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Direct product/sum of groups

  1. Apr 3, 2010 #1
    I have a pretty basic question about direct sum/product of groups.

    Say you were given the group (Z4 x Z2, +mod2). Now I know that Z4 x Z2 is given by { (0,0), (0,1), (1,0), (1,1), (2,0), (2,1), (3,0), (3,1) }. So now if you were going to add together two of the elements using the binary operation +mod2, e.g. doing (1,1) +mod2 (2,1). Does this give you:

    (1,1) +mod2 (2,1) = (1+2,1+1) = (3,2) = (1,0)?
    I'm pretty sure that this is correct, but I thought another possibility might have been that you add the first two elements in mod4 and the second two in mod 2

    e.g. (1,1) + (2,1) = (3,2) = (3,0).

    Help clarifying would be super.
     
  2. jcsd
  3. Apr 3, 2010 #2
    The definition of the direct product of two groups G and H is that you multiply (or add in the abelian case) componentwise using the operation in the given component. So in Z4 x Z2 the operation is the second one you did, e.g. (1,1) + (2,1) = (3,0). Of course to get the first thing you had (e.g. (1,1) + (2,1) = (1,0)) you could redefine the operation in Z4 to be addition mod 4, but then you would really have Z2 (so the direct product would be Z2 x Z2). For example with this operation on Z4 we are forced for 3 and 1 to be the same element since 3 - 1 = 2 = 0.
     
  4. Apr 3, 2010 #3
    thanks thats helpful.

    although i still dont quite get the meaning of the +mod2 part. When the group is defined as (Z4 x Z2, +mod2), doesn't the +mod2 tell you the binary operation by which you combine elements of the set? if thats the case, when you add componentwise how come you are adding the first two together in mod4 but the second two in mod2?
     
    Last edited: Apr 3, 2010
  5. Apr 5, 2010 #4
    never mind, i dont actually think it makes any sense to say (Z4 x Z2, +mod2)
     
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