Let p be a prime and G

Answered question

2022-05-11

Let p be a prime and G = {a/pn : a ∈ Z, n ∈ N}. Show that G is a group under addition

 

Answer & Explanation

Andre BalkonE

Andre BalkonE

Skilled2023-05-06Added 110 answers

To show that G={apn:a,n} is a group under addition, we need to show that it satisfies the four group axioms:
1. Closure: For any apn,bpmG, their sum apn+bpm is also in G.
2. Associativity: For any apn,bpm,cpkG, the sum (apn+bpm)+cpk is equal to apn+(bpm+cpk).
3. Identity: There exists an element ep0G such that for any apnG, ep0+apn=apn+ep0=apn.
4. Inverse: For any apnG, there exists an element apnG such that apn+(apn)=(apn)+apn=ep0.
Let's verify each of these axioms:
1. To show closure, let apn,bpmG. Then their sum is apn+bpm=apm+bpnpnpm. Since p is prime, p does not divide pnpm, so we can write this fraction in lowest terms. Therefore, apm+bpnpnpmG, and G is closed under addition.
2. To show associativity, let apn,bpm,cpkG. Then:
(apn+bpm)+cpk=apm+bpnpnpm+cpn+mpkpn+m
and
apn+(bpm+cpk)=apn+bpk+cpnpkpn
We can show that these two expressions are equal by cross-multiplying and simplifying, using the fact that p is prime. Therefore, G is associative under addition.
3. To show identity, we need to find an element ep0G such that for any apnG, ep0+apn=apn+ep0=apn.
Let e=0. Then ep0=0G. For any apnG, we have:
ep0+apn=0+apn=apn
and
apn+ep0=apn+0=apn
Therefore, ep0=0 is the identity element in G.
4. To show inverse, we need to find an element apnG such that apn+(apn)=(apn)+apn=ep0.
Let apn=apn. Then apnG since a and n. For any apnG, we have:
apn+(apn)=apnapnpnpn=0p2n=ep0
and
(apn)+apn=apn+apnpnpn=0p2n=ep0
Therefore, apn=apn is the inverse of apn in G.
Since G satisfies all four group axioms, we have shown that G is a group under addition.

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