# Direct Sum Problem

• October 1st 2008, 05:31 AM
Direct Sum Problem
Suppose that $G = C_6 \times C_6 \times C_6$, $C_6$ generated by an element r. Let $H = \{ (a,a,a) \in G : a \in C_6 \}$, $
I = \{ (a,a,1) \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]
• October 1st 2008, 03:43 PM
ThePerfectHacker
Quote:

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 $I$ ?
• October 2nd 2008, 05:08 AM
sorry, $I = \{ (a,a,1) \in G : a \in C_6 \}$
• October 2nd 2008, 07:28 AM
ThePerfectHacker
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.
• October 7th 2008, 10:11 AM
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?
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, $G\not = H\oplus I \oplus J_3$.