Exploring Subgroups of Finite Nonabelian Groups: Order and Cyclic Structure

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In summary, we are given a finite nonabelian group G and a subgroup H with an order of p, where p is a prime. We need to prove that H is abelian, which can be done by showing that H is cyclic. To solve (2), we can use Lagrange's Theorem to determine the number of elements in H with an order of p. (3) introduces a second subgroup K with an order of q, where q is a prime and not equal to p. To find the intersection of H and K, we can use divisibility arguments and consider the order of elements in both subgroups.
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let G be a finite nonabelian group and h be a subgroup of G. suppose that the order of H is p where p is a prime
(1)prove that h is abelian
(2)How many elements of h have order P?
(3)Suppose K is a subgroup of G with order q, where q is a prime and q doesn't equal p. what is H intersect K?

for(1) to prove h is abelian, I can show H is cyclic, but how to show H is cyclic?(because order of p is prime?)
I have no idea how to do (2) and (3)
 
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(1) consider the order of a non-identity element h of H. what orders are possible for h?

what does that tell us about the order of <h>?

(2) how many non-identity elements does H have? do they all have the same order?

(3) suppose x is in H, and x is in K. what can we say about the order of x? are there any divisibility arguments you can use?
 

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