1. ## 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!!!

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.