut}(N) and so at most four non-isomorphic groups of order 70. That said, the groups \mathbb{Z}_{70} (the only possible cyclic group), D_{7}\times \mathbb{Z}_5, D_5\times\mathbb{Z}_7, and D_{35} are four non-isomorphic groups of 35. Thus, these are all the possible groups of order 70.
    Last edited by Drexel28; Dec 3rd 2011 at 03:01 PM.
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    Re: Classifying Groups

    Why are you forming the semi-direct product of T and Aut(N)?
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    MHF Contributor Drexel28's Avatar
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    Re: Classifying Groups

    Quote Originally Posted by dwsmith View Post
    Why are you forming the semi-direct product of T and Aut(N)?
    Look here and see that all the necessary conditions for G to be a semidirect product of T by N are satisfied. This is a common technique for classifying groups of a given order--reduce the problem to classifying semidirect products of certain subgroups.
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    Re: Classifying Groups

    i believe he meant:

    G\cong N\rtimes_\varphi T, for which we need \varphi \in \text{Hom}_{\text{\bf{Grp}}}(T,\text{Aut}(N))

    the idea is, if you have two subgroups of G, H and N, but only one of them is normal, you can still use G = HN to "decompose" G (as long as H∩N = {e}).

    the classic example is D_{2n} the dihedral group with 2n elements. here, \langle r \rangle \cong \mathbb{Z}_n and \langle s \rangle \cong \mathbb{Z}_2.

    now we have two obvious choices for a homomorphism:

    \varphi:\mathbb{Z}_2 \to \text{Aut}(\mathbb{Z}_n), one is:

    \varphi(k) = \text{id}_{\mathbb{Z}_n} for k = 0,1 (the trivial homomorphism, where φ maps everything to the identity automorphism)

    this leads to the direct product \mathbb{Z}_n \times \mathbb{Z}_2 (which is abelian).

    the other "obvious" choice is:

    \varphi(0) = \text{id}_{\mathbb{Z}_n}\ (k \to k)
    \varphi(1) = (k \to -k) (the inversion automorphism).

    this leads to \mathbb{Z}_n \rtimes \mathbb{Z}_2, which has the multiplication:

    (k,m)*(k',m') = (k+(-1)^mk',m+m')

    in particular:

    (0,1)*(1,0) = (-1,1) = (-1,0)*(0,1) which is the "semi-direct product" form of the relation: sr = r^{-1}s.

    it turns out that this is equivalent to letting H act on N by conjugation (which works because/when N is normal). this tells us quite a bit about G from knowing things about H and N.

    ther is one caveat, however. it may be that G = HN (where H and N have trivial intersection), with N normal in G, and that G' = H'N' (with N' normal in G', and H' and N' have trivial intersection), and further: that H ≅ H', N ≅ N', but G is NOT isomorphic to G'. for example:

    \mathbb{Z}_n \times \mathbb{Z}_2 \not \cong D_{2n}

    in general, groups of order 2^n are quite difficult to classify. for example there are 49,487,365,422 non-isomorphic groups of order 1024 (i think there are 51 non-isomorphic groups of order 32, classifying groups is a "hard" problem (not impossible, though)).
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