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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?

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$\newcommand{\Span}[1]{\left\langle #1 \right\rangle}$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 $G$.

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  • $\begingroup$ Do you really need $p$ to be odd ? The counterexample I have given uses the same group for $p=2$, but a different $K$. $\endgroup$ – Arnaud D. Jun 19 at 9:41
  • $\begingroup$ Well, there are plenty of counterexamples, and we just happened to choose two different ones. Of course I agree that different $K$ have to be used in the even and odd case: in the odd case there is no characteristic cyclic subgroup of order $p^{2}$, while in the even case the elements of order a divisor of $2$ do not form a subgroup. $\endgroup$ – Andreas Caranti Jun 19 at 9:53
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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 $2$.

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