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Since H = {0,3,6,9} is cyclic of order 4 (generated by 3 in Z_12), H is isomorphic to Z_4.
Moreover, because K = {0,2,4,6,8,10,12,14,16} is cyclic of order 9 (generated by 2 in Z_18), K is isomorphic to Z_9.
Thus, K ⊕ H is the desired subgroup of Z_12 ⊕ Z_18 isomorphic to Z_9 ⊕ Z_4.
I hope this helps!
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Since H = {0,3,6,9} is cyclic of order 4 (generated by 3 in Z_12), H is isomorphic to Z_4.
Moreover, because K = {0,2,4,6,8,10,12,14,16} is cyclic of order 9 (generated by 2 in Z_18), K is isomorphic to Z_9.
Thus, K ⊕ H is the desired subgroup of Z_12 ⊕ Z_18 isomorphic to Z_9 ⊕ Z_4.
I hope this helps!