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..
Presentation by Andreas Schleicher Tackling the School Absenteeism Crisis 30 ...
G is abelian, k is a fixed positive integer, and H={a element of G, .pdf
1. 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.
