No group is the union of two proper subgroups

For any group $𝐺$ it is impossible to construct the entire group as the union two proper subgroups $𝐹, 𝐻 ⊊ 𝐺$ . group

Proof

Let $𝐹, 𝐻 ⊊ 𝐺$ be strict subgroups of $𝐺$ such that $𝐹 \cup 𝐻 = 𝐺$ . Clearly these subgroups cannot be identical, so without loss of generality assume that there exists some $𝑥 \in 𝐹$ such that $𝑥 \notin 𝐻$ . For any $𝑦 \in 𝐻$ it follows that $𝑥 𝑦 \in 𝐹 \cup 𝐻 = 𝐺$ . If $𝑥 𝑦 \in 𝐻$ then $𝑥 𝑦 𝑦 - 1 = 𝑥 \in 𝐻$ which is a contradiction. If $𝑥 𝑦 \in 𝐹$ then $𝑥 - 1 𝑥 𝑦 = 𝑦 \in 𝐹$ . Therefore for any $𝑦 \in 𝐻$ also $𝑦 \in 𝐹$ , hence $𝐻 ⊊ 𝐺$ and $𝐹 \cup 𝐻 = 𝐻 = 𝐺$ , contradicting the requirement that $𝐻 ⊊ 𝐺$ . Thus no group is the union of two proper subgroups.

For more than two proper subgroups it is possible. For example, consider the group $𝐺 = {𝑅 0, 𝑅 180, 𝐻, 𝑉}$ , itself a subgroup of $𝐷 4$ . In this case $𝐺 = {𝑅 0, 𝑅 180} \cup {𝑅 0, 𝐻} \cup {𝑅 0, 𝑉}$ , which are all proper subgroups.

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No group is the union of two proper subgroups

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