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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.

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3 Answers 3

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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.

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    $\begingroup$ You know, this is the answer I like more, but to be fair Kyle answer it first. You really demonstrated me that the answer wasn't difficult at all, thanks. $\endgroup$
    – Alvaro Fuentes
    Apr 28, 2015 at 0:27
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    $\begingroup$ @xndrme When you'll do products of groups, you'll realize that any group $G\times G$ (with non trivial $G$) produces an example: the map $(x,y)\mapsto(y,x)$ is an automorphism that moves $G\times\{1\}$ (both Kyle's examples are of this form). $\endgroup$
    – egreg
    Apr 28, 2015 at 8:20
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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.

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  • $\begingroup$ No doubt you are totally right? But I should mention that I'm only on the 2nd chapter of the book, and until now I only studied Group definition, Subgroup, Lagrange Theorem, and Homomorphisms, maybe that's why the book said it was difficult to find an example. Anyways I will be happy to read about Klein-four groups, inner and outer automorphisms. But currently I don't understand your answer. $\endgroup$
    – Alvaro Fuentes
    Apr 27, 2015 at 21:18
  • $\begingroup$ I've updated the question for more context, maybe could you show a more simple example? Anyways thanks for the answer. $\endgroup$
    – Alvaro Fuentes
    Apr 27, 2015 at 21:24
  • $\begingroup$ @xndme: If you know the definition of normal subgroups, the definition of inner automorphisms is only one tiny step further. A normal subgroup $N < G$ is a subgroup such that $g^{-1}Ng=N$ for all $g \in G$. An inner automorphism is an automorphism of the form $i_g(h) = g^{-1} h g$ for $g \in G$. Putting these together, a normal subgroup $N < G$ is a subgroup such that $i_g(N)=N$ for all $g \in G$, in other words $N$ is closed under all inner automorphisms. $\endgroup$
    – Lee Mosher
    Apr 27, 2015 at 21:55
  • $\begingroup$ @LeeMosher thanks for the explanation. $\endgroup$
    – Alvaro Fuentes
    Apr 28, 2015 at 0:22
  • $\begingroup$ Thanks for the example solution, at the end it wasn't that hard the answer to the problem :) $\endgroup$
    – Alvaro Fuentes
    Apr 28, 2015 at 0:25
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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.

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  • $\begingroup$ Yes but the problem is the example, maybe I was too impressed by the statement that it was hard to find :( $\endgroup$
    – Alvaro Fuentes
    Apr 27, 2015 at 20:59
  • $\begingroup$ @xndrme, what is the simplest abelian group you know which is not cyclic? Does that provide an example for what you want? $\endgroup$
    – Mariano Suárez-Álvarez
    Apr 27, 2015 at 21:03
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    $\begingroup$ @MarianoSuárez-Alvarez Hum... I've never thinking of it, thinking now...$U_8$ I suppose? $\endgroup$
    – Alvaro Fuentes
    Apr 27, 2015 at 21:12
  • $\begingroup$ @MarianoSuárez-Alvarez the problem is how do I prove that a subgroup of $U_8$ is not a characteristic? Given that I found a candidate. $\endgroup$
    – Alvaro Fuentes
    Apr 27, 2015 at 21:28
  • $\begingroup$ @xndrme you will need to construct or verify that there is an automorphism that sends that subgroup to a different subgroup $\endgroup$
    – Rolf Hoyer
    Apr 27, 2015 at 21:34

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