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Tuesday, March 12, 2013

Properties of Restricted Direct Products (5.1.17)

Dummit and Foote Abstract Algebra, section 5.1, exercise 17:

MathJax TeX Test Page Let I be any nonempty index set and Gi be a group for each iI. The restricted direct product or direct sum R of the groups Gi is the set of elements of the direct product that are the identity in all but finitely many components, i.e. are the nonidentity in finitely many coordinates, which is to say, the set of all elements aiiIGi such that ai=1i for all but a finite number of iI.
(a) Prove that RiIGi.
(b) Prove that RiIGi.
(c) Let I=Z+ and let pi be the ith prime. Show that if Gi=Z/piZ for all iZ+, then every element of R of iIGi has finite order, but the unabridged group has elements of infinite order as well. In this case, show that Tor(iIGi)=R (cf. 2.1.6).

Proof: (a) We have aibi=aibi has a nonidentity element in the jth coordinate only if either ai or bi has a nonidentity element in that coordinate, which limits the potential coordinates of their product for nonidentity elements to occur down to a finite number, and thus the product is within R. Furthermore, inverses clearly don't change nonidentity coordinates, the identity element is within R, and the associativity axiom is evident.

(b) For any ciiIGi and aiR, we have the jth coordinate of ciaic1i is equal to 1 if aj=1. This implies that there is only a finite number of coordinates for nonidentity coordinates to occur in, and therefore ciaic1iR and ciRc1i=R so that RiIGi.

(c) Let ai be the element such that ai=1 for all i. If |ai|=n<, then observe the (n+1)th coordinate; we should have an+1nn0modpn+1, even though 0<n<pn+1. Therefore |ai|=.

For any aiR, set n=iKpi, where K is the subset of I such that ai is nonidentity. For any nonidentity coordinate, we have amn0modpm. Clearly, it also fixes the identity coordinates. Therefore, |ai|n, and in particular the order of every element is finite. This proves RTor(iIGi).

Assume xiTor(iIGi) such that xi has infinitely many nonidentity coordinates. Observe its order n and the (n+1)th nonidentity coordinate. We have xmn0modpm with xm0, and we can see 0<n<pn+1pm. Therefore, pmxm and pmn, so that pmxmn, i.e. xmn0modpm, a contradiction. Therefore, Tor(iIGi)R and now Tor(iIGi)=R.

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