Exercise 4.4.8 (Results on characteristic subgroups)

Proof. Here $H\le K\le G$.

(a)

Every inner automorphism ${\gamma }_a$ of $G$, defined by ${\gamma }_a(x)=\mathit{𝑎𝑥}{a}^{-1}$ for every $x\in G$, gives by restriction an automorphism $\sigma$ of $K$, because $K⊴G$. So $\sigma \in \mathrm{Aut}(K)$, and $H$ is a characteristic subgroup of $K$, therefore $\sigma (H)=H$. This shows that $\mathit{𝑎𝐻}{a}^{-1}=H$. This is true for every $a\in G$, so $H⊴G$.

(b)

Now we suppose that $H$ is characteristic in $K$ and $K$ is characteristic in $G$. Let $\sigma$ be any automorphism of $G$. Since $K$ is characteristic in $G$, the restriction $\tilde{\sigma }:K\to K$ of $\sigma$ is an automorphism of $K$. Since $H$ is characteristic in $K$, $\tilde{\sigma }(H)=H$. Moreover, since $H\subseteq K$, $\tilde{\sigma }(H)=\sigma (H)$, thus $\sigma (H)=H$. This shows that $H$ is a characteristic subgroup of $G$.

(c)

Consider the following counterexample: $$\begin{array}{llll}\hfill G& =A_4,& \hfill & \\ \hfill K& =⟨(12)(34),(13)(24)⟩=\{(),(12)(34),(13)(24),(14)(23)\},& \hfill & \\ \hfill H& =⟨(12)(34)⟩=\{(),(12)(34)\}& \hfill & \end{array}$$

• $H⊴K$, since $K\simeq V_4$ is abelian.
• $K$ is a characteristic subgroup of $G$, because $K$ is the only subgroup of $A_4$ of order $4$ (see the lattice of subgroups of $A_4$ p. 111).

• $H$ is not normal in $A_4$: $(12)(34)\in H$, but

  $$(123)(12)(34){(123)}^{-1}=(23)(14)\notin H.$$

If $H$ is normal in $K$ and $K$ is characteristic in $G$, then $H$ is not necessarily normal in $G$.