Exercise 4.4.6 (Characteristic subgroups)

Proof. Let $H$ be a characteristic subgroup of a group $G$. By definition, if $\sigma$ is any automorphism of $G$, then $\sigma (H)=H$. In particular, if $\sigma$ is a inner automorphism of $G$, defined by $\sigma (x)=\mathit{𝑎𝑥}{a}^{-1}$ for some $a\in G$, then $\sigma (H)=H$, i.e., $\mathit{𝑎𝐻}{a}^{-1}=H$. This is true for every $a\in G$, so $H$ is a normal in G.

Consider the group $G=\mathbb{Z}∕2\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z}$, and $H=⟨(1,1)⟩=\{(0,0),(1,1)\}$. Then $H$ is a subgroup of $G$. Since $G$ is abelian, $H⊴G$.

But the map

is a linear bijective application from the vector space $\mathbb{Z}∕2\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z}$ on itself, of matrix $\left(\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 1\hfill \end{array}\right)$ in the natrural base of $\mathbb{Z}∕2\mathbb{Z}\times \mathbb{Z}∕2\mathbb{Z}$.

A fortiori $\sigma$ is a group automorphism of $G$.

But

so $\sigma ((1,1))=(1,0)\notin H$, where $(1,1)\in H$, so $\sigma (H)\ne H$. This shows that $H$ is normal in $G$, but is not a characteristic subgroup of $G$. □