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

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

Andre BalkonE · Accepted Answer

To show that G={apn:a∈ℤ,n∈ℕ} is a group under addition, we need to show that it satisfies the four group axioms:1. Closure: For any apn,bpm∈G, their sum apn+bpm is also in G.2. Associativity: For any apn,bpm,cpk∈G, the sum (apn+bpm)+cpk is equal to apn+(bpm+cpk).3. Identity: There exists an element ep0∈G such that for any apn∈G, ep0+apn=apn+ep0=apn.4. Inverse: For any apn∈G, there exists an element −apn∈G such that apn+(−apn)=(−apn)+apn=ep0.Let&#039;s verify each of these axioms:1. To show closure, let apn,bpm∈G. 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+bpnpnpm∈G, and G is closed under addition.2. To show associativity, let apn,bpm,cpk∈G. Then:(apn+bpm)+cpk=apm+bpnpnpm+cpn+mpkpn+mandapn+(bpm+cpk)=apn+bpk+cpnpkpnWe 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 ep0∈G such that for any apn∈G, ep0+apn=apn+ep0=apn.Let e=0. Then ep0=0∈G. For any apn∈G, we have:ep0+apn=0+apn=apnandapn+ep0=apn+0=apnTherefore, ep0=0 is the identity element in G.4. To show inverse, we need to find an element −apn∈G such that apn+(−apn)=(−apn)+apn=ep0.Let −apn=−apn. Then −apn∈G since a∈ℤ and n∈ℕ. For any apn∈G, we have:apn+(−apn)=apn−apnpnpn=0p2n=ep0and(−apn)+apn=−apn+apnpnpn=0p2n=ep0Therefore, −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.

Let p be a prime and G

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