Nonabelian group of order $2p, p>2$ has trivial center

Question

I'm stuck on an old algebra prelim problem. The problem is to prove that a nonabelian group of order $2p,p \text{ an odd prime}$ has trivial center. One thing I know is that any nonabelian group of order $2p,p>2$ is isomorphic to the dihedral group $D_{2p}$, but I think here we have to use the class equation

$$|G|= |Z(G)| + \sum_{i=1}^r [G: C_{G}(g_i)],$$ where the $g_i$ are the noncentral conjugacy classes of the group $G$, i.e. they are not in the center $Z(G)$. I haven't yet attempted it, but I think it boils down to assuming to the contrary that $|Z(G)| \neq 1$, and coming up with an argument that should lead to $p=2$. Not yet clear what that is, though. I would really appreciate some assistance here.

Answer 1

I think your problem is wrong: Since the center $Z(G)$ is a (normal) subgroup, Lagrange's Theorem gives

$$|Z(G)| \in \{1,2,p,2p \} $$

If $Z(G) \neq 1$, the quotient group $G/Z(G)$ will be cyclic, but that would imply that $G$ is abelian...

Answer 2

Let $D_{2k} = \langle r, s | r^k = 1, s^2 = 1, rs = sr^{-1} \rangle $ be the dihedral group of order $2k, k \geq 3$ with standard generators So every element of $D_{2k}$ can be written uniquely of the form $s^ir^j, i=0,2, 1\leq j \leq n.$

Claim: $Z :=Z(D_{2k}) = \{1 \},$ if $k$ is odd.

Let $x = sr^a \in Z$. Then $rx = xr \Rightarrow sr^{a-1} = sr^{a+1} \Rightarrow a-1=a+1$ mod $n.$ But then, $1 = -1$ mod $n$, which is not possible since $k \geq 3.$ Now let $x = r^a \in Z.$ Then $sx = xs \Rightarrow sr^a = r^as = sr^{-a}\Rightarrow k|2a,$ a contradiction.