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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 center is $Z(\mathfrak{g})=\{z:[z,x]=0\text{ for every }x\in \mathfrak{g}\}$.

A nontrivial normal subgroup $N$ of a finite p-group meets the center of a group nontrivially. The conjugation action of $G$ on $N$ has singleton orbits exactly at $N\cap Z(G)$. The orbit-stabilizer theorem makes every other orbit size divisible by $p$, so $|N\cap Z(G)|\equiv |N|\equiv 0(\mathrm{mod}p)$. Since the intersection contains the identity, its size is at least $p$.

The nontrivial center of a finite p-group theorem implies that the center of a group of order $p^2$ has order $p$ or $p^2$. If it has order $p$, the quotient by the center has prime order and is cyclic; a cyclic quotient by the center forces the whole group to be abelian, contradicting the assumed size of its center. Thus the group is abelian. An element of order $p^2$ makes it cyclic. Otherwise choose $a\ne 1$ and $b\notin ⟨a⟩$; their order-$p$ cyclic subgroups intersect trivially and generate the direct product of groups $C_p\times C_p$. The two possibilities are nonisomorphic because only the cyclic one has an element of order $p^2$.

The modular Heisenberg group consists of upper-unitriangular $3\times 3$ matrices over the finite field ${\mathbb{F}}_p$. In coordinates $M(x,y,z)$, multiplication is $M(x,y,z)M(x\prime ,y\prime ,z\prime )=M(x+x\prime ,y+y\prime +xz\prime ,z+z\prime )$, and its order is $p^3$. Its center of a group is $\{M(0,y,0)\}\cong C_p$; the quotient group by the centre is $C_p\times C_p$, represented by the $x,z$ coordinates. The elementary generators $a=M(1,0,0)$, $b=M(0,0,1)$, $c=M(0,1,0)$ satisfy $a^p=b^p=c^p=e$, $ab{a}^{-1}{b}^{-1}=c$, and $c$ central. For $p=2$, $(ab{)}^2=c$, so the whole group does not have exponent two.

Every nontrivial finite p-group has a nontrivial center. The conjugation action partitions the group into conjugacy classes whose noncentral sizes are positive powers of $p$; the class equation then implies $|Z(G)|\equiv |G|\equiv 0(\mathrm{mod}p)$.

If $N=\{1,z\}$ is a normal subgroup of a group $G$, conjugation by any element of $G$ preserves $N$ and fixes $1$. It must therefore fix $z$ as well. Thus $N$ lies in the center of a group. This fact can turn a small normal subgroup into a commuting factor without assuming the whole group is abelian.