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The question is to produce an element that has infinite order in $\prod_{i} \mathbb{Z}/p_i\mathbb{Z}$ such that $i$ is over $\mathbb{Z}^+$ and $p_i$ denotes the $i^{th}$ integer prime. I am having trouble doing so.

For more context, a previous part of this question asked us to prove that the subgroup of elements in the restricted direct product of $\prod_{i} \mathbb{Z}/p_i\mathbb{Z}$ contains only elements of finite order, and I was able to show that, so I assume we are looking for any/all elements that have identity components in infinitely many places.

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    $\begingroup$ How about $1{}$? $\endgroup$ – Lord Shark the Unknown Nov 1 '18 at 20:45
  • $\begingroup$ Since you have the direct product any element with all non-zero components in the tuple will have infinite order I believe. Since the order of any element in the tuple will be the product of the orders of the individual components. $\endgroup$ – Aaron Zolotor Nov 1 '18 at 20:52
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    $\begingroup$ Exercise: the torsion subgroup of this group is precisely the restricted direct product. $\endgroup$ – YCor Nov 1 '18 at 22:22
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So take the sequence $x=(1,1,1,1,\ldots)$. Here by $1$ I mean $1+p_i\mathbb{Z}$. Actually any generator of $\mathbb{Z}/p_i\mathbb{Z}$ will be ok.

Note that $x$ is of infinite order. Indeed, if $nx=0$ for some $n>0$ then by looking at each coordinate you get $n1\equiv 0\text{ mod }p_i$ for all $p_i$. This clearly is false if $p_i$ is an unbounded sequence (you don't even need $p_i$'s to be prime).

In case each $p_i$ is a distinct prime you can check that $(x_1,x_2,\ldots)$ is of finite order if and only if $x_i=0$ eventually.

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