(a) Prove that R≤∏i∈IGi.
(b) Prove that R⊴∏i∈IGi.
(c) Let I=Z+ and let pi be the ith prime. Show that if Gi=Z/piZ for all i∈Z+, then every element of R of ∏i∈IGi has finite order, but the unabridged group has elements of infinite order as well. In this case, show that Tor(∏i∈IGi)=R (cf. 2.1.6).
Proof: (a) We have ∏ai∏bi=∏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 ∏ci∈∏i∈IGi and ∏ai∈R, we have the jth coordinate of ∏ci∏ai∏c−1i 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 ∏ci∏ai∏c−1i∈R and ∏ciR∏c−1i=R so that R⊴∏i∈IGi.
(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+1n≡n≡0modpn+1, even though 0<n<pn+1. Therefore |∏ai|=∞.
For any ∏ai∈R, set n=∏i∈Kpi, where K is the subset of I such that ai is nonidentity. For any nonidentity coordinate, we have amn≡0modpm. Clearly, it also fixes the identity coordinates. Therefore, |∏ai|∣n, and in particular the order of every element is finite. This proves R≤Tor(∏i∈IGi).
Assume ∏xi∈Tor(∏i∈IGi) such that ∏xi has infinitely many nonidentity coordinates. Observe its order n and the (n+1)th nonidentity coordinate. We have xmn≡0modpm with xm≢0, and we can see 0<n<pn+1≤pm. Therefore, pm∤xm and pm∤n, so that pm∤xmn, i.e. xmn≢0modpm, a contradiction. Therefore, Tor(∏i∈IGi)≤R and now Tor(∏i∈IGi)=R. ◻
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