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Let $\mathbb{Z}^n,\mathbb{Z}^m$ be a free abelian group, $m<n.$

Prove $\mathbb{Z}^n \not\cong \mathbb{Z}^m$

Is it possible that the problem is trivial since $\mathbb{Z}^n$ has $n$ generators and $\mathbb{Z}^m$ has $m$ generators ,$n\neq m$ ,hence $\mathbb{Z}^n\not\cong\mathbb{Z}^m$?

edited Jun 21 '22 at 13:56

Eric Wofsey

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asked Jun 21 '22 at 13:54

Algo

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How do you know $\mathbb{Z}^n$ doesn't have $m$ generators? A group typically has many different generating sets. – Eric Wofsey Jun 21 '22 at 13:56
Besides having an $n$-element generating set, $\mathbb Z^n$ also has an $n+1$ element generating set. For instance, $\mathbb Z^1=\mathbb Z$ has a $2$ element generating set ${1,2}$. – Lee Mosher Jun 21 '22 at 13:56
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Tensor with $\mathbb Q$. – Pedro Jun 21 '22 at 13:59
The typical proof is tensoring with a field. – WhatsUp Jun 21 '22 at 14:00
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Based on the OP's other questions I would guess that this question is for someone in an intro to abstract algebra class, where they would not learn about tensor products. I believe the "correct" proof at this level is using the fundamental theorem of finitely generated abelian groups, often (?) taught in intro abstract algebra without proof. – fish Jun 21 '22 at 14:14
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You can also kill $2x$ for every $x$ in the group, and compare what you get. – Dustan Levenstein Jun 21 '22 at 14:21

Let $\mathbb{Z}^n,\mathbb{Z}^m$ be a free abelian group, $m

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