2
$\begingroup$

I want to find all the normal subgroups of $D_n$.

We have that $K$ is a normal subgroup of $D_n$ iff $$gkg^{-1}=k\in K, \forall g\in D_n \text{ and } \forall k\in K$$ right?

Could you give me some hints how we could find all these subgroups?

$\endgroup$
  • $\begingroup$ Are you familiar with the fact that $s$ and $r$ generate $D_{2n}$ where, $s^2=r^n=e$ and $srs^{-1}=r^{-1}$. $\endgroup$ – Anurag A Feb 23 '16 at 23:24
  • $\begingroup$ Yes. How can we use this fact? @AnuragA $\endgroup$ – Mary Star Feb 23 '16 at 23:36
2
$\begingroup$

A good place to start is to determine the conjugacy classes of $D_n$. Since a subgroup $H$ is normal in $G$ iff $H$ is the union of some conjugacy classes of $G$, you can try to union two or more conjugacy classes and see if the order of the resulting potential subgroup divides $|G|$. This approach usually narrows down your research quite a bit.

$\endgroup$
  • $\begingroup$ Why is a subgroup $H$ normal in $G$ iff $H$ is the union of some conjugacy classes of $G$ ? Is this equivalent to $$gkg^{-1}\in K, \forall g\in D_n \text{ and } \forall k\in K$$ ? $\endgroup$ – Mary Star Feb 25 '16 at 15:47
0
$\begingroup$

You will likely need to distinguish the cases when $n$ is even and when $n$ is odd. To start off, try finding the conjugacy classes of $D_n$, and then take unions of these classes to see if they form subgroups of $D_n$ (because normal subgroups are those which are the union of conjugacy classes).

Let us take $D_4$ as a concrete example. We want to find the conjugacy classes of the elements of $D_4$, which can be split into two sets: the set of elements of the form $r^a$ and the set of elements of the form $r^as$ (for $0 \leq a \leq 3$).

To find the conjugacy class of $r^a$, we need to take elements of $D_4$, conjugate them with $r^a$, and see what results you have. The set of results you obtain is your conjugacy class of $r^a$. There are two cases for this arbitrary element: either it is in the form $r^b$ or $r^bs$ (again, for $0 \leq b \leq 3$). If we take an element of form $r^b$, we have $(r^b)r^a(r^b)^{-1}=r^br^ar^{n-b}=r^{b+a+n-b}=r^a$, so $r^a$ conjugates with $r^b$ to give back $r^a$. If we take an element of form $r^bs$, we have $(r^bs)r^a(r^bs)^{-1}=r^b(sr^a)(sr^{n-b})=r^br^{n-a}(ss)r^{n-b}=r^{b+n-a+n-b}=r^{n-a}$. Therefore, our set of results is $\{r^a,r^{n-a}\}$, so the conjugacy class of $r^a$ is $\{r^a,r^{n-a}\}$. If we go over all values of $a$, our distinct conjugacy classes are $\{\varepsilon\}$, $\{r,r^3\}$, and $\{r^2\}$.

We repeat this process for the element $r^as$. We have $(r^b)r^as(r^b)^{-1}=r^br^asr^{n-b}=r^br^ar^bs=r^{2b+a}s$, and $(r^bs)r^as(r^bs)^{-1}=r^bsr^a(ss)r^{n-b}=r^br^{n-a}sr^{n-b}=r^br^{n-a}r^bs=r^{b+n-a+b}s=r^{2b-a}s$. This is a bit more complicated since we ended up with a conjugacy class $\{r^{2b+a}s,r^{2b-a}s\}$, where $b$ can take on multiple values. What you will find is that since $2b$ is even, $2b+a$ and $2b-a$ are even when $a$ is even, and odd when $a$ is odd. And since $0 \leq b \leq 3$, the possible values of $2b+a$ (mod 4) for even $a$ are {0,2}, and the possible values of $2b+a$ (mod 4) for odd $a$ are {1,3}. Therefore, the conjugacy class of $r^as$ where $a$ is odd is $\{rs,r^3s\}$, and the conjugacy class of $r^as$ where $a$ is even is $\{s,r^2s\}$.

Now, you have all conjugacy classes of $D_4$: $\{\varepsilon\}$, $\{r,r^3\}$, $\{r^2\}$, $\{rs,r^3s\}$, and $\{s,r^2s\}$. Now you can try to find which unions of these classes produce a subgroup, in which case it must be normal. In this particular example, there are precisely 6 normal subgroups: $\{\varepsilon\}$, $\{\varepsilon, r^2\}$, $\{\varepsilon, r, r^2, r^3\}$, $\{\varepsilon, r^2, rs, r^3s\}$, $\{\varepsilon, r^2, s, r^2s\}$ and $D_4$ itself.

The calculation is the same for odd $n$, but the subgroups will come out differently. Try repeating this procedure for a small odd $n$, and you will see that there is a pattern in the subgroups of $D_n$. Then, you will be able to generalize about the subgroups of $D_n$.

Hope this helped.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.