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Is infinite product of nilpotent (solvable) groups nilpotent (solvable)?

(I known they are true for finite cases)

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2 Answers 2

No to both. Take a product of groups $G_n$ which are $n$-step nilpotent (resp. $n$-step solvable) but not $n-1$-step nilpotent (resp. $n-1$-step solvable). However, this is true if (and only if!) the nilpotence (resp. solvability) degree is uniformly bounded.

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I don't have enough rep. to comment the following remark.

Each term of the lower central series (resp. derived series) commutes with an arbitrary cartesian product.

This also proves the last assertion of Qiaochu.

I think a correct argument is the following.

The cartesian product $\prod G_i$ satisfies a law $w$ if and only if each $G_i$ satisfies $w$.

Nilpotent groups of class at most $c$ are precisely the groups satisfying the law $[x_1,\ldots,x_{c+1}]=1$.

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This is false! In general you have $$[\prod G_i,\prod G_i]\subseteq \prod [G_i,G_i]$$but equality need not hold. Take $G_i$ to be free of countable rank, with free generating set $x_1,y_1,x_2,y_2,x_3,y_3,\ldots$. Take the element on the right hand side whose $n$th entry is $[x_1,y_1]\cdots[x_n,y_n]$. This element is not on the left hand side. –  Arturo Magidin Jun 22 '12 at 3:59
    
@Arturo: Oops, sorry! And thanks! Sometimes one can write very stupid things. –  APena Jun 22 '12 at 5:57
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No problem; I even reviewed a published paper once that made the same incorrect assertion. In general, for any set of words $W$, we have $W(\prod G_i)\subseteq \prod W(G_i)$. If all $G_i$ satisfy $W$, then this implies $\prod G_i$ satisfies $W$; the converse also holds, but it does not follow from the simple inclusion. –  Arturo Magidin Jun 22 '12 at 14:44

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