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Isomorphic Type

  1. Sep 5, 2010 #1
    1. The problem statement, all variables and given/known data

    [PLAIN]http://img541.imageshack.us/img541/9880/34132542.gif [Broken]

    3. The attempt at a solution

    For part (a), I think since the order of an element g is the smallest integer n such that gn=e, we will have:

    8n mod 65 = 1 => n=4

    64n mod 65 = 1 => n=2

    14n mod 65 =1 => n=2

    Am I right so far?

    Now for part (b), what is it meant by an "isomorphic type"? And what do I need to do?
     
    Last edited by a moderator: May 4, 2017
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  3. Sep 5, 2010 #2

    HallsofIvy

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    It means, I believe the "type" of group that G is isomorphic to. Since G has 8 elements, it is, of course, isomorphic to a group of order [itex]8= 2^3[/itex]. What can you say about groups with exactly 8 elements. How many different "kinds" are there?
     
  4. Sep 5, 2010 #3
    Do you mean the following direct products:

    [tex]\mathbb{Z}_{2^3}[/tex]

    [tex]\mathbb{Z}_{2^2} \oplus \mathbb{Z}_{2}[/tex]

    [tex]\mathbb{Z}_{2} \oplus \mathbb{Z}_{2} \oplus \mathbb{Z}_{2}[/tex]

    Is that right?
     
  5. Sep 6, 2010 #4

    HallsofIvy

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    Yes. Now which of those is your G isomorphic to?
     
  6. Sep 7, 2010 #5
    G is a subgroup of U(65), not all of U(65). So I think

    [tex]U(65) \cong U(5.13) \cong U(5) \oplus U(13) \cong \mathbb{Z}_4 \oplus \mathbb{Z}_{12}[/tex]

    But the problem is that none of the 3 groups in my previous post will then be isomorphic to [tex]\mathbb{Z}_4 \oplus \mathbb{Z}_{12}[/tex]. E.g. since lcm(12,4)=12, [tex]\mathbb{Z}_4 \oplus \mathbb{Z}_{12}[/tex] has elements of order 12, but [tex]\mathbb{Z}_{2^3}[/tex] has no elements of order 12. So they are not isomorphic.

    So what is wrong? I'm confused... :confused:
     
  7. Sep 26, 2010 #6
    HallsofIvy, "by finding the isomorphic type" they mean write G as an external product of cyclic groups of prime power order.

    So, G is an abelian group of order 8=23. The fundamental theorem of Finite Abelian Groups implies that either [tex]G \cong \mathbb{Z}_8[/tex] or [tex]\mathbb{Z}_4 \oplus \mathbb{Z}_2[/tex] or [tex]\mathbb{Z}_2 \oplus \mathbb{Z}_2 \oplus \mathbb{Z}_2[/tex].

    From my work in part (a) I know that G has 2 distinct subgroups of order 2: [tex]\left\langle14 \right\rangle = \{ 1,14 \}[/tex] and [tex]\left\langle 64 \right\rangle= \{ 1,64 \}[/tex], and one subgroup of order 4: [tex]\left\langle 8 \right\rangle = \{ 1,8,57,64 \}[/tex].

    Does it follow that G is non-cyclic and [tex]G \cong \mathbb{Z}_2 \oplus \mathbb{Z}_2 \oplus \mathbb{Z}_4[/tex]? :rolleyes:
     
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