Why is $\langle r\rangle$ characteristic in $D_n$?

I need to determinate if $$\langle r\rangle$$ is characteristic in $$D_n = \langle r \rangle_n \rtimes \langle s \rangle_2$$. This is trivial if I use the result that every cyclic group is characteristic (ALERT:This is wrong, see the comments to see why), but I found an automorphism $$\alpha$$ in $$D_n$$ where $$\alpha(\langle r\rangle)\neq \langle r\rangle$$.

In the special case $$D_3$$ I defined $$\alpha$$ to satisfy $$\alpha(r)=s$$ and $$\alpha(s)=s$$. I derived the value of $$\alpha$$ of the rest of elements from this two, and it looks like a valid automorphism. And of course the problem is that $$\alpha(\langle r\rangle)= \{ 1,s \} \neq \langle r\rangle$$

On the other hand I know that every morphism must satisfy that $$|\alpha(r)|$$ divides $$|s|$$, but this is not the case.

So, what is failing in $$\alpha$$?

• Why do you believe that “every cyclic group is characteristic”? First, “characteristic” is a contextual property (a (sub)group is characteristic in another group; it’s not an intrinsic property of groups). Second, not every cyclic subgroup of a group is characteristic in that group: none of the three nontrivial proper cyclic subgroups of the Klein $4$-group are characteristic in the Klein subgroup. – Arturo Magidin Apr 28 at 20:54
• 1) Not all cyclic subgroups are characteristic. Some of them are even not normal (take for example a subgroup generated by a single transposition in $S_3$). – Yanior Weg Apr 28 at 20:55
• 2) What is $r$? – Yanior Weg Apr 28 at 20:55
• You do NOT have an automorphism of $D_3$ with $\alpha(r)=s$. $r$ has order $3$, so the order of its image must be $3$; but $s$ has order $2$. Just because you write down $\alpha(r)=s$ and $\alpha(s)=s$ does not mean you have written down an automorphism with those properties. Your $\alpha$ is not group morphism at all. – Arturo Magidin Apr 28 at 20:56
• 3) If $\alpha(r) = s = \alpha(s)$, then either $s=r$ or $\alpha$ is not an automorphism (as all automorphisms are bijections). – Yanior Weg Apr 28 at 20:59

I assume that by $$D_n$$ you mean the dihedral group of order $$2n$$, that is, $$D_n = \Bigl\langle r,s\Bigm| r^n = s^2 = 1,\ sr=r^{-1}s\Bigr\rangle.$$
The statement you claim is false for $$n=2$$. If $$n=1$$, you have the cyclic group of order $$2$$; then $$\langle r\rangle$$ is the trivial subgroup, which is characteristic. However, when $$n=2$$, you get the Klein $$4$$-group, and none of its proper nontrivial subgroups are characteristic. The result does hold for $$n\gt 2$$.
Assume $$n\geq 3$$. Let $$\alpha$$ be an automorphismn of $$D_n$$; we want to show that $$\alpha(\langle r\rangle) = \langle r\rangle$$ It suffices to show that $$\alpha(r)\in\langle r\rangle$$, since $$r$$ is a generator of the subgroup, so this implies $$\alpha(\langle r\rangle)\subseteq\langle r\rangle$$, and $$\alpha(r)$$ has order $$|r|$$, we get equality.
The elements of $$D_n$$ are: the elements of $$\langle r\rangle$$, which have order dividing $$n$$; and the elements of the form $$r^is$$, $$0\leq i \lt n$$, which have order $$2$$. The image of $$r$$ under $$\alpha$$ must have order $$n\gt 2$$, and hence must lie in $$\langle r\rangle$$, the only place where elements of order $$n$$ exist at all. Thus, $$\alpha(r)\in\langle r\rangle$$, showing that $$\langle r\rangle$$ is characteristic.
There are multiple errors in what you write. First, “characteristic” is not an intrinsic property (groups are not “characteristic”), it is an extrinsic, contextual property: subgroups are characteristic in other groups. So it makes no sense to say “cyclic groups are characteristics” (perhaps you are thinking that if $$G$$ is a cyclic group, then all of its subgroups are characteristic in $$G$$?). Second, it is false that cyclic subgroups are always characteristic, as the Klein $$4$$-group example shows. Thirdly, your $$\alpha$$ is not a group morphism; just because you write $$\alpha(r)=s$$ does not mean you have defined a group morphism. Recall that to define a morphism using the presentation, you must not only specify what the image of $$r$$ and $$s$$ are, you must also show that the images satisfy the defining relations. So to see whether $$\alpha(r)=s$$ and $$\alpha(s)=s$$ defines a group homomorphism from $$D_n$$ to itself, you would need to check that $$\alpha(r)^n = \alpha(s)^2 = 1$$, and $$\alpha(s)\alpha(r) = \alpha(r)^{-1}\alpha(s)$$. But the first one fails; so you have not defined a group homomorphism. Your $$\alpha$$ doesn’t work at all.