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Math Help - Cyclic subgroups

  1. #1
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    Cyclic subgroups

    Let C_n denote the cyclic group of order n and let  G = C_6 \times C_6 \times C_6 , assume  C_6 is generated by the element r.

    1. Prove that H = \{ (a,a,a) \in G : a \in C_6 \} , and  J= \{ (1,a,a) \in G : a \in C_6 \} are cyclic subgroups of G of order 6.

    Proof.

    I think what is throwing me off is the notations here. Both H and J are cyclic since all of their elements are drew from a cyclic group (is this enough?).

    Pick (x,x,x),(y,y,y) \in H , we have  (x,x,x)(y,y,y)=(xy,xy,xy) \in H since xy \in C_6. The inverses, of course, would also have to be in there as well. Same thing for J.

    Is this okay?

    2. Let J_1 = \{ (a,1,1) \in G : a \in C_6 \}  , J_2 = \{ (1,a,1) \in G : a \in C_6 \} , J_3 = \{ (1,1,a) \in G : a \in C_6 \} , prove that G = J_1 \oplus J_2 \oplus J_3

    I can tell that this is true, but I'm having trouble trying to show it is. Since G = C_6 \times C_6 \times C_6 , each one of those cyclic subgroups would constitute a coordinate here, but I'm confused in what to write to prove it.

    Thanks!!!
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  2. #2
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    Quote Originally Posted by tttcomrader View Post
    1. Prove that H = \{ (a,a,a) \in G : a \in C_6 \} , and  J= \{ (1,a,a) \in G : a \in C_6 \} are cyclic subgroups of G of order 6.
    How many elements in H it is six because there are six elements in C_6. The next thing you need to do is just check the definition of being a subgroup. I start you off. First closure. If x\in H and y\in H then x=(a,a,a) and y=(b,b,b) then x+y = (a+b,a+b,a+b) but a+b\in C_6 and so x+y \in H. Thus, H is closed.
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