(i) Let G=Z12(sub12 don't know how to put it), the group of integers modulo 12. Prove that H= {0, 6} AND K= {0, 4, 8} are subgroups of G. Calculate the subset H+K formed by adding together all possible pairs of elements from H and K, i.e.
H+K= {h+kh is a subgroup of H, k is a subgroup of K}
Prove that this is also a subgroup of G.

(ii) If H and K are subgroups of an abelian group G, prove that the set HK={hkh is subgroup of H, k is subgroup of K} is always a subgroup of G.
Show, by considering the group G=S, or otherwise, that the corresponding result is false when G is not abelian.

(b)

(i) in each of the group Z12(sub 12 should be placed at the corner below 12), and S3(3 is placing at the corner, as sub 3), list the elements of order 1 or two. In each case, does this set of elements from a subgroup?

(ii) Prove that in any abelian group G, the set {g/g²=e} is a subgroup of G.
Does this result remain true if G is not abelian? Justify your answer.

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Problem #3
(a) Proof:
(1) In the group , we consider the element . Since in , then the order of is 2. Thus is a cyclic subgroup of . Similarly, the order of is in . So is a ...

Solution Summary

Abelian Groups and Subgroups are investigated. The solution is detailed and well presented.

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