I saw a theorem
For odd $p$, a $p$-group possesses a characteristic subgroup $D$ of class at most $2$ and of exponent $p$ such that every nontrivial $p’$-automorphism of $G$ induces a nontrivial automorphism of $D$.
The group $D$ is a $p$-group. If it’s of exponent $p$, namely the least common multiple of the orders of all elements of $D$ is $p$, then $D$ is elementary abelian and then the nilpotency class is just $1$ and cannot be $2$! There must be something wrong with my understanding but I don’t know what that is.
No, it is false that a group of exponent $p$ is abelian, let alone elementary abelian, when $p$ is odd.
The Heisenberg $p$-group, for example, is the group of order $p^3$ and exponent $p$ which is not abelian. It can be realized as the multiplicative group of all $3\times 3$ unitriangular matrices with entries in $\mathbb{Z}/p\mathbb{Z}$: $$\left(\begin{array}{ccc} 1 & a & c\\ 0 & 1 & b\\ 0 & 0 & 1 \end{array}\right),\qquad a,b,c\in\mathbb{Z}/p\mathbb{Z}.$$ Verify that it is a group, and that when $p$ is odd, the group has exponent $p$ and is not abelian.
It can also be given by generators and relators as $$\bigl\langle x,y,z\bigm| x^p=y^p=z^p=1,\ xy=yxz,\ xz=zx,\ yz=zy\bigr\rangle.$$ To connect both descriptions, identify $x$ with the matrix with $a=1$, $b=c=0$; $y$ with the matrix with $b=1$, $a=c=0$, and $z$ with the matrix with $a=b=0$ and $c=1$.