No finite group is the union of conjugates of a proper subgroup

Let \(G\) be a finite group and let \(H < G\) be a proper subgroup, so \([G : H] = k > 1\).

1. The number of distinct conjugate subgroups of \(H\) in \(G\) is \([G : N_G(H)] \leq [G : H] = k\). Let these conjugates be \(H_1, H_2, \ldots, H_m\), where \(m \leq k\).

2. Each conjugate \(H_i\) has order \(\abs{H_i} = \abs H\), and every conjugate contains the identity element \(e\).

3. Count the number of non-identity elements in \(\bigcup_{i=1}^m H_i\): \[ \abs{\bigcup_{i=1}^m (H_i \setminus \{e\})} \leq \sum_{i=1}^m (\abs{H_i} - 1) = m(\abs H - 1). \]

4. Therefore: \[ \abs{\bigcup_{i=1}^m H_i} \leq 1 + m(\abs H - 1) \leq 1 + k(\abs H - 1) = 1 + [G:H](\abs H - 1) = 1 + \abs G - [G:H]. \]

5. Since \(H < G\) is proper, \([G : H] \geq 2\), which implies: \[ 1 + \abs G - [G:H] \leq \abs G - 1 < \abs G. \]

Thus \(\bigcup_{g \in G} gHg^{-1} \subsetneq G\), so \(G\) cannot be the union of the conjugates of any proper subgroup.