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The full problem is as stated in the title. I am here to check if this is a valid proof. I thought it would be easiest using Linear Algebra.

Recall that an infinite cyclic group is isomorphic to $\mathbb{Z}$. We wish to show that we do not have an isomorphism between $\mathbb{ZxZ\;and\;Z}$. Note that $\mathbb{ZxZ}$ is an infinite group (under addition of course). Now, in order for there to even be potential for an isomorphism, two spaces must have equal dimension. Since the $\mathbb{dim(\mathbb{ZxZ})}=2>\mathbb{dim(\mathbb{Z})}=1$, we know that $\nexists$ an isomorphism between our spaces. Hence, $\mathbb{ZxZ}$ is not a cyclic group.

My question (besides a validity check): was there a better way to prove this? I just found this to be the easiest way.

EDIT Totally wrong with this one. Back to the cutting board.

user26857
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user146925
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2 Answers2

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Suppose to the contrary that $(a,b)$ is a generator. Since $(1,0)$ is in the group, we have $b=0$. But then $(a,b)$ cannot generate $(0,1)$.

Remark: This is in a sense a dimension argument. But we cannot borrow a dimension theorem, since our groups are not vector spaces over any field.

André Nicolas
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If $G\times K$ is cyclic then it is generated by some $(g,k)\in G\times K$ s.t $g,k$ are the generators of $G,K$ respectively.

In your case $\mathbb{Z}$ is generated by $\pm1$ but $(\pm1,\pm1)$ does not generate $\mathbb{Z}\times\mathbb{Z}$

Belgi
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