How to prove that $\mathbb{Z/2Z\times Z/2Z}$ is not cyclic?

Question

How to prove that $\mathbb{Z/2Z\times Z/2Z}$ is not cyclic?

Here is what I did, what is wrong with it if anything and is there a better way to prove it? A group $G$ is cyclic if $\exists\ a \in G$ such that $\forall g \in G, \exists\ k \in N$ such that $g=a^k$.

Here an element $g$ of $G=\mathbb{Z/2Z\times Z/2Z}$ writes $g=(x,y)$ with $x,y \in \{\bar{0},\bar{1}\}$, ie $g \in \{(0,0),(0,1),(1,0),(1,1)\}$

and there exists no $a \in G$ and $k,k',k'' \in N$ such that $a^k=(0,0), a^{k'}=(0,1)$ and $a^{k''}=(1,0)$

Answer 1

What you did was fine.

Another way would be to prove that $\mathbb{Z}_2 \times \mathbb{Z}_2$ is not isomorph with $\mathbb{Z}_4$.

Since the latter is the only cyclic group of order $4$ (up to isomorphism), the result follows.

To actually prove they are not isomorph, you can look at the orders of the elements of both groups.

Answer 2

Because the group is so small, the easiest elementary way is probably finding the order of each nonzero element:

• $2\cdot(1,0) = (0,0)$
• $2\cdot(0,1) = (0,0)$
• $2\cdot(1,1) = (0,0)$

So we can conclude that every element has at most order $2$, so no element has order $4$, therefore no element generates the whole group, therefore the group is not cyclic.