## Characteristic subgroups are normal

Prove that characteristic subgroups are normal. Give an example of a normal subgroup that is not characteristic.

Suppose $H \leq G$ is characteristic, and let $\varphi_g$ denote conjugation by $g \in G$. We know that $\varphi_g$ is an automorphism of $G$; since $H$ is characteristic, we have $gHg^{-1} = \varphi_g[H] = H$ for all $g \in G$. Hence $H$ is normal.

Now define $\varphi : Q_8 \rightarrow Q_8$ by $\varphi(i) = i$ and $\varphi(j) = k$, and extend homomorphically to all of $Q_8$. Note that $Q_8 = \langle \varphi(i), \varphi(j) \rangle$, so that $\varphi$ is surjective; since $Q_8$ is finite, $\varphi$ is an automorphism.

Recall that every subgroup of $Q_8$ is normal. Now $\varphi[\langle j \rangle] = \langle \varphi(j) \rangle = \langle k \rangle \neq \langle j \rangle$; thus $\langle j \rangle$ is normal but not characteristic in $Q_8$.

Another example: $Z_p \times 1 \unlhd Z_p \times Z_p$ is normal(since it is a subgroup of an abelain group) but it is not fixed under the automorphism $\varphi(x,y)=(y,x)$.