Show that $\mathbb{Z}_{2} \times \mathbb{Z}_{4}$ is not a cyclic group. This question is from the book 'Of Abstract Algebra' by Pinter. Now $\mathbb{Z}_{2} \times \mathbb{Z}_{4}$ containt 8 elements. I found them to be as follows:
$$(0,0),(0,1),(0,2),(0,3),(1,0),(1,1),(1,2),(1,3)$$
Now if $\mathbb{Z}_{2} \times \mathbb{Z}_{4}$ were cyclic, one of these elements should be a generator of the entire group. However, for every element listed we find that repeated repetitions of the group operation yields the identity element already before having to apply the group operation eight times. Therefore, using any of these elements as a single element would result in a subgroup of order less than 8, in this case it will have either order 4, order 2 or order 1. Therefore the group is not cyclic. Now I am quite sure this constitutes a correct proof but I am wondering if a more elegant way exist to show this result. This might be especially useful when investigating similar questions for much larger groups. Thanks in advance
P.S. I have been posting more questions from this book because I find them very interesting, but I am reading it on my own so I am sometimes unsure if the methods I use are the most elegant and quick. So far I have received great help and I am grateful to the stackexchange community for this 🙂
Best Answer
I think your approach of studying the order of the elements is the right one. If you want to do that in a more systematic way, you may study order of elements in a direct product. Namely, say we have $G_1, \ldots, G_r$ some groups, some element $g = (g_1, \ldots g_r) \in G_1 \times \ldots \times G_r$ in the direct product, and we want to know the order of $g$ inside $G_1 \times \ldots \times G_r$. It turns out that if $n_i$ denotes the order $g_i \in G_i$, then the order of $g$ is $\operatorname{lcm}(n_1,\ldots,n_r)$. In particular, since the order $g_i$ divides $|G_i|$ (the number of elements of the group), then order of $g$ divides $\operatorname{lcm}(|G_1|,\ldots,|G_r|)$.
In your example, take $G_1 = \mathbb{Z}_2$ and $G_2 = \mathbb{Z}_4$. We know the order of any element inside $G_1 \times G_2$ divides $\operatorname{lcm}(|G_1|,|G_2|) = \operatorname{lcm}(2,4) = 4$. So there is no element of order $8$.
Using the same principle we may see for example that $ \mathbb{Z}_2 \times \mathbb{Z}_3$ is cyclic. Indeed, we know $(1,1)$ has order $\operatorname{lcm}(2,3)= 6$.