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.
 A: 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 $a = 0$.
A: 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 $\tau\sigma^i\tau\tau=\tau\sigma^i\not\in\langle \sigma^i\tau\rangle$.
