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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 :)

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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 $\operatorname{lcm}(2,3)= 6$.

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  • $\begingroup$ I think you mean $G_2 = \mathbb Z_4$ $\endgroup$ – Karl Damgaard Asmussen Nov 3 '13 at 22:35
  • $\begingroup$ @Karl Daamgaard Asmussen : Indeed. Thanks :) $\endgroup$ – Joel Cohen Nov 3 '13 at 23:10
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Observe that $2$ and $4$ are not relatively prime. In general, assume that $(a,b)$ is an element in the group $\mathbb{Z}_m\times\mathbb{Z}_n$, and let $c=\operatorname{lcm}(m,n)$. Then $(a,b)^c=(0,0)$, so the order of $(a,b)$ is a divisor of $c$. But $c$ is strictly smaller than the number of elements $mn$ if $m$ and $n$ are not relatively prime, so the group cannot be cyclic.

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A good theorem to use in checking the direct product of cyclic groups is the following:

$$\mathbb Z_m\times \mathbb Z_n \cong \mathbb Z_{(mn)} \; \underbrace{\iff}_{\text{ IF AND ONLY IF }\;} \gcd(m, n) = 1$$

This holds for any number of factors: $\mathbb Z_{n_1} \times \mathbb Z_{n_2} \times \cdots \times \mathbb Z_{n_n} \cong \mathbb Z_{n_1\cdot n_2\cdots n_n}$ if and only if the $n_i$ are pairwise prime.

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    $\begingroup$ $\Bbb Z_{mn}$ is always cyclic, with generator $1$. (So your first equality is false) $\endgroup$ – Pedro Tamaroff Nov 3 '13 at 13:33
  • $\begingroup$ @Pedro I was just reminding the OP that $\mathbb Z_{mn}$ is always cyclic. Lighten up. $\endgroup$ – Namaste Nov 3 '13 at 13:38
  • $\begingroup$ What do you mean? $\endgroup$ – Pedro Tamaroff Nov 3 '13 at 13:38
  • $\begingroup$ @Pedro My earlier "aside" that $\mathbb Z_{mn}$ is cyclic was not intended to be dependent on $\gcd(m, n)$; it was simply a parenthetical remark that $\mathbb Z_{mn}$ is cyclic. $\endgroup$ – Namaste Nov 3 '13 at 13:41
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Your proof is perfectly fine, and it's probably the easiest and most direct way to show the result with such a small group. For a bigger group (say $\mathbb{Z}/12345\mathbb{Z} \times \mathbb{Z}/48274\mathbb{Z}$, you only need to slightly adapt the argument: you're not going to check each element one by one in this case.

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  • $\begingroup$ Could you please describe the symbols you have used? I'm not familiar with them. $\endgroup$ – polarise Dec 11 '14 at 4:47
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Lemma: A group $G$ is cyclic if and only if $G$ has a uniqe subgroup for each $d$ dividing $|G|$.

In our case, It has atleast two subgroup of order $2$, $\mathbb Z_2\times e$ and $e\times H$ where $H$ is subgroup of $\mathbb Z_4$ with order $2$ so it can not be cyclic.

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The Invariant factor decomposition of finitely-generated abelian groups directly tells you that the group is not cyclic. Though killing a fly with a rocket launcher may not be considered as "elegant", it gives a method for larger groups.

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If $G$ and $H$ are cyclic groups then $G \oplus H $ is cyclic if and only if $\gcd (|G|, |H|) =1$

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