# whether every normal subgroup is characteristic?

Let $G$ be a group. A subgroup $H$ of $G$ is called characteristic if $\varphi(H)\subset H$ for all automorphisms $\varphi$ of $G$.

Now it is easy to show that every characteristic group is normal, as for every inner automorphism $\tau_g$, $\tau_g(H)\subset H$, i.e. $g^{-1}Hg\subset H$ for all $g\in G$.

Is that converse true? That is, whether every normal subgroup is characteristic?

• Look at the Klein-four-group, all subgroups are normal but none of them is characteristic – Peter Melech Nov 24 '17 at 14:50
• @PeterMelech A little pedantic, but you might want to specify that this only holds for the nontrivial subgroups. – Sebastian Schoennenbeck Nov 24 '17 at 14:59
• $(x,y) \mapsto (y,z)$ permutes the two copies of $\mathbb{Z}$ in $\mathbb{Z} \times \mathbb{Z}$. – D_S Nov 24 '17 at 15:05
• Because linear transformations can change one vector subspace to another. To paint a mental picture, imagine rotating a line in the plane. For a sketch of an algebraic justification, note every subspace has a basis that can be extended to a basis for the whole space, and the general linear group acts transitively on the bases of the whole space (since, after applying a matrix to the coordinate basis, the new basis is just the set of column vectors). – anon Nov 24 '17 at 15:45
• has nothing to do with the group being finite. Beyond that, if $L$ is a one-dimensional subspace of a vector space $V$ of greater dimension over a field $k$, then $L=kv$ is mappable via an automorphism of $V$ to any other one-dimensional $kw$. – Lubin Nov 24 '17 at 15:46