The group product of elements in commuting subgroups generate a subgroup

Given two subgroups $𝐻, 𝐾 \subseteq 𝐺$ whose elements commute so that $ℎ 𝑘 = 𝑘 ℎ$ for any $ℎ \in 𝐻$ and $𝑘 \in 𝐾$ , the set $𝐻 𝐾 = {ℎ 𝑘 ∣ ℎ \in 𝐻, 𝑘 \in 𝐾}$ is a subgroup. group

Proof

Clearly $𝑒 \in 𝐻 𝐾$ . Let $𝑎, 𝑏 \in 𝐻 𝐾$ so that $𝑎 = ℎ 𝑥$ and $𝑏 = 𝑘 𝑦$ where $ℎ, 𝑘 \in 𝐻$ and $𝑥, 𝑦 \in 𝐾$ . Then $𝑎 𝑏 - 1 = (ℎ 𝑥) (𝑘 𝑦) - 1 = ℎ 𝑥 𝑦 - 1 𝑘 - 1$ . Since $𝑥 𝑦 - 1 \in 𝐾$ it commutes with $𝑘 - 1 \in 𝐻$ so that $𝑎 𝑏 - 1 - (ℎ 𝑘 - 1) (𝑥 𝑦 - 1)$ where $ℎ 𝑘 - 1 \in 𝐻$ and $𝑥 𝑦 - 1 \in 𝐾$ , hence $𝑎 𝑏 - 1 \in 𝐻 𝐾$ . Therefore $𝐻 𝐾$ is a subgroup by One step subgroup test.

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The group product of elements in commuting subgroups generate a subgroup

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