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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?

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    $\begingroup$ Look at the Klein-four-group, all subgroups are normal but none of them is characteristic $\endgroup$ – Peter Melech Nov 24 '17 at 14:50
  • $\begingroup$ @PeterMelech A little pedantic, but you might want to specify that this only holds for the nontrivial subgroups. $\endgroup$ – Sebastian Schoennenbeck Nov 24 '17 at 14:59
  • $\begingroup$ $(x,y) \mapsto (y,z)$ permutes the two copies of $\mathbb{Z}$ in $\mathbb{Z} \times \mathbb{Z}$. $\endgroup$ – D_S Nov 24 '17 at 15:05
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    $\begingroup$ 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). $\endgroup$ – anon Nov 24 '17 at 15:45
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    $\begingroup$ 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 $kw$. $\endgroup$ – Lubin Nov 24 '17 at 15:46

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