# prove that the order of the elements of restricted direct product is finite .

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

• 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. – André Nicolas May 29 '13 at 19:34
• You've written $Z\setminus pZ$ but I guess you mean $Z/pZ$? – rschwieb May 29 '13 at 19:35
• Made some minor changes at the beginning there to the TeX. Hope you find them useful/faithful to your intended quesiton. – rschwieb May 29 '13 at 19:45
• @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 . – Fawzy Hegab May 30 '13 at 5:57
• @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 :) – Fawzy Hegab May 30 '13 at 5:58

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.