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If an abelian subgroup $A$ has index $p^a$, the centralizer of any $1\ne g\in A$ contains $A$, so the class size of $g$ divides $p^a$. The prime-power conjugacy-class obstruction to simplicity rules this out. If $A=1$, the group is a p-group and has nontrivial center of a group, also ruling out nonabelian simplicity.

The centre of a group is $Z(G)=\{z\in G:zg=gz\text{ for every }g\in G\}$. It is a normal subgroup. A nontrivial finite p-group has nontrivial centre: in the class equation, each noncentral conjugacy class has size divisible by $p$, so $|Z(G)|\equiv |G|\equiv 0(\mathrm{mod}p)$.