1 problem found
Give a careful argument to show that, if \(G_{1}\) and \(G_{2}\) are subgroups of a finite group \(G\) such that every element of \(G\) is either in \(G_{1}\) or in \(G_{2},\) then either \(G_{1}=G\) or \(G_{2}=G\). Give an example of a group \(H\) which has three subgroups \(H_{1},H_{2}\) and \(H_{3}\) such that every element of \(H\) is either in \(H_{1},H_{2}\) or \(H_{3}\) and \(H_{1}\neq H,H_{2}\neq H,H_{3}\neq H\).
Solution: Suppose \(|G_1|, |G_2| < |G|\) for sake of contraction. Then by Lagrange's theorem \(|G_1| \mid |G|\) and \(|G_2| \mid |G|\), so \(|G_1|, |G_2| \leq \frac{|G|}{2}\). But \(|G_1 \cup G_2| = |G_1| + |G_2| - |G_1 \cap G_2|\). \(|G_1 \cup G_2| = |G|\) by assumption, and \(e \in G_1 \cap G_2\), so \(|G_1 \cap G_2| \geq 1\). Therefore \(|G| = |G_1| + |G_2| - |G_1 \cap G_2| \leq \frac{|G|}{2} + \frac{|G|}{2} - 1 = |G| - 1 < |G|\), contradiction! Let \(H = K_4 = \{e, a, b, c\}\) with \(a^2 = b^2 = c^2 = e\). then \(H = \{e, a\} \cup \{e, b\} \cup \{e, c\}\) with all subgroups distinct