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Groups: cyclic, abelian

  1. Mar 16, 2010 #1
    1. The problem statement, all variables and given/known data

    Let G1 and G2 be groups, let G = G1 x G2 and define the binary operation on G by
    Prove that this makes G into a group. Prove G is abelian iff G1 and G2 are abelian.

    Hence or otherwise give examples of a non-cyclic abelian group of order 8 and a non-abelian group of order 42.

    2. Relevant equations

    3. The attempt at a solution

    I have done the 1st part of this question and I'm just struggling with the examples. From reading around the subject I think a non-cyclic abelian group of order 8 would be Z2 x Z2 x Z2 where Z2 is the integers modulo 2 under addition. However, I don't really understand this.

    Also, I'm unsure how to tackle the non-abelian group of order 42.

    Thanks :)
  2. jcsd
  3. Mar 16, 2010 #2


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    I think you are right with Z2xZ2xZ2. Why isn't it cyclic? For the other one think of two groups whose order multiplies to 42, and at least one of which isn't abelian.
  4. Mar 16, 2010 #3
    I'm not sure I fully understand what cyclic actually means. If we have any element in the group, in our case it will look like (a,b,c) where a,b,c are either 0 or 1. When we square it we will get 1, will this be (1,1,1) for us? But this doesn't actually make sense, if we have (1,0,0)(1,0,0) = (1,0,0)? I think I'm very confused.

    For the order 42 group, could I use sym(3) x Z7 (Z7 is the integers modulo 7 under addition) and sym(3) is not abelian. Again, I really don't understand why this works, if it does.
  5. Mar 16, 2010 #4


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    When they define the group operation by (a1,a2)(b1,b2):=(a1b1,a2b2) a1b1 doesn't mean multiply, it means combine a1 and b1 according to whatever the group operation is. For Z2 that's addition mod 2. (1,1,1) is not the identity of Z2xZ2xZ2. Does that help?
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