# Thread: Cyclic subgroups

1. ## Cyclic subgroups

Let $\displaystyle C_n$ denote the cyclic group of order n and let $\displaystyle G = C_6 \times C_6 \times C_6$, assume $\displaystyle C_6$ is generated by the element r.

1. Prove that $\displaystyle H = \{ (a,a,a) \in G : a \in C_6 \}$, and $\displaystyle J= \{ (1,a,a) \in G : a \in C_6 \}$ 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 $\displaystyle (x,x,x),(y,y,y) \in H$, we have $\displaystyle (x,x,x)(y,y,y)=(xy,xy,xy) \in H$ since $\displaystyle xy \in C_6$. The inverses, of course, would also have to be in there as well. Same thing for J.

Is this okay?

2. Let $\displaystyle J_1 = \{ (a,1,1) \in G : a \in C_6 \}$$\displaystyle , J_2 = \{ (1,a,1) \in G : a \in C_6 \} , J_3 = \{ (1,1,a) \in G : a \in C_6 \}$, prove that $\displaystyle G = J_1 \oplus J_2 \oplus J_3$

I can tell that this is true, but I'm having trouble trying to show it is. Since $\displaystyle G = C_6 \times C_6 \times C_6$, each one of those cyclic subgroups would constitute a coordinate here, but I'm confused in what to write to prove it.

Thanks!!!

2. Originally Posted by tttcomrader
1. Prove that $\displaystyle H = \{ (a,a,a) \in G : a \in C_6 \}$, and $\displaystyle J= \{ (1,a,a) \in G : a \in C_6 \}$ are cyclic subgroups of G of order 6.
How many elements in $\displaystyle H$ it is six because there are six elements in $\displaystyle C_6$. The next thing you need to do is just check the definition of being a subgroup. I start you off. First closure. If $\displaystyle x\in H$ and $\displaystyle y\in H$ then $\displaystyle x=(a,a,a)$ and $\displaystyle y=(b,b,b)$ then $\displaystyle x+y = (a+b,a+b,a+b)$ but $\displaystyle a+b\in C_6$ and so $\displaystyle x+y \in H$. Thus, $\displaystyle H$ is closed.