## The number of elements in conjugates of a maximal subgroup is bounded

Recall that a proper subgroup is called maximal if it is not properly contained in any other proper subgroup. 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 nonnormal maximal subgroup of a finite group $G$ then the number of nonidentity elements that are contained in conjugates of $M$ is at most $(|M| - 1) \cdot [G : M]$.

Since $M$ is a subgroup, we have $M \leq N_G(M) \leq G$; clearly then $N_G(M) = M$ or $N_G(M) = G$. If $M$ is not normal, then $N_G(M) = M$.

The number of conjugates of $M$ is $|G \cdot M| = [G : N_G(M)] = [G:M]$. Now all conjugates of $M$ have the same cardinality as $M$, and we will have the largest number of nonidentity elements in the conjugates of $M$ precisely when these conjugates intersect trivially. In this case, the number of nonidentity elements in the conjugates of $M$ is at most $(|M|-1) \cdot [G : M]$.