Prove that if \(M\) is a maximal subgroup of \(G\) then either \(N_G(M) = M\) or \(N_G(M) = G\). Deduce that if \(M\) is a maximal subgroup of \(G\) that is not normal in \(G\) then the number of nonidentity elements in \(G\) that are contained in conjugates of \(M\) is at most \((\left|M\right|-1)\left|G : M\right|\).
The first statement is clearly true since \(M \le N_G(M) \le G\). As for the second part, we shall first prove the following lemma: if \(g_1M = g_2M\), then \(g_1Mg_1^{-1} = g_2Mg_2^{-1}\).
Let \(m \in M\). Since \(g_2^{-1}g_1\) and \(g_1^{-1}g_2\) are both in \(M\), it follows that for some \(m' \in M\), \((g_2^{-1}g_1)m(g_1^{-1}g_2) = m'\), or \(g_1mg_1^{-1} = g_2m'g_2^{-1}\). So \(g_1Mg_1^{-1} \subseteq g_2Mg_2^{-1}\). By symmetry, \(g_2Mg_2^{-1} \subseteq g_1Mg_1^{-1}\) as well, giving us the desired result.
The lemma tells us that there are only \(\left|G : M\right|\) distinct conjugates of \(M\). Furthermore, the element \(g1g^{-1} = 1 \in gMg^{-1}\) is not counted, so at most \(\left|M\right|-1\) nonidentity elements are contained in conjugates of \(M\). The total is as its maximum when all conjugates are distinct; then the total is precisely \((\left|M\right|-1)\left|G : M\right|\).
Note that if \(M\) is a normal subgroup of \(G\), then the maximum total is instead \(\left|M\right|-1\), since \(gMg^{-1} = M\).
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