# Finding the center and normal subgroup of $D_{10}$

Consider $D_{10}$, the dihedral group of order $10$. This is the group of symmetries of the regular pentagon. Label the vertices of the pentagon clockwise as $1,2,3,4,5$. Let $\sigma = (12345)$ and $\tau = (13)(45)$.

How do I find the center and normal subgroup of $D_{10}$? I have subgroups of $D_{10}$, but I am not sure how that will help me to find the normal subgroup of $D_{10}$. Please help me.

• An idea to find the normal subgroups is to use Lagrange's Theorem and the fact that a normal subgroup has always index 2. For instance, if H is a normal subgroup of $D_{10}$, then $|D_{10}|=2|H|$. – Franciele Daltoé Nov 27 '15 at 0:44
• It's not true that a normal subgroup always has index $2$. For example, take any abelian group: all its subgroups are normal regardless of their index. It IS true that a subgroup of index $2$ is always normal. – Ethan Alwaise Nov 27 '15 at 0:45

## 2 Answers

One way is to use the class equation for $D_{10}$. Any element in the center must be in its own conjugacy class, and a normal subgroup must be a union of conjugacy classes. However, I think producing the class equation is actually more laborious in this case than just directly inspecting the group.

$D_{10}$ has the presentation $$D_{10} = \left<r,f : r^5 = f^2 = e, rf = fr^{-1}\right>.$$ Now $rf = fr^{-1}$ implies that $r,r^2,r^3,r^4,f$ are not in the center of $D_{10}$. Now consider an element $r^af$ for $1 \leq a \leq 4$. We have $(r^af)f = r^a$, but $f(r^a)f = ffr^{-a} = r^{-a}$. So $r^af$ is not in the center, and we conclude that the center is just $\{e\}$.

Now for normal subgroups. If you include any element $r^a$ for $1 \leq a \leq 4$, you get all of $\left<r\right>$. So then if you also add any of the elements $r^af$, where $0 \leq a \leq 4$, you will get $f$ and thus get the entire group. Suppose you include two elements $r^af, r^bf$ where $0 \leq a,b \leq 4$. Then you get $r^afr^bf = r^{a - b}$, but then you get $\left<r\right>$ and since you already included $r^af$ you get the whole group. We conclude that the only subgroups of $D_{10}$ are $\{e\}, D_{10}, \left<r\right>$, and $\left<r^af\right>$ for each $0 \leq a \leq 4$. Only the first three are normal ($\left<r\right>$ is normal because it has index $2$). You can easily show $\left<r^af\right>$ is not normal by multiplying on the left and right by $f$ if $a \neq 0$ and by $r$ if $a = 0$.

The center is trivial. See here.

Since $$\langle \sigma\rangle$$ has index $$2$$, it is normal. There are no other subgroups of order $$5$$, because only $$\sigma^i\,,1\le i\le4$$ have order $$5$$.

The only other options for normal subgroups are subgroups of order $$2$$, namely $$\langle \sigma^i\tau\rangle$$. But these aren't normal. For if $$i=0$$, we have $$\sigma\tau\sigma^{-1}\not\in\langle\tau\rangle$$. And if $$0\lt i\le4$$, we have $$\tau\sigma^i\tau\tau=\tau\sigma^i\not\in\langle \sigma^i\tau\rangle$$.