A normal subgroup that is not a characteristic

In the book I'm study is written:

A normal subgroup of a group need not be characteristic.

And as an exercise I'm supposed to find an example, it also said that is pretty hard to find one. After trying for two days I wasn't able to find one example.

So, I'm asking for an example of a group $G$ with a normal subgroup $H$ that is not a characteristic of $G$.

I will add some context because maybe it will clarify why the book said it is difficult to find an example: the main problem with the exercise is that until it the book had only covered Group Definition, Subgroups, Lagrange's Theorem and Homomorphisms. So I'm supposed to find a example with such lack of advanced tools.

Solution 1:

Consider the additive group $\mathbb{Q}$ of rational numbers. The map $\varphi\colon\mathbb{Q}\to\mathbb{Q}$ defined by $\varphi(x)=x/2$ is readily seen to be an automorphism.

The subgroup $\mathbb{Z}$ is not sent into itself by $\varphi$, because $\varphi(1)=1/2\notin\mathbb{Z}$.

Note that $\mathbb{Q}$ is abelian, so every subgroup is normal.

Solution 2:

There are two kinds of automorphisms, inner automorphisms and outer automorphisms. Normal subgroups are closed under inner automorphisms, so the task is to find a group with outer automorphisms which has a normal subgroup not fixed by one.

Consider the Klein-four group $\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}$, generated by $a$ and $b$. The outer automorphism $\tau$ which swaps the roles of $a$ and $b$ does not fix the normal subgroup $\langle a\rangle$.

Another related example is the group $\mathbb{Z}\times\mathbb{Z}$ of pairs of integers under addition (so, for instance, $(1,2) + (-3,4)=(-2,6)$). Consider the infinite cyclic subgroups generated by $a=(1,0)$ and $b=(0,1)$, and define the homomorphism $\varphi(x,y)=(y,x)$. It is easy to see that $\varphi$ is also an automorphism and that it carries $\langle a\rangle$ to $\langle b\rangle$. These subgroups are normal because the group is abelian, and $\varphi$ demonstrates they are not characteristic.

Solution 3:

One idea might be to look at abelian $G$. Then, all subgroups are normal, so you need to find an example where at least one subgroup is not characteristic. Finite cyclic groups won't work, since in that case each subgroup has a different order. Therefore, you need to take a look at examples at least slightly more complicated.