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The operation here is taken to be addition. Clearly $\mathbb Z$ is cyclic since $\mathbb Z = \langle 1 \rangle = \langle -1 \rangle.$ I was then looking at a question that asked if $\mathbb Z^4 \times \mathbb Z^5$ was cyclic, which I think is equivalent to the question of whether $\mathbb Z^9$ is cyclic (correct me if I'm wrong). My thinking is no, since the only choice for a potential generator would be $(1,1,\ldots, 1)$ or $(-1, -1, \ldots, -1)$, but this only generates the cyclic subgroup of $\mathbb Z^n$ of the form $\{(a,a,\ldots, a): a \in \mathbb Z\}$. Then I think this argument could extend to saying that $\mathbb Z^n$ is not cyclic for $n>1.$ Is that correct?

Update: As was suggested in a comment, the original question I looked at actually was asking if $\mathbb Z_4\times \mathbb Z_5,$ also written $\mathbb Z/4\mathbb Z\times \mathbb Z/5\mathbb Z$, is cyclic. It is, since $\gcd(4,5)=1$. One generator is simply $(1,1).$

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    $\begingroup$ You have not explained why you are only considering two specific elements as potential generators. It is up to you to prove that no element at all can be a generator, not just assert that some can't. $\endgroup$ – Slade Sep 28 '14 at 6:16
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    $\begingroup$ Is it possible the question was asking if $(\mathbb{Z}/4\mathbb{Z})\times(\mathbb{Z}/5\mathbb{Z})$ was cyclic? This is sometimes written as $\mathbb{Z}_4\times\mathbb{Z}_5$, and would make for a more interesting question. $\endgroup$ – Mike Earnest Sep 28 '14 at 6:17
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If $\Bbb{Z}^n$ is cyclic for $n\ge 2$, then $\Bbb{Z}^2$ is also cyclic since $\Bbb{Z}^n$ contains a subgroup isomorphic to $\Bbb{Z}^2$. So we only check that $\Bbb{Z}^2$ is not cyclic. But if $(a,b)$ is nonzero element of $\Bbb{Z}^2$ then $(-b,a)$ is not generated by $(a,b)$.

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If $\mathbb{Z}^n$ is generated by the element $x$, then the vector space $\mathbb{Q}\otimes \mathbb{Z}^n \cong \mathbb{Q}^n$ is generated by the element $1\otimes x$.

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A little hand-wavy but yeah it's correct. Just finish with the fact that if $(z_1,\ldots,z_n)$ generates $\mathbb{Z}^n$ and $z_1=0$ then $(1,0,\ldots,0)$ is not in the subgroup generated by $(z_1,\ldots,z_n)$.

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Suppose $\mathbb{Z}^n$ is cyclic, generated by an element $(m_1,…,m_n) \in \mathbb{Z}^n$. This means that every element of $\mathbb{Z}^n$ has the form $(km_1,\ldots,km_n)$ for some integer $k$. So all you have to do is write down one element of $\mathbb{Z}^n$ which does not have that form. That's easy enough if each of $m_1,\ldots,m_n$ equals zero. So assume that some $m_j$ is nonzero, let $m_i$ be any other coordinate, and replace $m_i$ by $m_i+m_j$.

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