How could we know that every automorphism of $G$ would map $H \le G $ to itself?

Question

I'm reading the definition of characteristic subgroup, but I wonder how could we know that every automorphism of $G$ would map $H \le G $ to itself?

Answer 1

I will answer the question properly! As I said in my comment, there are essentially two ways of going about this. The easiest is to use know results that certain subgroups, such as $Z(G)$ (and also all members of the upper central series of $G$), $[G,G]$ (and also all members of the derived series and of the lower central series of $G$), and $O_p(G)$ (the largest normal $p$-subgroup for a prime $p$, which generalizes to sets of primes) are all characteristic.

You can also use these in combination with each other, remembering that characteristic subgroups of characteristic subgroups (such as $Z([G,G])$) are characteristic, as are products of characteristic subgroups, such as $Z(G)O_p(G)$.

A variant of this method can be used if $N$ is the unique normal subgroup of $G$ with a certain property, so for example the cyclic subgroup of index $2$ in a dihedral group of order at least $6$ is the unique such subgroup, so it must be characteristic.

The second method is to compute ${\rm Aut}(G)$ (possibly using computer software such as GAP) and check directly that $\alpha(N) = N$ for all $\alpha \in {\rm Aut}(G)$, and to do that it is sufficient to check it for all $\alpha$ in a generatong set of ${\rm Aut}(G)$. Since all characteristic subgroups are normal, you should of course only do this for normal subgroups.