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I came up with the following question reading this(Finite Abelian Groups question).

Let $G$ be an abelian group. Suppose there is an integer $n \ge 1$ such that $nG = 0$. Let $m$ be the smallest integer $\ge 1$ such that $m G = 0$.

Is there an element in $G$ of the order $m$?

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  • There is no reason to downvote this. It is an easy to answer but nevertheless a well-posed and natural arising question. – MooS Jul 15 '15 at 06:09
  • @vadim123 Why do you ask? Do you think that a finite abelian group and an infinite one are not different? –  Jul 15 '15 at 20:27
  • @vadim123 The accepted answer does use the finiteness condition. Or do you think that all abelian groups are finitely generated? –  Jul 15 '15 at 21:21

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The answer is positive. All elements have finite order divisible by $m$, so your proof (https://math.stackexchange.com/a/1361378/211913) works fine even in that case. Note that you have never used the finiteness of the group, but only the (bounded) finiteness of the order of each element.

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