Let X be a prime

Proof:
First, we should know the definition of Dp.
Dp=<a,b> is generated by two elements a and b, where
a^2=b^p=1, ab=b^(-1)a.
So K=<b> is a normal subgroup of Dp. Because aba^(-1)=b^(-1) is in K.
Now I claim that K is the unique normal subgroup of Dp if p>=3 is a prime.
We know, |Dp|=2p, |K|=p, so [Dp:K]=2 and Dp = K union aK. So each element
in Dp has form b^n or ab^n.
If x=b^n not equal to 1, then x is in K. since p is a prime, then any element
not equal to 1 has order p. So x has order p.
If x=ab^n, then x^2=(ab^n)^2= b^(-n)a * ab^n = b^(-n) * (a^2) * b^n = 1
because a^2=1. So ab^n has order 2 for any n.
Suppose H is a normal subgroup of Dp, then |H| divides |Dp|=2p. But p is an
odd prime, so |H| divides 2 or |H| divides p.
If |H| divides 2 and H is not trivial, then H=<x> where x is an element of order 2.
From above analysis, we must have H=<ab^n> for some integer n<p.
But since H is normal, then b(ab^n)b^(-1)=ab^(-1)*b^n*b^(-1)=ab^(n-2) is in H.
So ab^(n-2)=1 or ab^(n-2)=ab^n.
If ab^(n-2)=ab^n, then b^2=1. So p=2. But p>=3. This is a contradiction.
If ab^(n-2)=1, a=b^(2-n) and thus a is in H. This is impossible.
Thus H can not have order 2. So H must have order p.
From above analysis, we know, only elements in K has order p. So H=k.
Therefore, K=<b> is the unique normal subgroup of Dp.
Thus Dp/K is isomorphic to Z2, which is a cyclic group.