Every proper abelian subgroup of a finite simple non-abelian group has order less than its square root

Statement

Suppose $G$ is a Finite simple non-abelian group (?) and $A$ is a proper abelian subgroup of $G$. Then, $|G|>|A|^{2}$.

Facts used

1. Finite simple non-abelian group has order greater than product of order of proper subgroup and its centralizer

Proof

The proof follows from fact (1), and the fact that if $A$ is an abelian subgroup, $|C_{G}(A)|\geq |A|$.

References

Journal references

• Theorem 2.1 (Page 909) of A measuring argument for finite groups by Andrew Chermak and Alberto Delgado, Proceedings of the American Mathematical Society, Volume 107,Number 4, Page 907 - 914(Year 1989): ^{Official copyMore info}