More information for the question,,, I.3.7 Let be prime and H a subgroup of Z()

Question

More information for the question,,, I.3.7 Let be prime and H a subgroup of Z() (Exercise 1.10). (a) Every element of Z() has finite order pn for some n 0. (b) If at least one element of H has order pk and no element of H has order greater than pk, then H is the cyclic subgroup generated by , whence H (c) If there is no upper bound on the orders of elements of H, then H = Z()); [see Exercise 2. 1 6]. (d) The only proper subgroups of Z()) are the finite cyclic groups Cn = (n = 1 ,2, . . . ) . Furthermore, = C0 < C1 < C2 < C3 < … (e) Let x1,x2, . . . be elements of an abelian group G such that , px2 = x1 , px3 = x2, . . . , pxn+l = xn ,… The subgroup generated by the xi (i 1) is isomorphic to Z(). [Hint: Verify that the map induced by xi is a well-defined isomorphism.] More information for the question,,, I.3.7 Let be prime and H a subgroup of Z() (Exercise 1.10). (a) Every element of Z() has finite order pn for some n 0. (b) If at least one element of H has order pk and no element of H has order greater than pk, then H is the cyclic subgroup generated by , whence H (c) If there is no upper bound on the orders of elements of H, then H = Z()); [see Exercise 2. 1 6]. (d) The only proper subgroups of Z()) are the finite cyclic groups Cn = (n = 1 ,2, . . . ) . Furthermore, = C0 < C1 < C2 < C3 < … (e) Let x1,x2, . . . be elements of an abelian group G such that , px2 = x1 , px3 = x2, . . . , pxn+l = xn ,… The subgroup generated by the xi (i 1) is isomorphic to Z(). [Hint: Verify that the map induced by xi is a well-defined isomorphism.] 13. For each prime p the group Z(p) satisfies the descending but not the ascending chain condition on subgroups [see Exercise 1.3.7).

Explanation / Answer

Fix a prime number p and put

H = { m/p^n | m Z , n 0 } . Then clearly H is a subgroup of Q and

0 Z H H/Z = G 0

is exact, where the second mapping is inclusion and G is the group. ( where G = { a/p^n + Z | n 0 , a Z } and,

for i 0 , G_i = { a/p^i + Z | a Z } . Clearly G is a subgroup of Q/Z and each G_i is a subgroup of G

Moreover, G0 G1 . . . Gn . . . () is a strictly increasing sequence, so, regarded as a Z-module G does not satisfy the a.c.c. )

Thus H doesn’t satisfy the d.c.c. because it has a subgroup Z which doesn’t, and

H doesn’t satisfy the a.c.c. because it has a quotient G which doesn’t

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