- Prove that if \(H \operatorname{char} G\) (i.e. \(H\) is a characteristic subgroup of \(G\)), then \(H \trianglelefteq G\) (\(H\) is a normal subgroup of \(G\)).
- Give an example showing that normality does not imply characteristic (so “characteristic” is strictly stronger than “normal”).
Solution
If \(H\operatorname{char}G\), then \(H\) is fixed by every automorphism of \(G\). In particular it is fixed by every inner automorphism \[ \iota_g(x)=gxg^{-1},\qquad g\in G. \] Hence \(gHg^{-1}=H\) for every \(g\in G\), so \(H\trianglelefteq G\).
The converse fails. In the Klein four-group \[ V_4=\{1,a,b,c\}, \] every subgroup is normal because \(V_4\) is abelian. But \(\langle a\rangle\) is not characteristic: an automorphism exchanging \(a\) and \(b\) sends \(\langle a\rangle\) to \(\langle b\rangle\).