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I Is A4 isomorphic to D3xZ2

  1. May 2, 2016 #1
    I have a hard time trying to figure out these types of problems.
    Usually, we are supposed to compare their orders, compare if they are both abelian, or see if one group has element of order n which the other group does not have.

    So in this case, A4 has order 12 and D3xZ2 also has order 12. So this doesn't help.
    An alternating group is non-abelian for n<=3 so A4 is non-abelian. D3 is non-abelian as well and the product of non-abelian to a group is non-abelian (?). So this doesn't help.
    So what I'm thinking is, it's either "A4 has an element of order 4, but D3xZ2 does not", or "D3xZ2 has element order 6 but A4 does not".
    So my question is, how do we find that D3xZ2 does or does not contain element of order 6? What if it was Z3xZ2 or S4xD3?

    Are there any tricks to figuring out if a product of two groups has element of order n so that I can compare it with the other group, in this case A4?

    Thank you
     
  2. jcsd
  3. May 2, 2016 #2

    fresh_42

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    ##A_4## doesn't have a subgroup of order 6.
     
  4. May 2, 2016 #3
    Okay, but how do we show that D3xZ2 has an element of order 6?
     
  5. May 2, 2016 #4

    fresh_42

    Staff: Mentor

    I assumed ##D_3## to denote the dihedral group. Then ##D_3## is a subgroup of ##D_3 \times \mathbb{Z}_2## per definition of the direct product. An isomorphism would map one subgroup to another without changing the number of elements. So ##D_3## (with six elements) cannot be isomorphically mapped into ##A_4##.
     
  6. May 3, 2016 #5

    mathwonk

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    in a product group, the two factors act independently of each other, so if x is an element of D3 of order 3, and y is an element of Z2 of order 2, then (x,y) has order 6. try multiplying it by itself and see what happens. i.e. (x,y)^n = (x^n,y^n)......so.....

    one can also see what order are the elements of A4 by representing it as the rotations of a tetrahedron, (which permute the four vertices). Then one sees there are eight elements of order 3 (each fixing some vertex) and three elements of order 2, (each intertchanging two disjoint pairs of vertices).
     
  7. May 3, 2016 #6
    I understand that the order of product group is the product of the individual orders.

    But what I'm confused about is, say, why is Z12 not isomorphic to Z6xZ2? They both have order 24, but why can we say that Z6xZ2 does not have element of order 12? I thought possible element of order divides the order of the group by Lagrange's theorem, so since |Z6xZ2|=24. doesn't Z6xZ2 have element of order 12?
    Thanks
     
  8. May 3, 2016 #7

    fresh_42

    Staff: Mentor

    No it doesn't. If we have an element ##(a,b) \in ℤ_6 \times ℤ_2## and addition as group operation then we get
    $$(a,b)+(a,b)+(a,b)+(a,b)+(a,b)+(a,b) = ((a+a+a+a+a+a),(b+b)+(b+b)+(b+b)) = (0, 0+0+0) = (0,0) $$
    that is, six is the highest possible order and hence no isomorphism to ##ℤ_{12}## can exist.

    You can have different groups with the same number of elements. The smallest examples are ##ℤ_4## and ##ℤ_2^2 = ℤ_2 \times ℤ_2 ≅ D_2 ≅ V_4##. The last one, the dihedral group ##D_2## is also known as Klein's four-group ##V_4##. All elements are of order 2 there. The formal definition is, e.g. ##V_4 = <a,b | a^2 = b^2 = (ab)^2 = 1>##. You could write down both multiplication (or addition) tables of both groups with four elements as an exercise and look for the differences.

    Edit: Here you can find a list of all small groups: https://en.wikipedia.org/wiki/List_of_small_groups

    Edit 2: ##ℤ_{12} ≅ ℤ_3 \times ℤ_4## Here you get an element of order 12, because 3 and 4 are coprime. Therefore 12 is the least common multiple to get both components simultaneously and for sure to zero (as neutral element).
     
    Last edited: May 3, 2016
  9. May 4, 2016 #8
    Thank you. Does the coprime rule saying Z12 is isomorphic to Z4xZ3 work with groups other than Z (I can't really think of an example)?
     
  10. May 4, 2016 #9

    fresh_42

    Staff: Mentor

    For cyclic groups ##ℤ_p## and ##ℤ_q## with coprime ##p,q## holds ##ℤ_p \times ℤ_q ≅ ℤ_{pq}## because they have single generators of appropriate orders.
    I'm not sure what you meant by "work with other groups". There are many non-abelian finite groups or semidirect products of finite groups to build all possible variations of finite groups. The basic property of cyclic groups is that there is only one generator. Things are much different if there are more than one.
     
  11. May 5, 2016 #10

    micromass

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