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give an example of a sequence for which the set of values x1,x2...i.pdf
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G is abelian, k is a fixed positive integer, and H={a element of G, .pdf
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G is abelian, k is a fixed positive integer, and H={a element of G, |a|divides k}. Prove that H is a subgroup of G Solution Firstly note that if a,b are in H, then |a| | k and |b| | k as well. In particular, ak = bk = 1. Since the group G is abelian, (ab)k=akbk=1 since both the terms there are one. In particular, |ab| divides k. Hence H is closed under the grop multiplication in G. We now need to show that if a is in H then so is a-1. But this is equivalent to stating that (a-1)k = 1. But note that (a-1)k = (ak)-1 = 1 since ak = 1. Hence H is closed under inverses. Finally, the group identity is clearly in H. Hence H is a subgroup of G..
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G is abelian, k is a fixed positive integer, and H={a element of G, .pdf
- G is abelian, k is a fixed positive integer, and H={a element of G, |a|divides k}. Prove that H is a subgroup of G Solution Firstly note that if a,b are in H, then |a| | k and |b| | k as well. In particular, ak = bk = 1. Since the group G is abelian, (ab)k=akbk=1 since both the terms there are one. In particular, |ab| divides k. Hence H is closed under the grop multiplication in G. We now need to show that if a is in H then so is a-1. But this is equivalent to stating that (a-1)k = 1. But note that (a-1)k = (ak)-1 = 1 since ak = 1. Hence H is closed under inverses. Finally, the group identity is clearly in H. Hence H is a subgroup of G.
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