For one such subgroup, observe that if H is a subgroup of order 4 in Z_40 and K is a subgroup of order 3 in Z_30, then H ⊕ K will be a subgroup of order 12 in Z_40 ⊕ Z_30.
In Z_40, we can set H = {0, 10, 20, 30}, since 10 is of order 4 in H.
In Z_30, we can set K = {0, 10, 20}, since 10 is of order 3 in K.
This H ⊕ K is one such subgroup of order 12 (being isomorphic to Z_4 ⊕ Z_3).
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For another such subgroup, observe that if H is a subgroup of order 2 in Z_40 and K is a subgroup of order 6 in Z_30, then H ⊕ K will be a subgroup of order 12 in Z_40 ⊕ Z_30.
In Z_40, set H = {0, 20}, since 20 is of order 2 in H.
In Z_30, set K = {0, 5, 10, 15, 20, 25}, since 5 is of order 6 in K.
This H ⊕ K is of order 12 (being isomorphic to Z_2 ⊕ Z_6).
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For one such subgroup, observe that if H is a subgroup of order 4 in Z_40 and K is a subgroup of order 3 in Z_30, then H ⊕ K will be a subgroup of order 12 in Z_40 ⊕ Z_30.
In Z_40, we can set H = {0, 10, 20, 30}, since 10 is of order 4 in H.
In Z_30, we can set K = {0, 10, 20}, since 10 is of order 3 in K.
This H ⊕ K is one such subgroup of order 12 (being isomorphic to Z_4 ⊕ Z_3).
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For another such subgroup, observe that if H is a subgroup of order 2 in Z_40 and K is a subgroup of order 6 in Z_30, then H ⊕ K will be a subgroup of order 12 in Z_40 ⊕ Z_30.
In Z_40, set H = {0, 20}, since 20 is of order 2 in H.
In Z_30, set K = {0, 5, 10, 15, 20, 25}, since 5 is of order 6 in K.
This H ⊕ K is of order 12 (being isomorphic to Z_2 ⊕ Z_6).
I hope this helps!