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if $G_i = \Bbb Z/p_i \Bbb Z$
($\Bbb Z$ means integers)

where $p_i$ is the ith integer prime , I=the positive integers

show that , every element of the restricted direct product of the $G_i$'s has finite order


my trial to solve it ,

i supposed that $S\subset I$ is the indes of the restricted direct product .

let $S=${$i,i+1,i+2,...,m-1,m$}

$ G = G_i \times G_{i+1} \times G_{i+2} \times ... \times G_m \times G_1 \times G_2 \times ... \times G_{i-1} \times G_{m+1} \times ... $

let $g=(1_i ,1_{i+1} , 1_{i+1} , ... ,1_m , g_1 ,g_2 ,...,g_{i-1} , g_{m+1} , ... ) \in G$

where , $1_i$ is the identity of the ith group , $ g_1 \in G_i$

now the problem is that to compute the order of this element g , i have to calculate l.c.m of the orders of the elements $g_1 , g_2 , ... ,g_{i-1} , g_{m+1} , ... $ and those orders are infinte numbers of primes so the l.c.m can't be finite ! so the order must be infinte .

but i'm asked to prove it's finite !

so i think i made a error in my definitions " may be i understand the definition of restricted wrongly , i'm not sure "

any help plz , thanx

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  • $\begingroup$ Elements of the direct product are "sequences." Restricted direct product here means that for any sequence, the number of places that it differs from the identity is finite. $\endgroup$ – André Nicolas May 29 '13 at 19:34
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    $\begingroup$ You've written $Z\setminus pZ$ but I guess you mean $Z/pZ$? $\endgroup$ – rschwieb May 29 '13 at 19:35
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    $\begingroup$ Made some minor changes at the beginning there to the TeX. Hope you find them useful/faithful to your intended quesiton. $\endgroup$ – rschwieb May 29 '13 at 19:45
  • $\begingroup$ @AndréNicolas , Wow , i have understood it as that the number of the elements which IS the identity is finite , than you for your correcting my understanding . $\endgroup$ – Fawzy Hegab May 30 '13 at 5:57
  • $\begingroup$ @rschwieb , thank you , your correction is what i really mean , i'm not professional with latex , so i do my best typing the latex code here , when someone make a correction i check the changes to learn from them , thank you so much :) $\endgroup$ – Fawzy Hegab May 30 '13 at 5:58
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A "tuple" in the restricted direct product only differs from the identity on finitely many coordinates. So, you will be able to find the LCM of orders of coordinates to find the order of an element, as you began to do.


I think a good way to understand the restricted direct product is as a subgroup of the full direct product. The direct product, as you know, is just all "vectors" of elements where the $i$'th entry is from $G_i$, and you make this into a group with pointwise operations.

The restricted direct product is a subgroup of that consisting of elements which are the identity except possibly on finitely many coordinates. You can easily see that the two products share the identity, and that the elements I described are closed under multiplication and inversion.

In fact, I think you will have a good time showing that the restricted direct product is a normal subgroup of the direct product :)

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  • $\begingroup$ i think it's So easy to show it's normal ! i found it that saying the restricted direct product is normal subgroup of the direct product is as clear as 1+1=2 ! but i will make it into symbolic notation ! $\endgroup$ – Fawzy Hegab May 30 '13 at 6:03

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