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Topic: **Example of abstract writing****Question:**
I have so far:
We have the the order of G = p^k and by a theorem (I don't want to write it out, it's Theorem 3.5.2 in Beachy and Blair, Abstract Algebra) G is isomorphic to Z_p^k. Because isomorphisms preserve the order of elements, it suffices to show that Z_p^k has a subgroup of order p. (Because it may be easier to work with a specific group instead of a general one.)
What now though? Any help would be amazing!!

June 25, 2019 / By Ocean

You're incorrect - G need not be isomorphic to Z_p^k. For example Z_2 x Z_2 is a group with four elements but has no element of order 4 and so cannot be isomorphic to Z_4. Let g be any nonidentity element of G. Since the order of G must divide the order of the group, |g| = p^l with 1 < l < or = k. Consider h = g^{p^{l-1}}. It is not hard to see that |h| = p, and so the subgroup generated by h has order p.

👍 234 | 👎 3

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You're incorrect - G need not be isomorphic to Z_p^k. For example Z_2 x Z_2 is a group with four elements but has no element of order 4 and so cannot be isomorphic to Z_4. Let g be any nonidentity element of G. Since the order of G must divide the order of the group, |g| = p^l with 1 < l < or = k. Consider h = g^{p^{l-1}}. It is not hard to see that |h| = p, and so the subgroup generated by h has order p.

i imagine the reply has something to do with the incontrovertible truth that a subgroup of G must have an order divisible with information from p because the purely ingredient of p^ok could be p as p^ok is the unique factorization of a few integer in accordance to the needed theorem of mathematics. yet, i'm no longer useful how a suitable evidence of this will be written. the project with the previous solutions is that we are no longer attempting to discover an ingredient of order p, yet we favor a subgroup of order p.

i imagine the reply has something to do with the incontrovertible truth that a subgroup of G must have an order divisible with information from p because the purely ingredient of p^ok could be p as p^ok is the unique factorization of a few integer in accordance to the needed theorem of mathematics. yet, i'm no longer useful how a suitable evidence of this will be written. the project with the previous solutions is that we are no longer attempting to discover an ingredient of order p, yet we favor a subgroup of order p.

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Since the second part refers to "the number you used to justify part (a)", it sounds like it's perfectly acceptable to simply demonstrate that 3 satisfies a. In any case, the best you could do would be to "formalize" the proof as follows: Let p = 3; then p + 2 = 5 and p + 4 = 7. 3, 5, and 7 are prime; therefore, p, p+2 and p+4 are prime. Therefore, there exists a prime number p such that p+2 and p+4 are also prime. Part b is the harder part. Let p = a prime number other than 3. Every integer is either divisible by 3, one greater than a number that is divisible by 3, or two greater than a number that is divisible by 3. So, p must be equal to 3k, 3k+1, or 3k+2 for some integer k. If p = 3k, it is divisible by 3, so is not prime. If p = 3k+1, p+2 = 3k+3; p+2 is divisible by 3, so it is not prime. If p = 3k+2, p+4 = 3k + 6; p+4 is divisible by 3, so it is not prime. Therefore, there is no number p that satisfies the conditions.

Actually, you can just say that 3 would satisfy part (a). 3, 5, and 7 are all prime, so 3 works. The second part requires proof. You need to show that at least one of the p, p+2, p+4 is divisible by 3. One of three cases follows: (1) p=3, in which case we have the prime from (a). (2) p=1, which isn't prime (3) p=-1, which also isn't prime.

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