Question #242447

Let G be an abelian group and n be a positive integer. Then prove that the set
{a ∈ G : θ(a) divides n} is a subgroup of G�

Expert's answer

Let A={a∈G:O(a)∣n}A=\{a\in G : O(a)|n\}\\

We shall show that A is a subgroup of G using two steps subgroup test.\\

Let a,b∈A  ⟹  a,b∈G:O(a)∣n and O(b)∣na,b \in A \implies a,b\in G: O(a)|n \text{ and } O(b)|n \\

  ⟹  ab∈GO(ab)=lcm(O(a),O(b)){Since G is an Abelian group}  ⟹  O(ab)∣n  ⟹  ab∈A\implies ab \in G\\ O(ab)=lcm(O(a),O(b)) \{\text{Since G is an Abelian group}\}\\ \implies O(ab)|n\\ \implies ab\in A

Also,

a−1∈G since a∈GO(a−1)=O(a){Element of a group and its inverse have same order}  ⟹  O(a−1)∣na^{-1}\in G \text{ since } a \in G\\ O(a^{-1})=O(a) \{\text{Element of a group and its inverse have same order}\}\\ \implies O(a^{-1})|n

This shows that A is a subgroup of G.


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