Fixed order subgroup of an abelian group

Given an Abelian group $𝐺$ , we may construct a subgroup $𝐻$ containing only those elements whose order divides a given integer $𝑛$ , i.e. $𝐻 = {𝑥 \in 𝐺 ∣ 𝑥 𝑛 = 𝑒}$ . #m/thm/group

Proof

Let $𝑛 \in ℤ$ Clearly $𝑒 𝑛 = 𝑒$ so $𝐻$ is inhabited. Let $𝑥, 𝑦 \in 𝐻$ . Since $𝐺$ is abelian $(𝑥 𝑦 - 1) 𝑛 = 𝑥 𝑛 (𝑦 𝑛) - 1 = 𝑒$ , it follows $𝑥 𝑦 - 1 \in 𝐻$ . Therefore $𝐻$ is a subgroup by One step subgroup test.

This construction can fail for non-abelian groups, for example in $𝐷 4$ the set ${𝑒, 𝑟 2, 𝑠 1, 𝑠 2, 𝑠 3, 𝑠 4}$ is not closed.

#state/tidy | #lang/en | #SemBr