On group quotient by a characteristic subgroup My question is: Let $H,K\leq G$ be two characteristic subgroups and assume $H\leq K$. Do we have $K/H$ is characteristic in $G/H$?
We know that any characteristic subgroup of $G/H$ must be of the form $K/H$ for some characteristic subgroup $K$ of $G$ since any automorphism of $G$ induces an automorphism of $G/H$.
But is the converse true? Or any counterexample?
 A: $\newcommand{\Span}[1]{\left\langle #1 \right\rangle}$Let $p$ be an odd prime, and $G$ be the  group of order $p^{3}$ and exponent $p^{2}$, which is then given by the presentation
$$
\Span{ a, b : a^{p^{2}} = b^p = 1, b^{-1} a b = a^{1+p}}.
$$
Then 


*

*$H = \Span{a^{p}} = Z(G)$ is a characteristic subgroup of order $p$ of $G$, 

*$K = \Span{a^{p}, b}$ is a characteristic subgroup of $G$, as it consists of the elements of $G$ of order a divisor of $p$, and has order $p^{2}$,

*$H \le K$.


But $G/H$ is elementary abelian of order $p^{2}$, and thus characteristically simple, so that $K/H$ (which has order $p$) is not characteristic in $G/H$.
The point here is that not all automorphisms of $G/H$ lift to automorphisms of $G$.
A: Take $G$ to be the dihedral group $D_4$. It has a presentation $\langle r,s\mid r^4=1=s^2,sr=r^3s\rangle$. Let $K=\langle r\rangle$ the subgroup of rotations (which is thus cyclic of order $4$), and $H=\langle r^2\rangle$. Any automorphism of $G$ must preserve $K$, since it must map $r$ to an element of order $4$, and has a consequence it also fixes $r^2$, thus $K$ and $H$ are characteristic. But $G/H$ is the Klein group, which has no non-trivial characteristic subgroup, and $K/H$ is the cyclic group of order $2$.
