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Thread: Direct Sum Problem

  1. #1
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    Direct Sum Problem

    Suppose that $\displaystyle G = C_6 \times C_6 \times C_6$, $\displaystyle C_6$ generated by an element r. Let $\displaystyle H = \{ (a,a,a) \in G : a \in C_6 \}$, $\displaystyle
    I = \{ (a,a,1) \in G : a \in C_6 \}
    $$\displaystyle J_1 = \{ (a,1,1) \in G : a \in C_6 \}$, $\displaystyle J_2 = \{ (1,a,1) \in G : a \in C_6 \}$, $\displaystyle J_3 = \{ (1,1,a) \in G : a \in C_6 \}$

    Prove that $\displaystyle G=H \oplus I \oplus J_i \ \ \ \ i=1,2 $, but $\displaystyle G \neq H \oplus I \oplus J_3 $

    Proof.

    Suppose that $\displaystyle g \in G$, then $\displaystyle g = abc \ \ \ a,b,c \in C_6 $, I want to get g to somehow be in the form [tex]
    Last edited by tttcomrader; Oct 2nd 2008 at 05:08 AM.
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  2. #2
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    Quote Originally Posted by tttcomrader View Post
    Suppose that $\displaystyle G = C_6 \times C_6 \times C_6$, $\displaystyle C_6$ generated by an element r. Let $\displaystyle H = \{ (a,a,a) \in G : a \in C_6 \}$, $\displaystyle J_1 = \{ (a,1,1) \in G : a \in C_6 \}$, $\displaystyle J_2 = \{ (1,a,1) \in G : a \in C_6 \}$, $\displaystyle J_3 = \{ (1,1,a) \in G : a \in C_6 \}$

    Prove that $\displaystyle G=H \oplus I \oplus J_i \ \ \ \ i=1,2 $, but $\displaystyle G \neq H \oplus I \oplus J_3 $

    Proof.

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

    Similarly with the other cases.
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    In the case of $\displaystyle G = H \oplus I \oplus J_3 $, it doesn't work because it is not unique? I have $\displaystyle 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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  6. #6
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    Quote Originally Posted by tttcomrader View Post
    In the case of $\displaystyle G = H \oplus I \oplus J_3 $, it doesn't work because it is not unique? I have $\displaystyle 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 $\displaystyle g=(x,y,z)$ we can be written in this form is $\displaystyle x=y$. So if $\displaystyle x\not = y$ then we cannot write $\displaystyle (x,y,z)$ in the form $\displaystyle (a,a,a)(b,b,1)(1,1,c)$. Therefore, $\displaystyle G\not = H\oplus I \oplus J_3$.
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