# 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?

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

• Do you really need $p$ to be odd ? The counterexample I have given uses the same group for $p=2$, but a different $K$. – Arnaud D. Jun 19 at 9:41
• 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. – Andreas Caranti Jun 19 at 9:53

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 $$2$$.