1. ## 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]

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 $\displaystyle I$ ?

3. sorry, $\displaystyle I = \{ (a,a,1) \in G : a \in C_6 \}$

4. 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.

5. 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.

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, $\displaystyle G\not = H\oplus I \oplus J_3$.