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Math Help - Direct Sum Problem

  1. #1
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    Direct Sum Problem

    Suppose that G = C_6 \times C_6 \times C_6, C_6 generated by an element r. Let H = \{ (a,a,a) \in G : a \in C_6 \}, <br />
I = \{ (a,a,1) \in G : a \in C_6 \}<br />
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=H \oplus I \oplus J_i \ \ \ \ i=1,2 , but G \neq H \oplus I \oplus J_3

    Proof.

    Suppose that g \in G, then g = abc \ \ \ a,b,c \in C_6 , I want to get g to somehow be in the form [tex]
    Last edited by tttcomrader; October 2nd 2008 at 06:08 AM.
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  2. #2
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    Quote Originally Posted by tttcomrader View Post
    Suppose that G = C_6 \times C_6 \times C_6, C_6 generated by an element r. Let H = \{ (a,a,a) \in G : a \in C_6 \}, 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=H \oplus I \oplus J_i \ \ \ \ i=1,2 , but G \neq H \oplus I \oplus J_3

    Proof.

    Suppose that g \in G, then g = abc \ \ \ a,b,c \in C_6 , I want to get g to somehow be in the form [tex]
    What is I ?
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  3. #3
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    sorry, I = \{ (a,a,1) \in G : a \in C_6 \}
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    Note that (a,a,a)(b,b,1)(c,1,1) = (abc,ab,a).
    Now given (x,y,z) \in G we need to find unique a,b,c so that (abc,ab,a) = (x,y,z). This is of course true since we must have a=z,b=z^{-1}y, c= y^{-1}x.

    Similarly with the other cases.
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    In the case of G = H \oplus I \oplus J_3 , it doesn't work because it is not unique? I have g = (x,y,z) = (a,a,a)(b,b,1)(1,1,c)=(ab,ab,ac) So I have x=y, means it doesn't work?

    Is this right? thanks.
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    Quote Originally Posted by tttcomrader View Post
    In the case of G = H \oplus I \oplus J_3 , it doesn't work because it is not unique? I have g = (x,y,z) = (a,a,a)(b,b,1)(1,1,c)=(ab,ab,ac) So I have x=y, means it doesn't work?
    What this shows that the only element g=(x,y,z) we can be written in this form is x=y. So if x\not = y then we cannot write (x,y,z) in the form (a,a,a)(b,b,1)(1,1,c). Therefore, G\not = H\oplus I \oplus J_3.
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